Proving the interval $(0, 1)$ and $(1, 3)$ have the same cardinality.
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Prop: Show that the intervals $(0, 1)$ and $(1, 3)$ have the same cardinality.
What I have tried: I showed that the intervals $[2, 4]$ and $[0, 5]$ have equal cardinality by creating a function $F(x) = frac52x - 5$. I did this by getting the length of both intervals, however, I don't understand why this works. In my notes, it shows that the length of $[2, 4]$ is $2$ but if we are including both endpoints shouldn't it be $3$? Same goes for the interval $[0, 5]$ we should have a length of $6$, not $5$, but when using $5/2$ and shifting everything by $-5$ it seems to work just fine.
Trying to solve this current prop:
Because it is an open interval we cant use $0$ and $1$ from what I understand so I don't quite understand what to use as the length of this interval.
I have tried reading other questions similar to this one however I am still stumped.
Thanks for the support!
discrete-mathematics interval-arithmetic
$endgroup$
add a comment |
$begingroup$
Prop: Show that the intervals $(0, 1)$ and $(1, 3)$ have the same cardinality.
What I have tried: I showed that the intervals $[2, 4]$ and $[0, 5]$ have equal cardinality by creating a function $F(x) = frac52x - 5$. I did this by getting the length of both intervals, however, I don't understand why this works. In my notes, it shows that the length of $[2, 4]$ is $2$ but if we are including both endpoints shouldn't it be $3$? Same goes for the interval $[0, 5]$ we should have a length of $6$, not $5$, but when using $5/2$ and shifting everything by $-5$ it seems to work just fine.
Trying to solve this current prop:
Because it is an open interval we cant use $0$ and $1$ from what I understand so I don't quite understand what to use as the length of this interval.
I have tried reading other questions similar to this one however I am still stumped.
Thanks for the support!
discrete-mathematics interval-arithmetic
$endgroup$
$begingroup$
Just use linear interpolation.
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– Sean Roberson
Dec 15 '18 at 3:59
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The length of an interval is like measuring it with a tape measure. You seem to be thinking of the number of integral points in the interval, which is where you say $[0,5]$ has length $6$. The corresponding claim would be that $(0,5)$ has length $4$, but we can find points within the interval that are further than $4$ units apart. Since a point has zero length, adding it to an interval cannot change the length.
$endgroup$
– Ross Millikan
Dec 15 '18 at 4:27
$begingroup$
Find a bijection between $(0,1)$ and $(1,3)$.
$endgroup$
– copper.hat
Dec 15 '18 at 5:11
add a comment |
$begingroup$
Prop: Show that the intervals $(0, 1)$ and $(1, 3)$ have the same cardinality.
What I have tried: I showed that the intervals $[2, 4]$ and $[0, 5]$ have equal cardinality by creating a function $F(x) = frac52x - 5$. I did this by getting the length of both intervals, however, I don't understand why this works. In my notes, it shows that the length of $[2, 4]$ is $2$ but if we are including both endpoints shouldn't it be $3$? Same goes for the interval $[0, 5]$ we should have a length of $6$, not $5$, but when using $5/2$ and shifting everything by $-5$ it seems to work just fine.
Trying to solve this current prop:
Because it is an open interval we cant use $0$ and $1$ from what I understand so I don't quite understand what to use as the length of this interval.
I have tried reading other questions similar to this one however I am still stumped.
Thanks for the support!
discrete-mathematics interval-arithmetic
$endgroup$
Prop: Show that the intervals $(0, 1)$ and $(1, 3)$ have the same cardinality.
What I have tried: I showed that the intervals $[2, 4]$ and $[0, 5]$ have equal cardinality by creating a function $F(x) = frac52x - 5$. I did this by getting the length of both intervals, however, I don't understand why this works. In my notes, it shows that the length of $[2, 4]$ is $2$ but if we are including both endpoints shouldn't it be $3$? Same goes for the interval $[0, 5]$ we should have a length of $6$, not $5$, but when using $5/2$ and shifting everything by $-5$ it seems to work just fine.
Trying to solve this current prop:
Because it is an open interval we cant use $0$ and $1$ from what I understand so I don't quite understand what to use as the length of this interval.
I have tried reading other questions similar to this one however I am still stumped.
Thanks for the support!
discrete-mathematics interval-arithmetic
discrete-mathematics interval-arithmetic
edited Dec 15 '18 at 3:53
Tianlalu
3,08621038
3,08621038
asked Dec 15 '18 at 3:49
Zdravstvuyte94Zdravstvuyte94
465
465
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Just use linear interpolation.
$endgroup$
– Sean Roberson
Dec 15 '18 at 3:59
$begingroup$
The length of an interval is like measuring it with a tape measure. You seem to be thinking of the number of integral points in the interval, which is where you say $[0,5]$ has length $6$. The corresponding claim would be that $(0,5)$ has length $4$, but we can find points within the interval that are further than $4$ units apart. Since a point has zero length, adding it to an interval cannot change the length.
$endgroup$
– Ross Millikan
Dec 15 '18 at 4:27
$begingroup$
Find a bijection between $(0,1)$ and $(1,3)$.
$endgroup$
– copper.hat
Dec 15 '18 at 5:11
add a comment |
$begingroup$
Just use linear interpolation.
