Prove that the real interval (0,10) is equipotent to its subset (0,1)
$begingroup$
I have a question regarding the following solution to the problem:
"Every infinite decimal is of the form $0.a_1a_2a_3...$ (where $a_i in {0,1,2,...9}$ and the digits are not all $0$ and not all $9$) corresponds bijectively to the infinite decimal $a_1.a_2a_3...$ this shows that there is a one-to-one correspondence between the elements of (0,1) and (0,10)"
My question:
For an element in $(0,1)$ say $0.a_1a_2a_3...$ are there not two corresponding elements in $(0,10)$ namely $0.a_1a_2a_3...$ and $a_1.a_2a_3...$ ?
Why in the solution does $0.a_1a_2a_3...$ correspond only to $a_1.a_2a_3... $?
real-analysis elementary-set-theory
asked Jan 8 at 19:06
stochasticmrfoxstochasticmrfox
967
$endgroup$
add a comment |
$begingroup$
I have a question regarding the following solution to the problem:
"Every infinite decimal is of the form $0.a_1a_2a_3...$ (where $a_i in {0,1,2,...9}$ and the digits are not all $0$ and not all $9$) corresponds bijectively to the infinite decimal $a_1.a_2a_3...$ this shows that there is a one-to-one correspondence between the elements of (0,1) and (0,10)"
My question:
For an element in $(0,1)$ say $0.a_1a_2a_3...$ are there not two corresponding elements in $(0,10)$ namely $0.a_1a_2a_3...$ and $a_1.a_2a_3...$ ?
Why in the solution does $0.a_1a_2a_3...$ correspond only to $a_1.a_2a_3... $?
real-analysis elementary-set-theory
asked Jan 8 at 19:06
stochasticmrfoxstochasticmrfox
967
$endgroup$
$begingroup$
Multiplication by $10$ is invertible (by multiplication by $1/10$).
$endgroup$
– Randall
Jan 8 at 19:11
2
$begingroup$
They are defining the relationship as $phi(0.a_1 a_2...) = a_1.a_2...$. So, $0.1$ will correspond to $1.0$ , $0.01$ will correspond to $0.1$, etc.
$endgroup$
– copper.hat
Jan 8 at 19:14
add a comment |
$begingroup$
I have a question regarding the following solution to the problem:
"Every infinite decimal is of the form $0.a_1a_2a_3...$ (where $a_i in {0,1,2,...9}$ and the digits are not all $0$ and not all $9$) corresponds bijectively to the infinite decimal $a_1.a_2a_3...$ this shows that there is a one-to-one correspondence between the elements of (0,1) and (0,10)"
My question:
For an element in $(0,1)$ say $0.a_1a_2a_3...$ are there not two corresponding elements in $(0,10)$ namely $0.a_1a_2a_3...$ and $a_1.a_2a_3...$ ?
Why in the solution does $0.a_1a_2a_3...$ correspond only to $a_1.a_2a_3... $?
real-analysis elementary-set-theory
asked Jan 8 at 19:06
stochasticmrfoxstochasticmrfox
967
$endgroup$
I have a question regarding the following solution to the problem:
"Every infinite decimal is of the form $0.a_1a_2a_3...$ (where $a_i in {0,1,2,...9}$ and the digits are not all $0$ and not all $9$) corresponds bijectively to the infinite decimal $a_1.a_2a_3...$ this shows that there is a one-to-one correspondence between the elements of (0,1) and (0,10)"
My question:
For an element in $(0,1)$ say $0.a_1a_2a_3...$ are there not two corresponding elements in $(0,10)$ namely $0.a_1a_2a_3...$ and $a_1.a_2a_3...$ ?
Why in the solution does $0.a_1a_2a_3...$ correspond only to $a_1.a_2a_3... $?
real-analysis elementary-set-theory
real-analysis elementary-set-theory
asked Jan 8 at 19:06
stochasticmrfoxstochasticmrfox
967
asked Jan 8 at 19:06
stochasticmrfoxstochasticmrfox
967
asked Jan 8 at 19:06
stochasticmrfoxstochasticmrfox
967
asked Jan 8 at 19:06
stochasticmrfoxstochasticmrfox
967
asked Jan 8 at 19:06
stochasticmrfoxstochasticmrfox
967
967
$begingroup$
Multiplication by $10$ is invertible (by multiplication by $1/10$).
$endgroup$
– Randall
Jan 8 at 19:11
2
$begingroup$
They are defining the relationship as $phi(0.a_1 a_2...) = a_1.a_2...$. So, $0.1$ will correspond to $1.0$ , $0.01$ will correspond to $0.1$, etc.
$endgroup$
– copper.hat
Jan 8 at 19:14
add a comment |
$begingroup$
Multiplication by $10$ is invertible (by multiplication by $1/10$).
$endgroup$
– Randall
Jan 8 at 19:11
2
$begingroup$
They are defining the relationship as $phi(0.a_1 a_2...) = a_1.a_2...$. So, $0.1$ will correspond to $1.0$ , $0.01$ will correspond to $0.1$, etc.
$endgroup$
– copper.hat
Jan 8 at 19:14
$begingroup$
Multiplication by $10$ is invertible (by multiplication by $1/10$).
