What is the difference between a vector and a countable, nowhere dense, totally ordered set?
$begingroup$
It seems like any countable, non-dense subset of the reals is a vector by any other name. Don't get me wrong, a set is a set, and a vector is a vector - it just seems like there isn't much to distinguish the two aside from name and context.
Consider, for example, the set of natural numbers $mathbb{N}={1,2,ldots}$. In what way is the vector $textbf{v}inleft{mathbb{N}^inftymid v_n=nright}, textbf{v}=(1,2,ldots)$ not the same as $mathbb{N}$? Both have the same number of elements, satisfy total order, contain the natural numbers, and it is possible to substitute one for the other in a great many cases with little to no effort (ex: $sum_{ninmathbb{N}}f(v_n)=sum_{ninmathbb{N}}f(n)$, $a_n=g(v_n)=g(n)=left(a_nright)_{ninmathbb{N}}$, etc.)
I suppose that the representation of a vector, in writing, is understood to imply its order (whereas the order of the terms in a set do not change the intended meaning) - but I feel like there should be something more substantial than this. After all, the written representation of a mathematical object is distinct from the object itself, and I could easily create a notation where the written order of terms in a vector need not correspond to that of the actual vector.
I hope this isn't a silly question.
To clarify: This is not to say that a set and a vector are the same thing. Rather, the information needed to describe a vector is the same information needed to describe a particular type of set (i.e. the ordering between any two members)
linear-algebra elementary-set-theory definition
asked Dec 30 '18 at 0:57
R. BurtonR. Burton
670110
$endgroup$
add a comment |
$begingroup$
It seems like any countable, non-dense subset of the reals is a vector by any other name. Don't get me wrong, a set is a set, and a vector is a vector - it just seems like there isn't much to distinguish the two aside from name and context.
Consider, for example, the set of natural numbers $mathbb{N}={1,2,ldots}$. In what way is the vector $textbf{v}inleft{mathbb{N}^inftymid v_n=nright}, textbf{v}=(1,2,ldots)$ not the same as $mathbb{N}$? Both have the same number of elements, satisfy total order, contain the natural numbers, and it is possible to substitute one for the other in a great many cases with little to no effort (ex: $sum_{ninmathbb{N}}f(v_n)=sum_{ninmathbb{N}}f(n)$, $a_n=g(v_n)=g(n)=left(a_nright)_{ninmathbb{N}}$, etc.)
I suppose that the representation of a vector, in writing, is understood to imply its order (whereas the order of the terms in a set do not change the intended meaning) - but I feel like there should be something more substantial than this. After all, the written representation of a mathematical object is distinct from the object itself, and I could easily create a notation where the written order of terms in a vector need not correspond to that of the actual vector.
I hope this isn't a silly question.
To clarify: This is not to say that a set and a vector are the same thing. Rather, the information needed to describe a vector is the same information needed to describe a particular type of set (i.e. the ordering between any two members)
linear-algebra elementary-set-theory definition
asked Dec 30 '18 at 0:57
R. BurtonR. Burton
670110
$endgroup$
$begingroup$
I don't understand. Do you know what a vector is?
$endgroup$
– lulu
Dec 30 '18 at 1:06
6
$begingroup$
Reminds me of the old riddle: "What is the difference between Marilyn Monroe and a submarine?" "Marilyn Monroe is a movie actress, and a submarine is a boat that goes under water."
$endgroup$
– bof
Dec 30 '18 at 1:44
$begingroup$
What does "nowhere dense" have to do with it? Why is a countable dense set like the rational numbers any less like a vector than a countable nowhere dense set?
$endgroup$
– bof
Jan 1 at 3:32
$begingroup$
@bof Some orderings of a dense set cannot be encoded by a vector. Take, for example, $mathbb{Q}cap[0,1]$, ordered by $i<jiff a_i<a_j$. The first element of the vector $(a_1,ldots,a_infty)$, would obviously be $0$, but since the next element can be arbitrarily close to $0$, there cannot be a unique $a_2$. Thus, there is no unambiguous vector which encodes both the specified, order and all elements of the rationals in $[0,1]$. Granted, it is possible to create a vector to represent any countable set. But only non-dense sets can be represented this way for any order.