$endgroup$
– Sean Roberson
Dec 15 '18 at 3:59
$begingroup$
The length of an interval is like measuring it with a tape measure. You seem to be thinking of the number of integral points in the interval, which is where you say $[0,5]$ has length $6$. The corresponding claim would be that $(0,5)$ has length $4$, but we can find points within the interval that are further than $4$ units apart. Since a point has zero length, adding it to an interval cannot change the length.
$endgroup$
– Ross Millikan
Dec 15 '18 at 4:27
$begingroup$
Find a bijection between $(0,1)$ and $(1,3)$.
$endgroup$
– copper.hat
Dec 15 '18 at 5:11
$begingroup$
Just use linear interpolation.
$endgroup$
– Sean Roberson
Dec 15 '18 at 3:59
$begingroup$
Just use linear interpolation.
$endgroup$
– Sean Roberson
Dec 15 '18 at 3:59
$begingroup$
The length of an interval is like measuring it with a tape measure. You seem to be thinking of the number of integral points in the interval, which is where you say $[0,5]$ has length $6$. The corresponding claim would be that $(0,5)$ has length $4$, but we can find points within the interval that are further than $4$ units apart. Since a point has zero length, adding it to an interval cannot change the length.
$endgroup$
– Ross Millikan
Dec 15 '18 at 4:27
$begingroup$
The length of an interval is like measuring it with a tape measure. You seem to be thinking of the number of integral points in the interval, which is where you say $[0,5]$ has length $6$. The corresponding claim would be that $(0,5)$ has length $4$, but we can find points within the interval that are further than $4$ units apart. Since a point has zero length, adding it to an interval cannot change the length.
$endgroup$
– Ross Millikan
Dec 15 '18 at 4:27
$begingroup$
Find a bijection between $(0,1)$ and $(1,3)$.
$endgroup$
– copper.hat
Dec 15 '18 at 5:11
$begingroup$
Find a bijection between $(0,1)$ and $(1,3)$.
$endgroup$
– copper.hat
Dec 15 '18 at 5:11
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
1) I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
2) I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
You are right, the cardinality is "the number of elements within a set" (see here). For example, for the finite sets: $|{1,2,3}|=|{5,6,7}|=3$, because there are three elements in each set. Also, you can say the two sets have equal cardinality because there is a bijection from the first set to the second. Both of these sets are countable, because their cardinalities are finite. The term "denumerable" usually refers to "countably infinite" or $aleph_0$ (the cardinality of the natural numbers), which does not apply here. The set of rational numbers is also denumerable (see here). See also this post for distinguishing the concepts.
The idea is the same for infinite sets. Note that $(0,1)$ and $(1,3)$ are uncountable, hence not denumerable. Yet, we can establish the equality of their cardinalities if we can find a bijection from one set to another. And we can, it is $f(x)=2x+1$, because every element of the first set can be injectively and surjectively matched with the elements of the second set. See the graph:
$hspace{5cm}$
How do we find the linear relation? Just find the line passing through the two points:
$$frac{y-1}{3-1}=frac{x-0}{1-0} Rightarrow y=2x+1.$$
$endgroup$
$begingroup$
After I have successfully solved for 2x + 1, if I create a diagram of 2 sets one with 0, 1 and the other 1, 3 when putting either 0 or 1 into the equation we get 1 and 3 respectively. Is this enough to prove this theorem even though there is an infinite amount of numbers in between? Also, because this is an open interval aren't we not supposed to include 0, 1, or 3 in the set.
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– Zdravstvuyte94
Dec 15 '18 at 18:29
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Yes, it is enough, because by definition equal cardinality implies bijective relation between the two sets. Both are open intervals, so you can ignore the border numbers. For each number inside $(0,1)$, there is a unique number inside $(1,3)$, that is all
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– farruhota
Dec 15 '18 at 18:37
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But what I mean is that I added the +1 onto the 2x for the function definition in order to make f(0) = 1 and f(1) = 3 but can I not use those 2 numbers to test the function or is that allowed? Otherwise, I wouldn't know how much to shift left or right in order to know my function is correct. I really appreciate your help as I am finally understanding this concept and I find it to be very interesting.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 19:25
$begingroup$
You must use given numbers $(0,1)$ and $(1,3)$ to show equal cardinality. Linear function is the easiest to establish the bijective relation between the two sets. You can find other functions, e.g. quadratic, rational, etc.
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– farruhota
Dec 16 '18 at 11:00
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@Zdravstvuyte94 to check injectivity just prove that if $f(x)=f(y)$, then $x=y$, in this step you really do not need to use your domain or co-domain, since this function is injective in the real field. To check surjectivity, just take $x in (1,3)$ and see if you have a pre-image in (0,1). (i.e $f(a)=x$ iff $2a+1=x$, then $a=frac{x-1}{2}$, then check that this number must be at (0,1)).
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– Bajo Fondo
Dec 17 '18 at 14:03
add a comment |
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The length of an interval that starts at $a$ and ends at $b$ is $b-a$ regardless of whether the interval is open or closed at the endpoints (or half open). Adding one or both of $a$ and $b$ does not change the length. The simple linear transformation will not change the open or closed status of the endpoints. The length of $(0,1)$ is $1$ and of $(1,3)$ is $2$. The transformation from $(0,1)$ to $(1,3)$ given by $F(x)=2x+1$ works fine.