$endgroup$
– Randall
Jan 8 at 19:11
$begingroup$
Multiplication by $10$ is invertible (by multiplication by $1/10$).
$endgroup$
– Randall
Jan 8 at 19:11
2
2
$begingroup$
They are defining the relationship as $phi(0.a_1 a_2...) = a_1.a_2...$. So, $0.1$ will correspond to $1.0$ , $0.01$ will correspond to $0.1$, etc.
$endgroup$
– copper.hat
Jan 8 at 19:14
$begingroup$
They are defining the relationship as $phi(0.a_1 a_2...) = a_1.a_2...$. So, $0.1$ will correspond to $1.0$ , $0.01$ will correspond to $0.1$, etc.
$endgroup$
– copper.hat
Jan 8 at 19:14
add a comment |
1 Answer
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$begingroup$
The author of the solution is defining a specific bijection $f:(0,1)to(0,10)$ by the rule $f(0.a_1a_2a_3ldots)=a_1.a_2a_3ldots$. In other words, for a given $0.a_1a_2a_3ldotsin(0,1)$, there is one and only one $f(0.a_1a_2a_3ldots)in(0,10)$, which is defined to be $a_1.a_2a_3ldots$.
It is also true that for $a_1.a_2a_3ldotsin(0,10)$, there is one and only one $f^{-1}(a_1.a_2a_3ldots)in(0,1)$, which is simply $0.a_1a_2a_3ldots$.
answered Jan 8 at 20:12
Chris CulterChris Culter
21.5k43888
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The author of the solution is defining a specific bijection $f:(0,1)to(0,10)$ by the rule $f(0.a_1a_2a_3ldots)=a_1.a_2a_3ldots$. In other words, for a given $0.a_1a_2a_3ldotsin(0,1)$, there is one and only one $f(0.a_1a_2a_3ldots)in(0,10)$, which is defined to be $a_1.a_2a_3ldots$.
It is also true that for $a_1.a_2a_3ldotsin(0,10)$, there is one and only one $f^{-1}(a_1.a_2a_3ldots)in(0,1)$, which is simply $0.a_1a_2a_3ldots$.
answered Jan 8 at 20:12
Chris CulterChris Culter
21.5k43888
$endgroup$
add a comment |
$begingroup$
The author of the solution is defining a specific bijection $f:(0,1)to(0,10)$ by the rule $f(0.a_1a_2a_3ldots)=a_1.a_2a_3ldots$. In other words, for a given $0.a_1a_2a_3ldotsin(0,1)$, there is one and only one $f(0.a_1a_2a_3ldots)in(0,10)$, which is defined to be $a_1.a_2a_3ldots$.
It is also true that for $a_1.a_2a_3ldotsin(0,10)$, there is one and only one $f^{-1}(a_1.a_2a_3ldots)in(0,1)$, which is simply $0.a_1a_2a_3ldots$.
answered Jan 8 at 20:12
Chris CulterChris Culter
21.5k43888
$endgroup$
add a comment |
$begingroup$
The author of the solution is defining a specific bijection $f:(0,1)to(0,10)$ by the rule $f(0.a_1a_2a_3ldots)=a_1.a_2a_3ldots$. In other words, for a given $0.a_1a_2a_3ldotsin(0,1)$, there is one and only one $f(0.a_1a_2a_3ldots)in(0,10)$, which is defined to be $a_1.a_2a_3ldots$.
It is also true that for $a_1.a_2a_3ldotsin(0,10)$, there is one and only one $f^{-1}(a_1.a_2a_3ldots)in(0,1)$, which is simply $0.a_1a_2a_3ldots$.
answered Jan 8 at 20:12
Chris CulterChris Culter
21.5k43888
$endgroup$
The author of the solution is defining a specific bijection $f:(0,1)to(0,10)$ by the rule $f(0.a_1a_2a_3ldots)=a_1.a_2a_3ldots$. In other words, for a given $0.a_1a_2a_3ldotsin(0,1)$, there is one and only one $f(0.a_1a_2a_3ldots)in(0,10)$, which is defined to be $a_1.a_2a_3ldots$.
It is also true that for $a_1.a_2a_3ldotsin(0,10)$, there is one and only one $f^{-1}(a_1.a_2a_3ldots)in(0,1)$, which is simply $0.a_1a_2a_3ldots$.
answered Jan 8 at 20:12
Chris CulterChris Culter
21.5k43888
answered Jan 8 at 20:12
Chris CulterChris Culter
21.5k43888
answered Jan 8 at 20:12
Chris CulterChris Culter
21.5k43888
answered Jan 8 at 20:12
Chris CulterChris Culter
21.5k43888
21.5k43888
add a comment |
add a comment |
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$begingroup$
Multiplication by $10$ is invertible (by multiplication by $1/10$).
$endgroup$
– Randall
Jan 8 at 19:11
2
$begingroup$
They are defining the relationship as $phi(0.a_1 a_2...) = a_1.a_2...$. So, $0.1$ will correspond to $1.0$ , $0.01$ will correspond to $0.1$, etc.
$endgroup$
– copper.hat
Jan 8 at 19:14