$endgroup$
– R. Burton
Jan 1 at 17:59
$begingroup$
How do you represent $mathbb Z$ with the natural order as a vector? Better yet, how about the nowhere dense countable set consisting of the midpoints of the "deleted intervals" of the Cantor set? This can be described as the set of numbers whose base $3$ representation has a $1$ in it, and all digits after the first $1$ are $1$'s, i.e., $0.overline1$, $0.0overline1$, $0.2overline1$, $0.00overline1$, $0.02overline1$, $0.20overline1$, $0.22overline1$, $0.000overline1$, $0.002overline1$, $dots$, This is a countable nowhere dense set, and it is order-isomorphic to $mathbb Q$.
$endgroup$
– bof
Jan 2 at 3:16
add a comment |
$begingroup$
It seems like any countable, non-dense subset of the reals is a vector by any other name. Don't get me wrong, a set is a set, and a vector is a vector - it just seems like there isn't much to distinguish the two aside from name and context.
Consider, for example, the set of natural numbers $mathbb{N}={1,2,ldots}$. In what way is the vector $textbf{v}inleft{mathbb{N}^inftymid v_n=nright}, textbf{v}=(1,2,ldots)$ not the same as $mathbb{N}$? Both have the same number of elements, satisfy total order, contain the natural numbers, and it is possible to substitute one for the other in a great many cases with little to no effort (ex: $sum_{ninmathbb{N}}f(v_n)=sum_{ninmathbb{N}}f(n)$, $a_n=g(v_n)=g(n)=left(a_nright)_{ninmathbb{N}}$, etc.)
I suppose that the representation of a vector, in writing, is understood to imply its order (whereas the order of the terms in a set do not change the intended meaning) - but I feel like there should be something more substantial than this. After all, the written representation of a mathematical object is distinct from the object itself, and I could easily create a notation where the written order of terms in a vector need not correspond to that of the actual vector.
I hope this isn't a silly question.
To clarify: This is not to say that a set and a vector are the same thing. Rather, the information needed to describe a vector is the same information needed to describe a particular type of set (i.e. the ordering between any two members)
linear-algebra elementary-set-theory definition
asked Dec 30 '18 at 0:57
R. BurtonR. Burton
670110
$endgroup$
It seems like any countable, non-dense subset of the reals is a vector by any other name. Don't get me wrong, a set is a set, and a vector is a vector - it just seems like there isn't much to distinguish the two aside from name and context.
Consider, for example, the set of natural numbers $mathbb{N}={1,2,ldots}$. In what way is the vector $textbf{v}inleft{mathbb{N}^inftymid v_n=nright}, textbf{v}=(1,2,ldots)$ not the same as $mathbb{N}$? Both have the same number of elements, satisfy total order, contain the natural numbers, and it is possible to substitute one for the other in a great many cases with little to no effort (ex: $sum_{ninmathbb{N}}f(v_n)=sum_{ninmathbb{N}}f(n)$, $a_n=g(v_n)=g(n)=left(a_nright)_{ninmathbb{N}}$, etc.)
I suppose that the representation of a vector, in writing, is understood to imply its order (whereas the order of the terms in a set do not change the intended meaning) - but I feel like there should be something more substantial than this. After all, the written representation of a mathematical object is distinct from the object itself, and I could easily create a notation where the written order of terms in a vector need not correspond to that of the actual vector.
I hope this isn't a silly question.
To clarify: This is not to say that a set and a vector are the same thing. Rather, the information needed to describe a vector is the same information needed to describe a particular type of set (i.e. the ordering between any two members)
linear-algebra elementary-set-theory definition
linear-algebra elementary-set-theory definition
asked Dec 30 '18 at 0:57
R. BurtonR. Burton
670110
asked Dec 30 '18 at 0:57
R. BurtonR. Burton
670110
edited Dec 31 '18 at 23:20
R. Burton
asked Dec 30 '18 at 0:57
R. BurtonR. Burton
670110
asked Dec 30 '18 at 0:57
R. BurtonR. Burton
670110
asked Dec 30 '18 at 0:57
R. BurtonR. Burton
670110
670110
$begingroup$
I don't understand. Do you know what a vector is?