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add a comment |
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Define $f:(0,1) to (1,3)$ as $f(x)=2x+1$ this is a bijective function (I will leave you the proof), hence they have the same cardinality. I think you are confusing cardinality with measure (length). You should check that out.
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I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
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– Zdravstvuyte94
Dec 15 '18 at 4:19
add a comment |
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Note that cardinally is different from length or measure (in a more general sense). Cardinally is a generalisation of number of elements in a set. So by setting a 1-1 correspondence between two sets $A$ and $B$, we can show that the cardinality of the two sets are the same. This does not necessarily imply the 'length' (or measure) of both sets are the same.
Edit:
In showing a function $F: A to B$ is a 1-1 correspondence (or bijective), one shall verify $F$ satisfies the following two properties:
- (Injectivity) $quad x = y iff F(x) = F(y) ; forall x,y in A$
- (Surjectivity) $quad forall ,y in B, exists , x in A $ such that $F(x) = y$
Now if $A$ and $B$ are intervals of $mathbb{R}$, then we can construct a linear function (you can verify by yourself that linear function indeed construct a 1-1 correspondence between the two intervals). To construct that linear function, one shall consider informally as if first translate interval $A$ for certain amount and then 'dilate' or 'shrink' the interval for a certain amount to get 'B'. That 'dilating' or 'shrinking' factor should be the ratio of lengths of the two intervals.
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I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
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– Zdravstvuyte94
Dec 15 '18 at 3:56
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Please refer to the editing I have just made, which should provide you some idea or motivation behind the construction of the function.
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– Ken Hung
Dec 15 '18 at 4:19
add a comment |
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To show equal cardinality of sets, you basically have to show a bijection, or one - to - one relationship exists between them.
For finite sets, you can just "count the elements", but that's actually establishing a bijection of both sets to a finite subset of the natural numbers.
For infinite sets, you have to come up with a one to one function that maps one set to the other.
It should be quite obvious a linear relationship will suffice here. That's a relationship of the form "$y=mx+c$". The relative lengths are simply to help you decide the "scaling factor" to go from one to the other. That decides $m$, the gradient or slope of the line. The value of $c$ (the "$y$ intercept) should be thought of as an offset or" zero error" to line up the respective initial points of the intervals.
So in the case of $(0,1) rightarrow (1,3)$, the respective lengths are $1$ and $2$ (with a ratio of $2$) while the offset is $1$. So the function is $f(x) = 2x +1$, and that's the bijection that proves equal cardinality.
The lengths of the interval are only important in finding that $m$ value. They don't directly impact cardinality.
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5 Answers
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5 Answers
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$begingroup$
1) I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
2) I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
You are right, the cardinality is "the number of elements within a set" (see here). For example, for the finite sets: $|{1,2,3}|=|{5,6,7}|=3$, because there are three elements in each set. Also, you can say the two sets have equal cardinality because there is a bijection from the first set to the second. Both of these sets are countable, because their cardinalities are finite. The term "denumerable" usually refers to "countably infinite" or $aleph_0$ (the cardinality of the natural numbers), which does not apply here. The set of rational numbers is also denumerable (see here). See also this post for distinguishing the concepts.
The idea is the same for infinite sets. Note that $(0,1)$ and $(1,3)$ are uncountable, hence not denumerable. Yet, we can establish the equality of their cardinalities if we can find a bijection from one set to another. And we can, it is $f(x)=2x+1$, because every element of the first set can be injectively and surjectively matched with the elements of the second set. See the graph:
$hspace{5cm}$
How do we find the linear relation? Just find the line passing through the two points:
$$frac{y-1}{3-1}=frac{x-0}{1-0} Rightarrow y=2x+1.$$
$endgroup$
$begingroup$
After I have successfully solved for 2x + 1, if I create a diagram of 2 sets one with 0, 1 and the other 1, 3 when putting either 0 or 1 into the equation we get 1 and 3 respectively. Is this enough to prove this theorem even though there is an infinite amount of numbers in between? Also, because this is an open interval aren't we not supposed to include 0, 1, or 3 in the set.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 18:29
$begingroup$
Yes, it is enough, because by definition equal cardinality implies bijective relation between the two sets. Both are open intervals, so you can ignore the border numbers. For each number inside $(0,1)$, there is a unique number inside $(1,3)$, that is all
$endgroup$
– farruhota
Dec 15 '18 at 18:37
$begingroup$
But what I mean is that I added the +1 onto the 2x for the function definition in order to make f(0) = 1 and f(1) = 3 but can I not use those 2 numbers to test the function or is that allowed? Otherwise, I wouldn't know how much to shift left or right in order to know my function is correct. I really appreciate your help as I am finally understanding this concept and I find it to be very interesting.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 19:25
$begingroup$
You must use given numbers $(0,1)$ and $(1,3)$ to show equal cardinality. Linear function is the easiest to establish the bijective relation between the two sets. You can find other functions, e.g. quadratic, rational, etc.
$endgroup$
– farruhota
Dec 16 '18 at 11:00
$begingroup$
@Zdravstvuyte94 to check injectivity just prove that if $f(x)=f(y)$, then $x=y$, in this step you really do not need to use your domain or co-domain, since this function is injective in the real field. To check surjectivity, just take $x in (1,3)$ and see if you have a pre-image in (0,1). (i.e $f(a)=x$ iff $2a+1=x$, then $a=frac{x-1}{2}$, then check that this number must be at (0,1)).