$endgroup$
– lulu
Dec 30 '18 at 1:06
6
$begingroup$
Reminds me of the old riddle: "What is the difference between Marilyn Monroe and a submarine?" "Marilyn Monroe is a movie actress, and a submarine is a boat that goes under water."
$endgroup$
– bof
Dec 30 '18 at 1:44
$begingroup$
What does "nowhere dense" have to do with it? Why is a countable dense set like the rational numbers any less like a vector than a countable nowhere dense set?
$endgroup$
– bof
Jan 1 at 3:32
$begingroup$
@bof Some orderings of a dense set cannot be encoded by a vector. Take, for example, $mathbb{Q}cap[0,1]$, ordered by $i<jiff a_i<a_j$. The first element of the vector $(a_1,ldots,a_infty)$, would obviously be $0$, but since the next element can be arbitrarily close to $0$, there cannot be a unique $a_2$. Thus, there is no unambiguous vector which encodes both the specified, order and all elements of the rationals in $[0,1]$. Granted, it is possible to create a vector to represent any countable set. But only non-dense sets can be represented this way for any order.
$endgroup$
– R. Burton
Jan 1 at 17:59
$begingroup$
How do you represent $mathbb Z$ with the natural order as a vector? Better yet, how about the nowhere dense countable set consisting of the midpoints of the "deleted intervals" of the Cantor set? This can be described as the set of numbers whose base $3$ representation has a $1$ in it, and all digits after the first $1$ are $1$'s, i.e., $0.overline1$, $0.0overline1$, $0.2overline1$, $0.00overline1$, $0.02overline1$, $0.20overline1$, $0.22overline1$, $0.000overline1$, $0.002overline1$, $dots$, This is a countable nowhere dense set, and it is order-isomorphic to $mathbb Q$.
$endgroup$
– bof
Jan 2 at 3:16
add a comment |
$begingroup$
I don't understand. Do you know what a vector is?
$endgroup$
– lulu
Dec 30 '18 at 1:06
6
$begingroup$
Reminds me of the old riddle: "What is the difference between Marilyn Monroe and a submarine?" "Marilyn Monroe is a movie actress, and a submarine is a boat that goes under water."
$endgroup$
– bof
Dec 30 '18 at 1:44
$begingroup$
What does "nowhere dense" have to do with it? Why is a countable dense set like the rational numbers any less like a vector than a countable nowhere dense set?
$endgroup$
– bof
Jan 1 at 3:32
$begingroup$
@bof Some orderings of a dense set cannot be encoded by a vector. Take, for example, $mathbb{Q}cap[0,1]$, ordered by $i<jiff a_i<a_j$. The first element of the vector $(a_1,ldots,a_infty)$, would obviously be $0$, but since the next element can be arbitrarily close to $0$, there cannot be a unique $a_2$. Thus, there is no unambiguous vector which encodes both the specified, order and all elements of the rationals in $[0,1]$. Granted, it is possible to create a vector to represent any countable set. But only non-dense sets can be represented this way for any order.
$endgroup$
– R. Burton
Jan 1 at 17:59
$begingroup$
How do you represent $mathbb Z$ with the natural order as a vector? Better yet, how about the nowhere dense countable set consisting of the midpoints of the "deleted intervals" of the Cantor set? This can be described as the set of numbers whose base $3$ representation has a $1$ in it, and all digits after the first $1$ are $1$'s, i.e., $0.overline1$, $0.0overline1$, $0.2overline1$, $0.00overline1$, $0.02overline1$, $0.20overline1$, $0.22overline1$, $0.000overline1$, $0.002overline1$, $dots$, This is a countable nowhere dense set, and it is order-isomorphic to $mathbb Q$.
$endgroup$
– bof
Jan 2 at 3:16
$begingroup$
I don't understand. Do you know what a vector is?
$endgroup$
– lulu
Dec 30 '18 at 1:06
$begingroup$
I don't understand. Do you know what a vector is?