$endgroup$
– Bajo Fondo
Dec 17 '18 at 14:03
add a comment |
$begingroup$
1) I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
2) I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
You are right, the cardinality is "the number of elements within a set" (see here). For example, for the finite sets: $|{1,2,3}|=|{5,6,7}|=3$, because there are three elements in each set. Also, you can say the two sets have equal cardinality because there is a bijection from the first set to the second. Both of these sets are countable, because their cardinalities are finite. The term "denumerable" usually refers to "countably infinite" or $aleph_0$ (the cardinality of the natural numbers), which does not apply here. The set of rational numbers is also denumerable (see here). See also this post for distinguishing the concepts.
The idea is the same for infinite sets. Note that $(0,1)$ and $(1,3)$ are uncountable, hence not denumerable. Yet, we can establish the equality of their cardinalities if we can find a bijection from one set to another. And we can, it is $f(x)=2x+1$, because every element of the first set can be injectively and surjectively matched with the elements of the second set. See the graph:
$hspace{5cm}$
How do we find the linear relation? Just find the line passing through the two points:
$$frac{y-1}{3-1}=frac{x-0}{1-0} Rightarrow y=2x+1.$$
$endgroup$
$begingroup$
After I have successfully solved for 2x + 1, if I create a diagram of 2 sets one with 0, 1 and the other 1, 3 when putting either 0 or 1 into the equation we get 1 and 3 respectively. Is this enough to prove this theorem even though there is an infinite amount of numbers in between? Also, because this is an open interval aren't we not supposed to include 0, 1, or 3 in the set.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 18:29
$begingroup$
Yes, it is enough, because by definition equal cardinality implies bijective relation between the two sets. Both are open intervals, so you can ignore the border numbers. For each number inside $(0,1)$, there is a unique number inside $(1,3)$, that is all
$endgroup$
– farruhota
Dec 15 '18 at 18:37
$begingroup$
But what I mean is that I added the +1 onto the 2x for the function definition in order to make f(0) = 1 and f(1) = 3 but can I not use those 2 numbers to test the function or is that allowed? Otherwise, I wouldn't know how much to shift left or right in order to know my function is correct. I really appreciate your help as I am finally understanding this concept and I find it to be very interesting.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 19:25
$begingroup$
You must use given numbers $(0,1)$ and $(1,3)$ to show equal cardinality. Linear function is the easiest to establish the bijective relation between the two sets. You can find other functions, e.g. quadratic, rational, etc.
$endgroup$
– farruhota
Dec 16 '18 at 11:00
$begingroup$
@Zdravstvuyte94 to check injectivity just prove that if $f(x)=f(y)$, then $x=y$, in this step you really do not need to use your domain or co-domain, since this function is injective in the real field. To check surjectivity, just take $x in (1,3)$ and see if you have a pre-image in (0,1). (i.e $f(a)=x$ iff $2a+1=x$, then $a=frac{x-1}{2}$, then check that this number must be at (0,1)).
$endgroup$
– Bajo Fondo
Dec 17 '18 at 14:03
add a comment |
$begingroup$
1) I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
2) I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
You are right, the cardinality is "the number of elements within a set" (see here). For example, for the finite sets: $|{1,2,3}|=|{5,6,7}|=3$, because there are three elements in each set. Also, you can say the two sets have equal cardinality because there is a bijection from the first set to the second. Both of these sets are countable, because their cardinalities are finite. The term "denumerable" usually refers to "countably infinite" or $aleph_0$ (the cardinality of the natural numbers), which does not apply here. The set of rational numbers is also denumerable (see here). See also this post for distinguishing the concepts.
The idea is the same for infinite sets. Note that $(0,1)$ and $(1,3)$ are uncountable, hence not denumerable. Yet, we can establish the equality of their cardinalities if we can find a bijection from one set to another. And we can, it is $f(x)=2x+1$, because every element of the first set can be injectively and surjectively matched with the elements of the second set. See the graph:
$hspace{5cm}$
How do we find the linear relation? Just find the line passing through the two points:
$$frac{y-1}{3-1}=frac{x-0}{1-0} Rightarrow y=2x+1.$$
$endgroup$
1) I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
2) I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
You are right, the cardinality is "the number of elements within a set" (see here). For example, for the finite sets: $|{1,2,3}|=|{5,6,7}|=3$, because there are three elements in each set. Also, you can say the two sets have equal cardinality because there is a bijection from the first set to the second. Both of these sets are countable, because their cardinalities are finite. The term "denumerable" usually refers to "countably infinite" or $aleph_0$ (the cardinality of the natural numbers), which does not apply here. The set of rational numbers is also denumerable (see here). See also this post for distinguishing the concepts.
The idea is the same for infinite sets. Note that $(0,1)$ and $(1,3)$ are uncountable, hence not denumerable. Yet, we can establish the equality of their cardinalities if we can find a bijection from one set to another. And we can, it is $f(x)=2x+1$, because every element of the first set can be injectively and surjectively matched with the elements of the second set. See the graph:
$hspace{5cm}$
How do we find the linear relation? Just find the line passing through the two points:
$$frac{y-1}{3-1}=frac{x-0}{1-0} Rightarrow y=2x+1.$$
answered Dec 15 '18 at 7:02
farruhotafarruhota
20.4k2739
20.4k2739
$begingroup$
After I have successfully solved for 2x + 1, if I create a diagram of 2 sets one with 0, 1 and the other 1, 3 when putting either 0 or 1 into the equation we get 1 and 3 respectively. Is this enough to prove this theorem even though there is an infinite amount of numbers in between? Also, because this is an open interval aren't we not supposed to include 0, 1, or 3 in the set.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 18:29
$begingroup$
Yes, it is enough, because by definition equal cardinality implies bijective relation between the two sets. Both are open intervals, so you can ignore the border numbers. For each number inside $(0,1)$, there is a unique number inside $(1,3)$, that is all
$endgroup$
– farruhota
Dec 15 '18 at 18:37
$begingroup$
But what I mean is that I added the +1 onto the 2x for the function definition in order to make f(0) = 1 and f(1) = 3 but can I not use those 2 numbers to test the function or is that allowed? Otherwise, I wouldn't know how much to shift left or right in order to know my function is correct. I really appreciate your help as I am finally understanding this concept and I find it to be very interesting.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 19:25
$begingroup$
You must use given numbers $(0,1)$ and $(1,3)$ to show equal cardinality. Linear function is the easiest to establish the bijective relation between the two sets. You can find other functions, e.g. quadratic, rational, etc.
$endgroup$
– farruhota
Dec 16 '18 at 11:00
$begingroup$
@Zdravstvuyte94 to check injectivity just prove that if $f(x)=f(y)$, then $x=y$, in this step you really do not need to use your domain or co-domain, since this function is injective in the real field. To check surjectivity, just take $x in (1,3)$ and see if you have a pre-image in (0,1). (i.e $f(a)=x$ iff $2a+1=x$, then $a=frac{x-1}{2}$, then check that this number must be at (0,1)).
$endgroup$
– Bajo Fondo
Dec 17 '18 at 14:03
add a comment |
$begingroup$
After I have successfully solved for 2x + 1, if I create a diagram of 2 sets one with 0, 1 and the other 1, 3 when putting either 0 or 1 into the equation we get 1 and 3 respectively. Is this enough to prove this theorem even though there is an infinite amount of numbers in between? Also, because this is an open interval aren't we not supposed to include 0, 1, or 3 in the set.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 18:29
$begingroup$
Yes, it is enough, because by definition equal cardinality implies bijective relation between the two sets. Both are open intervals, so you can ignore the border numbers. For each number inside $(0,1)$, there is a unique number inside $(1,3)$, that is all
$endgroup$
– farruhota
Dec 15 '18 at 18:37
$begingroup$
But what I mean is that I added the +1 onto the 2x for the function definition in order to make f(0) = 1 and f(1) = 3 but can I not use those 2 numbers to test the function or is that allowed? Otherwise, I wouldn't know how much to shift left or right in order to know my function is correct. I really appreciate your help as I am finally understanding this concept and I find it to be very interesting.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 19:25
$begingroup$
You must use given numbers $(0,1)$ and $(1,3)$ to show equal cardinality. Linear function is the easiest to establish the bijective relation between the two sets. You can find other functions, e.g. quadratic, rational, etc.
$endgroup$
– farruhota
Dec 16 '18 at 11:00
$begingroup$
@Zdravstvuyte94 to check injectivity just prove that if $f(x)=f(y)$, then $x=y$, in this step you really do not need to use your domain or co-domain, since this function is injective in the real field. To check surjectivity, just take $x in (1,3)$ and see if you have a pre-image in (0,1). (i.e $f(a)=x$ iff $2a+1=x$, then $a=frac{x-1}{2}$, then check that this number must be at (0,1)).
$endgroup$
– Bajo Fondo
Dec 17 '18 at 14:03
$begingroup$
After I have successfully solved for 2x + 1, if I create a diagram of 2 sets one with 0, 1 and the other 1, 3 when putting either 0 or 1 into the equation we get 1 and 3 respectively. Is this enough to prove this theorem even though there is an infinite amount of numbers in between? Also, because this is an open interval aren't we not supposed to include 0, 1, or 3 in the set.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 18:29
$begingroup$
After I have successfully solved for 2x + 1, if I create a diagram of 2 sets one with 0, 1 and the other 1, 3 when putting either 0 or 1 into the equation we get 1 and 3 respectively. Is this enough to prove this theorem even though there is an infinite amount of numbers in between? Also, because this is an open interval aren't we not supposed to include 0, 1, or 3 in the set.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 18:29
$begingroup$
Yes, it is enough, because by definition equal cardinality implies bijective relation between the two sets. Both are open intervals, so you can ignore the border numbers. For each number inside $(0,1)$, there is a unique number inside $(1,3)$, that is all
$endgroup$
– farruhota
Dec 15 '18 at 18:37
$begingroup$
Yes, it is enough, because by definition equal cardinality implies bijective relation between the two sets. Both are open intervals, so you can ignore the border numbers. For each number inside $(0,1)$, there is a unique number inside $(1,3)$, that is all
$endgroup$
– farruhota
Dec 15 '18 at 18:37
$begingroup$
But what I mean is that I added the +1 onto the 2x for the function definition in order to make f(0) = 1 and f(1) = 3 but can I not use those 2 numbers to test the function or is that allowed? Otherwise, I wouldn't know how much to shift left or right in order to know my function is correct. I really appreciate your help as I am finally understanding this concept and I find it to be very interesting.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 19:25
$begingroup$
But what I mean is that I added the +1 onto the 2x for the function definition in order to make f(0) = 1 and f(1) = 3 but can I not use those 2 numbers to test the function or is that allowed? Otherwise, I wouldn't know how much to shift left or right in order to know my function is correct. I really appreciate your help as I am finally understanding this concept and I find it to be very interesting.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 19:25
$begingroup$
You must use given numbers $(0,1)$ and $(1,3)$ to show equal cardinality. Linear function is the easiest to establish the bijective relation between the two sets. You can find other functions, e.g. quadratic, rational, etc.