$endgroup$
– lulu
Dec 30 '18 at 1:06
6
6
$begingroup$
Reminds me of the old riddle: "What is the difference between Marilyn Monroe and a submarine?" "Marilyn Monroe is a movie actress, and a submarine is a boat that goes under water."
$endgroup$
– bof
Dec 30 '18 at 1:44
$begingroup$
Reminds me of the old riddle: "What is the difference between Marilyn Monroe and a submarine?" "Marilyn Monroe is a movie actress, and a submarine is a boat that goes under water."
$endgroup$
– bof
Dec 30 '18 at 1:44
$begingroup$
What does "nowhere dense" have to do with it? Why is a countable dense set like the rational numbers any less like a vector than a countable nowhere dense set?
$endgroup$
– bof
Jan 1 at 3:32
$begingroup$
What does "nowhere dense" have to do with it? Why is a countable dense set like the rational numbers any less like a vector than a countable nowhere dense set?
$endgroup$
– bof
Jan 1 at 3:32
$begingroup$
@bof Some orderings of a dense set cannot be encoded by a vector. Take, for example, $mathbb{Q}cap[0,1]$, ordered by $i<jiff a_i<a_j$. The first element of the vector $(a_1,ldots,a_infty)$, would obviously be $0$, but since the next element can be arbitrarily close to $0$, there cannot be a unique $a_2$. Thus, there is no unambiguous vector which encodes both the specified, order and all elements of the rationals in $[0,1]$. Granted, it is possible to create a vector to represent any countable set. But only non-dense sets can be represented this way for any order.
$endgroup$
– R. Burton
Jan 1 at 17:59
$begingroup$
@bof Some orderings of a dense set cannot be encoded by a vector. Take, for example, $mathbb{Q}cap[0,1]$, ordered by $i<jiff a_i<a_j$. The first element of the vector $(a_1,ldots,a_infty)$, would obviously be $0$, but since the next element can be arbitrarily close to $0$, there cannot be a unique $a_2$. Thus, there is no unambiguous vector which encodes both the specified, order and all elements of the rationals in $[0,1]$. Granted, it is possible to create a vector to represent any countable set. But only non-dense sets can be represented this way for any order.
$endgroup$
– R. Burton
Jan 1 at 17:59
$begingroup$
How do you represent $mathbb Z$ with the natural order as a vector? Better yet, how about the nowhere dense countable set consisting of the midpoints of the "deleted intervals" of the Cantor set? This can be described as the set of numbers whose base $3$ representation has a $1$ in it, and all digits after the first $1$ are $1$'s, i.e., $0.overline1$, $0.0overline1$, $0.2overline1$, $0.00overline1$, $0.02overline1$, $0.20overline1$, $0.22overline1$, $0.000overline1$, $0.002overline1$, $dots$, This is a countable nowhere dense set, and it is order-isomorphic to $mathbb Q$.
$endgroup$
– bof
Jan 2 at 3:16
$begingroup$
How do you represent $mathbb Z$ with the natural order as a vector? Better yet, how about the nowhere dense countable set consisting of the midpoints of the "deleted intervals" of the Cantor set? This can be described as the set of numbers whose base $3$ representation has a $1$ in it, and all digits after the first $1$ are $1$'s, i.e., $0.overline1$, $0.0overline1$, $0.2overline1$, $0.00overline1$, $0.02overline1$, $0.20overline1$, $0.22overline1$, $0.000overline1$, $0.002overline1$, $dots$, This is a countable nowhere dense set, and it is order-isomorphic to $mathbb Q$.
$endgroup$
– bof
Jan 2 at 3:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Unfortunately, this probably is not a satisfying answer, but the key difference between the set and the vector is that the set has no order to its terms, while the vector does. You say you could change the written representation of the vector and change the order, but that change in the written representation does not alter the fact that the inherent mathematical object which that representation represents—that is, the vector—still has an order in its underlying structure.
I think a better way to explain this is by trying to better understand what $Bbb{N}^infty$ is. For me, I like to think of $Bbb{N}^infty$ as the set of all functions from $Bbb{N}rightarrow Bbb{N}$. Thus, here, $v$ is the vector such that $v(n)=n$ for all $nin Bbb{N}$, which is like the identity function on $Bbb{N}$. However, even though $v(n)$ can always just be replaced with $n$ and therefore the identity function may seem redundant, that does not change the underlying structure behind $v(n)$, which is that it is an infinite-dimensional vector, and not a set.