$endgroup$
– farruhota
Dec 16 '18 at 11:00
$begingroup$
You must use given numbers $(0,1)$ and $(1,3)$ to show equal cardinality. Linear function is the easiest to establish the bijective relation between the two sets. You can find other functions, e.g. quadratic, rational, etc.
$endgroup$
– farruhota
Dec 16 '18 at 11:00
$begingroup$
@Zdravstvuyte94 to check injectivity just prove that if $f(x)=f(y)$, then $x=y$, in this step you really do not need to use your domain or co-domain, since this function is injective in the real field. To check surjectivity, just take $x in (1,3)$ and see if you have a pre-image in (0,1). (i.e $f(a)=x$ iff $2a+1=x$, then $a=frac{x-1}{2}$, then check that this number must be at (0,1)).
$endgroup$
– Bajo Fondo
Dec 17 '18 at 14:03
$begingroup$
@Zdravstvuyte94 to check injectivity just prove that if $f(x)=f(y)$, then $x=y$, in this step you really do not need to use your domain or co-domain, since this function is injective in the real field. To check surjectivity, just take $x in (1,3)$ and see if you have a pre-image in (0,1). (i.e $f(a)=x$ iff $2a+1=x$, then $a=frac{x-1}{2}$, then check that this number must be at (0,1)).
$endgroup$
– Bajo Fondo
Dec 17 '18 at 14:03
add a comment |
$begingroup$
The length of an interval that starts at $a$ and ends at $b$ is $b-a$ regardless of whether the interval is open or closed at the endpoints (or half open). Adding one or both of $a$ and $b$ does not change the length. The simple linear transformation will not change the open or closed status of the endpoints. The length of $(0,1)$ is $1$ and of $(1,3)$ is $2$. The transformation from $(0,1)$ to $(1,3)$ given by $F(x)=2x+1$ works fine.
$endgroup$
add a comment |
$begingroup$
The length of an interval that starts at $a$ and ends at $b$ is $b-a$ regardless of whether the interval is open or closed at the endpoints (or half open). Adding one or both of $a$ and $b$ does not change the length. The simple linear transformation will not change the open or closed status of the endpoints. The length of $(0,1)$ is $1$ and of $(1,3)$ is $2$. The transformation from $(0,1)$ to $(1,3)$ given by $F(x)=2x+1$ works fine.
$endgroup$
add a comment |
$begingroup$
The length of an interval that starts at $a$ and ends at $b$ is $b-a$ regardless of whether the interval is open or closed at the endpoints (or half open). Adding one or both of $a$ and $b$ does not change the length. The simple linear transformation will not change the open or closed status of the endpoints. The length of $(0,1)$ is $1$ and of $(1,3)$ is $2$. The transformation from $(0,1)$ to $(1,3)$ given by $F(x)=2x+1$ works fine.
$endgroup$
The length of an interval that starts at $a$ and ends at $b$ is $b-a$ regardless of whether the interval is open or closed at the endpoints (or half open). Adding one or both of $a$ and $b$ does not change the length. The simple linear transformation will not change the open or closed status of the endpoints. The length of $(0,1)$ is $1$ and of $(1,3)$ is $2$. The transformation from $(0,1)$ to $(1,3)$ given by $F(x)=2x+1$ works fine.
answered Dec 15 '18 at 4:21
Ross MillikanRoss Millikan
297k23198371
297k23198371
add a comment |
add a comment |
$begingroup$
Define $f:(0,1) to (1,3)$ as $f(x)=2x+1$ this is a bijective function (I will leave you the proof), hence they have the same cardinality. I think you are confusing cardinality with measure (length). You should check that out.
$endgroup$
$begingroup$
I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 4:19
add a comment |
$begingroup$
Define $f:(0,1) to (1,3)$ as $f(x)=2x+1$ this is a bijective function (I will leave you the proof), hence they have the same cardinality. I think you are confusing cardinality with measure (length). You should check that out.
$endgroup$
$begingroup$
I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 4:19
add a comment |
$begingroup$
Define $f:(0,1) to (1,3)$ as $f(x)=2x+1$ this is a bijective function (I will leave you the proof), hence they have the same cardinality. I think you are confusing cardinality with measure (length). You should check that out.
$endgroup$
Define $f:(0,1) to (1,3)$ as $f(x)=2x+1$ this is a bijective function (I will leave you the proof), hence they have the same cardinality. I think you are confusing cardinality with measure (length). You should check that out.
answered Dec 15 '18 at 4:15
Bajo FondoBajo Fondo
410315
410315
$begingroup$
I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 4:19
add a comment |
$begingroup$
I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 4:19
$begingroup$
I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 4:19
$begingroup$
I don't believe I am confusing cardinality with the length I'm not sure if anyone understanding my question. Cardinality is the number of elements within a set and is used to check if a set is denumerable as an example. I don't understand what length is in this case if 0 or 1 is not included and that's what I'm trying to figure out, I have no idea how you got 2x with the open intervals of 1, 3 and 0, 1.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 4:19
add a comment |
$begingroup$
Note that cardinally is different from length or measure (in a more general sense). Cardinally is a generalisation of number of elements in a set. So by setting a 1-1 correspondence between two sets $A$ and $B$, we can show that the cardinality of the two sets are the same. This does not necessarily imply the 'length' (or measure) of both sets are the same.
Edit:
In showing a function $F: A to B$ is a 1-1 correspondence (or bijective), one shall verify $F$ satisfies the following two properties:
- (Injectivity) $quad x = y iff F(x) = F(y) ; forall x,y in A$
- (Surjectivity) $quad forall ,y in B, exists , x in A $ such that $F(x) = y$
Now if $A$ and $B$ are intervals of $mathbb{R}$, then we can construct a linear function (you can verify by yourself that linear function indeed construct a 1-1 correspondence between the two intervals). To construct that linear function, one shall consider informally as if first translate interval $A$ for certain amount and then 'dilate' or 'shrink' the interval for a certain amount to get 'B'. That 'dilating' or 'shrinking' factor should be the ratio of lengths of the two intervals.
$endgroup$
$begingroup$
I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 3:56
$begingroup$
Please refer to the editing I have just made, which should provide you some idea or motivation behind the construction of the function.
$endgroup$
– Ken Hung
Dec 15 '18 at 4:19
add a comment |
$begingroup$
Note that cardinally is different from length or measure (in a more general sense). Cardinally is a generalisation of number of elements in a set. So by setting a 1-1 correspondence between two sets $A$ and $B$, we can show that the cardinality of the two sets are the same. This does not necessarily imply the 'length' (or measure) of both sets are the same.
Edit:
In showing a function $F: A to B$ is a 1-1 correspondence (or bijective), one shall verify $F$ satisfies the following two properties:
- (Injectivity) $quad x = y iff F(x) = F(y) ; forall x,y in A$
- (Surjectivity) $quad forall ,y in B, exists , x in A $ such that $F(x) = y$
Now if $A$ and $B$ are intervals of $mathbb{R}$, then we can construct a linear function (you can verify by yourself that linear function indeed construct a 1-1 correspondence between the two intervals). To construct that linear function, one shall consider informally as if first translate interval $A$ for certain amount and then 'dilate' or 'shrink' the interval for a certain amount to get 'B'. That 'dilating' or 'shrinking' factor should be the ratio of lengths of the two intervals.
$endgroup$
$begingroup$
I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 3:56
$begingroup$
Please refer to the editing I have just made, which should provide you some idea or motivation behind the construction of the function.
$endgroup$
– Ken Hung
Dec 15 '18 at 4:19
add a comment |
$begingroup$
Note that cardinally is different from length or measure (in a more general sense). Cardinally is a generalisation of number of elements in a set. So by setting a 1-1 correspondence between two sets $A$ and $B$, we can show that the cardinality of the two sets are the same. This does not necessarily imply the 'length' (or measure) of both sets are the same.
Edit:
In showing a function $F: A to B$ is a 1-1 correspondence (or bijective), one shall verify $F$ satisfies the following two properties:
- (Injectivity) $quad x = y iff F(x) = F(y) ; forall x,y in A$
- (Surjectivity) $quad forall ,y in B, exists , x in A $ such that $F(x) = y$
Now if $A$ and $B$ are intervals of $mathbb{R}$, then we can construct a linear function (you can verify by yourself that linear function indeed construct a 1-1 correspondence between the two intervals). To construct that linear function, one shall consider informally as if first translate interval $A$ for certain amount and then 'dilate' or 'shrink' the interval for a certain amount to get 'B'. That 'dilating' or 'shrinking' factor should be the ratio of lengths of the two intervals.
$endgroup$
Note that cardinally is different from length or measure (in a more general sense). Cardinally is a generalisation of number of elements in a set. So by setting a 1-1 correspondence between two sets $A$ and $B$, we can show that the cardinality of the two sets are the same. This does not necessarily imply the 'length' (or measure) of both sets are the same.
Edit:
In showing a function $F: A to B$ is a 1-1 correspondence (or bijective), one shall verify $F$ satisfies the following two properties:
- (Injectivity) $quad x = y iff F(x) = F(y) ; forall x,y in A$
- (Surjectivity) $quad forall ,y in B, exists , x in A $ such that $F(x) = y$
Now if $A$ and $B$ are intervals of $mathbb{R}$, then we can construct a linear function (you can verify by yourself that linear function indeed construct a 1-1 correspondence between the two intervals). To construct that linear function, one shall consider informally as if first translate interval $A$ for certain amount and then 'dilate' or 'shrink' the interval for a certain amount to get 'B'. That 'dilating' or 'shrinking' factor should be the ratio of lengths of the two intervals.
edited Dec 15 '18 at 4:17
answered Dec 15 '18 at 3:55
Ken HungKen Hung
613
613
$begingroup$
I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 3:56
$begingroup$
Please refer to the editing I have just made, which should provide you some idea or motivation behind the construction of the function.
$endgroup$
– Ken Hung
Dec 15 '18 at 4:19
add a comment |
$begingroup$
I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 3:56
$begingroup$
Please refer to the editing I have just made, which should provide you some idea or motivation behind the construction of the function.