Another way to realize the difference is by examining the vector $w(1)=2$, $w(2)=1$ and $w(n)=n$ for all $n > 3$. Clearly, the set of components of this vector are the natural numbers. However, this is also true for $v$, yet $vneq w$ because $v$ and $w$ have their components ordered differently: $v$ has the order $1$, then $2$, while $w$ has the order $2$, then $1$. On the other hand, $Bbb{N}$ has no such underlying order to it, so it is fundamentally distinct from vectors.
answered Dec 30 '18 at 1:08
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
1
$begingroup$
Good answer. I would add that a vector is only a vector in virtue of the larger structure of a vector space.
$endgroup$
– Ben W
Dec 30 '18 at 1:13
$begingroup$
For each vector whose entries are (for continuity's sake) natural numbers, would there be a unique function $Sort:mathbb{N}tomathbb{N}$ which assigns a unique position in $v$ to each $nin mathbb{N}$?
$endgroup$
– R. Burton
Dec 30 '18 at 1:14
1
$begingroup$
@R.Burton I'm not sure if I understand your comment, but I think what you're talking about is called the inverse function and is denoted as $v^{-1}$. Obviously, for $v$, $v^{-1}(n)=n$ since the position of each element $n$ in $v$ is simply $n$. However, the inverse function does not exist for all elements of $Bbb{N}^infty$. See $zeta(n)=0$ for all $ninBbb{N}$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:15
$begingroup$
@NobleMushtak Correction: for each totally ordered $Xsubseteqmathbb{N}$ there is a unique function $Sort:2^mathbb{N}tobigcup_{i=0}^inftymathbb{N}^i$ such that $Sort(X)=textbf{v}$, where $textbf{v}$ is the equivalent vector. The converse, that every vector is the image of a set under $Sort$ is only true if $Sort$ admits multisets as inputs, or is multivalued, otherwise, only vectors with unique entries can be described this way.
$endgroup$
– R. Burton
Jan 1 at 18:08
$begingroup$
@R.Burton I see what you mean now. Yes, for any subset of $X subset Bbb{N}$, there is a corresponding vector $mathbf{v}$ such that the set of all components in $mathbf{v}$ is $X$. However, this does not mean $mathbf{v}$ is the "equivalent" vector or sequence to $X$, because there are multiple vectors for which the set of all components is $X$.
$endgroup$
– Noble Mushtak
Jan 1 at 20:06
|
show 1 more comment
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056416%2fwhat-is-the-difference-between-a-vector-and-a-countable-nowhere-dense-totally%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Unfortunately, this probably is not a satisfying answer, but the key difference between the set and the vector is that the set has no order to its terms, while the vector does. You say you could change the written representation of the vector and change the order, but that change in the written representation does not alter the fact that the inherent mathematical object which that representation represents—that is, the vector—still has an order in its underlying structure.
I think a better way to explain this is by trying to better understand what $Bbb{N}^infty$ is. For me, I like to think of $Bbb{N}^infty$ as the set of all functions from $Bbb{N}rightarrow Bbb{N}$. Thus, here, $v$ is the vector such that $v(n)=n$ for all $nin Bbb{N}$, which is like the identity function on $Bbb{N}$. However, even though $v(n)$ can always just be replaced with $n$ and therefore the identity function may seem redundant, that does not change the underlying structure behind $v(n)$, which is that it is an infinite-dimensional vector, and not a set.
Another way to realize the difference is by examining the vector $w(1)=2$, $w(2)=1$ and $w(n)=n$ for all $n > 3$. Clearly, the set of components of this vector are the natural numbers. However, this is also true for $v$, yet $vneq w$ because $v$ and $w$ have their components ordered differently: $v$ has the order $1$, then $2$, while $w$ has the order $2$, then $1$. On the other hand, $Bbb{N}$ has no such underlying order to it, so it is fundamentally distinct from vectors.
answered Dec 30 '18 at 1:08
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
1
$begingroup$
Good answer. I would add that a vector is only a vector in virtue of the larger structure of a vector space.