$endgroup$
– Ken Hung
Dec 15 '18 at 4:19
$begingroup$
I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 3:56
$begingroup$
I understand that but I don't understand how we get the two length values from both intervals in order to create a function to show this one to one relationship.
$endgroup$
– Zdravstvuyte94
Dec 15 '18 at 3:56
$begingroup$
Please refer to the editing I have just made, which should provide you some idea or motivation behind the construction of the function.
$endgroup$
– Ken Hung
Dec 15 '18 at 4:19
$begingroup$
Please refer to the editing I have just made, which should provide you some idea or motivation behind the construction of the function.
$endgroup$
– Ken Hung
Dec 15 '18 at 4:19
add a comment |
$begingroup$
To show equal cardinality of sets, you basically have to show a bijection, or one - to - one relationship exists between them.
For finite sets, you can just "count the elements", but that's actually establishing a bijection of both sets to a finite subset of the natural numbers.
For infinite sets, you have to come up with a one to one function that maps one set to the other.
It should be quite obvious a linear relationship will suffice here. That's a relationship of the form "$y=mx+c$". The relative lengths are simply to help you decide the "scaling factor" to go from one to the other. That decides $m$, the gradient or slope of the line. The value of $c$ (the "$y$ intercept) should be thought of as an offset or" zero error" to line up the respective initial points of the intervals.
So in the case of $(0,1) rightarrow (1,3)$, the respective lengths are $1$ and $2$ (with a ratio of $2$) while the offset is $1$. So the function is $f(x) = 2x +1$, and that's the bijection that proves equal cardinality.
The lengths of the interval are only important in finding that $m$ value. They don't directly impact cardinality.
$endgroup$
add a comment |
$begingroup$
To show equal cardinality of sets, you basically have to show a bijection, or one - to - one relationship exists between them.
For finite sets, you can just "count the elements", but that's actually establishing a bijection of both sets to a finite subset of the natural numbers.
For infinite sets, you have to come up with a one to one function that maps one set to the other.
It should be quite obvious a linear relationship will suffice here. That's a relationship of the form "$y=mx+c$". The relative lengths are simply to help you decide the "scaling factor" to go from one to the other. That decides $m$, the gradient or slope of the line. The value of $c$ (the "$y$ intercept) should be thought of as an offset or" zero error" to line up the respective initial points of the intervals.
So in the case of $(0,1) rightarrow (1,3)$, the respective lengths are $1$ and $2$ (with a ratio of $2$) while the offset is $1$. So the function is $f(x) = 2x +1$, and that's the bijection that proves equal cardinality.
The lengths of the interval are only important in finding that $m$ value. They don't directly impact cardinality.
$endgroup$
add a comment |
$begingroup$
To show equal cardinality of sets, you basically have to show a bijection, or one - to - one relationship exists between them.
For finite sets, you can just "count the elements", but that's actually establishing a bijection of both sets to a finite subset of the natural numbers.
For infinite sets, you have to come up with a one to one function that maps one set to the other.
It should be quite obvious a linear relationship will suffice here. That's a relationship of the form "$y=mx+c$". The relative lengths are simply to help you decide the "scaling factor" to go from one to the other. That decides $m$, the gradient or slope of the line. The value of $c$ (the "$y$ intercept) should be thought of as an offset or" zero error" to line up the respective initial points of the intervals.
So in the case of $(0,1) rightarrow (1,3)$, the respective lengths are $1$ and $2$ (with a ratio of $2$) while the offset is $1$. So the function is $f(x) = 2x +1$, and that's the bijection that proves equal cardinality.
The lengths of the interval are only important in finding that $m$ value. They don't directly impact cardinality.
$endgroup$
To show equal cardinality of sets, you basically have to show a bijection, or one - to - one relationship exists between them.
For finite sets, you can just "count the elements", but that's actually establishing a bijection of both sets to a finite subset of the natural numbers.
For infinite sets, you have to come up with a one to one function that maps one set to the other.
It should be quite obvious a linear relationship will suffice here. That's a relationship of the form "$y=mx+c$". The relative lengths are simply to help you decide the "scaling factor" to go from one to the other. That decides $m$, the gradient or slope of the line. The value of $c$ (the "$y$ intercept) should be thought of as an offset or" zero error" to line up the respective initial points of the intervals.
So in the case of $(0,1) rightarrow (1,3)$, the respective lengths are $1$ and $2$ (with a ratio of $2$) while the offset is $1$. So the function is $f(x) = 2x +1$, and that's the bijection that proves equal cardinality.
The lengths of the interval are only important in finding that $m$ value. They don't directly impact cardinality.
edited Dec 15 '18 at 23:35
answered Dec 15 '18 at 4:51
DeepakDeepak
17k11536
17k11536
add a comment |
add a comment |
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Just use linear interpolation.
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– Sean Roberson
Dec 15 '18 at 3:59
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The length of an interval is like measuring it with a tape measure. You seem to be thinking of the number of integral points in the interval, which is where you say $[0,5]$ has length $6$. The corresponding claim would be that $(0,5)$ has length $4$, but we can find points within the interval that are further than $4$ units apart. Since a point has zero length, adding it to an interval cannot change the length.
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– Ross Millikan
Dec 15 '18 at 4:27
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Find a bijection between $(0,1)$ and $(1,3)$.
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– copper.hat
Dec 15 '18 at 5:11