$endgroup$
– Ben W
Dec 30 '18 at 1:13
$begingroup$
For each vector whose entries are (for continuity's sake) natural numbers, would there be a unique function $Sort:mathbb{N}tomathbb{N}$ which assigns a unique position in $v$ to each $nin mathbb{N}$?
$endgroup$
– R. Burton
Dec 30 '18 at 1:14
1
$begingroup$
@R.Burton I'm not sure if I understand your comment, but I think what you're talking about is called the inverse function and is denoted as $v^{-1}$. Obviously, for $v$, $v^{-1}(n)=n$ since the position of each element $n$ in $v$ is simply $n$. However, the inverse function does not exist for all elements of $Bbb{N}^infty$. See $zeta(n)=0$ for all $ninBbb{N}$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:15
$begingroup$
@NobleMushtak Correction: for each totally ordered $Xsubseteqmathbb{N}$ there is a unique function $Sort:2^mathbb{N}tobigcup_{i=0}^inftymathbb{N}^i$ such that $Sort(X)=textbf{v}$, where $textbf{v}$ is the equivalent vector. The converse, that every vector is the image of a set under $Sort$ is only true if $Sort$ admits multisets as inputs, or is multivalued, otherwise, only vectors with unique entries can be described this way.
$endgroup$
– R. Burton
Jan 1 at 18:08
$begingroup$
@R.Burton I see what you mean now. Yes, for any subset of $X subset Bbb{N}$, there is a corresponding vector $mathbf{v}$ such that the set of all components in $mathbf{v}$ is $X$. However, this does not mean $mathbf{v}$ is the "equivalent" vector or sequence to $X$, because there are multiple vectors for which the set of all components is $X$.
$endgroup$
– Noble Mushtak
Jan 1 at 20:06
|
show 1 more comment
$begingroup$
Unfortunately, this probably is not a satisfying answer, but the key difference between the set and the vector is that the set has no order to its terms, while the vector does. You say you could change the written representation of the vector and change the order, but that change in the written representation does not alter the fact that the inherent mathematical object which that representation represents—that is, the vector—still has an order in its underlying structure.
I think a better way to explain this is by trying to better understand what $Bbb{N}^infty$ is. For me, I like to think of $Bbb{N}^infty$ as the set of all functions from $Bbb{N}rightarrow Bbb{N}$. Thus, here, $v$ is the vector such that $v(n)=n$ for all $nin Bbb{N}$, which is like the identity function on $Bbb{N}$. However, even though $v(n)$ can always just be replaced with $n$ and therefore the identity function may seem redundant, that does not change the underlying structure behind $v(n)$, which is that it is an infinite-dimensional vector, and not a set.
Another way to realize the difference is by examining the vector $w(1)=2$, $w(2)=1$ and $w(n)=n$ for all $n > 3$. Clearly, the set of components of this vector are the natural numbers. However, this is also true for $v$, yet $vneq w$ because $v$ and $w$ have their components ordered differently: $v$ has the order $1$, then $2$, while $w$ has the order $2$, then $1$. On the other hand, $Bbb{N}$ has no such underlying order to it, so it is fundamentally distinct from vectors.
answered Dec 30 '18 at 1:08
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
1
$begingroup$
Good answer. I would add that a vector is only a vector in virtue of the larger structure of a vector space.
$endgroup$
– Ben W
Dec 30 '18 at 1:13
$begingroup$
For each vector whose entries are (for continuity's sake) natural numbers, would there be a unique function $Sort:mathbb{N}tomathbb{N}$ which assigns a unique position in $v$ to each $nin mathbb{N}$?
$endgroup$
– R. Burton
Dec 30 '18 at 1:14
1
$begingroup$
@R.Burton I'm not sure if I understand your comment, but I think what you're talking about is called the inverse function and is denoted as $v^{-1}$. Obviously, for $v$, $v^{-1}(n)=n$ since the position of each element $n$ in $v$ is simply $n$. However, the inverse function does not exist for all elements of $Bbb{N}^infty$. See $zeta(n)=0$ for all $ninBbb{N}$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:15
$begingroup$
@NobleMushtak Correction: for each totally ordered $Xsubseteqmathbb{N}$ there is a unique function $Sort:2^mathbb{N}tobigcup_{i=0}^inftymathbb{N}^i$ such that $Sort(X)=textbf{v}$, where $textbf{v}$ is the equivalent vector. The converse, that every vector is the image of a set under $Sort$ is only true if $Sort$ admits multisets as inputs, or is multivalued, otherwise, only vectors with unique entries can be described this way.
$endgroup$
– R. Burton
Jan 1 at 18:08
$begingroup$
@R.Burton I see what you mean now. Yes, for any subset of $X subset Bbb{N}$, there is a corresponding vector $mathbf{v}$ such that the set of all components in $mathbf{v}$ is $X$. However, this does not mean $mathbf{v}$ is the "equivalent" vector or sequence to $X$, because there are multiple vectors for which the set of all components is $X$.
$endgroup$
– Noble Mushtak
Jan 1 at 20:06
|
show 1 more comment
$begingroup$
Unfortunately, this probably is not a satisfying answer, but the key difference between the set and the vector is that the set has no order to its terms, while the vector does. You say you could change the written representation of the vector and change the order, but that change in the written representation does not alter the fact that the inherent mathematical object which that representation represents—that is, the vector—still has an order in its underlying structure.
I think a better way to explain this is by trying to better understand what $Bbb{N}^infty$ is. For me, I like to think of $Bbb{N}^infty$ as the set of all functions from $Bbb{N}rightarrow Bbb{N}$. Thus, here, $v$ is the vector such that $v(n)=n$ for all $nin Bbb{N}$, which is like the identity function on $Bbb{N}$. However, even though $v(n)$ can always just be replaced with $n$ and therefore the identity function may seem redundant, that does not change the underlying structure behind $v(n)$, which is that it is an infinite-dimensional vector, and not a set.
Another way to realize the difference is by examining the vector $w(1)=2$, $w(2)=1$ and $w(n)=n$ for all $n > 3$. Clearly, the set of components of this vector are the natural numbers. However, this is also true for $v$, yet $vneq w$ because $v$ and $w$ have their components ordered differently: $v$ has the order $1$, then $2$, while $w$ has the order $2$, then $1$. On the other hand, $Bbb{N}$ has no such underlying order to it, so it is fundamentally distinct from vectors.
answered Dec 30 '18 at 1:08
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
Unfortunately, this probably is not a satisfying answer, but the key difference between the set and the vector is that the set has no order to its terms, while the vector does. You say you could change the written representation of the vector and change the order, but that change in the written representation does not alter the fact that the inherent mathematical object which that representation represents—that is, the vector—still has an order in its underlying structure.
I think a better way to explain this is by trying to better understand what $Bbb{N}^infty$ is. For me, I like to think of $Bbb{N}^infty$ as the set of all functions from $Bbb{N}rightarrow Bbb{N}$. Thus, here, $v$ is the vector such that $v(n)=n$ for all $nin Bbb{N}$, which is like the identity function on $Bbb{N}$. However, even though $v(n)$ can always just be replaced with $n$ and therefore the identity function may seem redundant, that does not change the underlying structure behind $v(n)$, which is that it is an infinite-dimensional vector, and not a set.
Another way to realize the difference is by examining the vector $w(1)=2$, $w(2)=1$ and $w(n)=n$ for all $n > 3$. Clearly, the set of components of this vector are the natural numbers. However, this is also true for $v$, yet $vneq w$ because $v$ and $w$ have their components ordered differently: $v$ has the order $1$, then $2$, while $w$ has the order $2$, then $1$. On the other hand, $Bbb{N}$ has no such underlying order to it, so it is fundamentally distinct from vectors.
answered Dec 30 '18 at 1:08
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 30 '18 at 1:08
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 30 '18 at 1:08
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 30 '18 at 1:08
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
1
$begingroup$
Good answer. I would add that a vector is only a vector in virtue of the larger structure of a vector space.
$endgroup$
– Ben W
Dec 30 '18 at 1:13
$begingroup$
For each vector whose entries are (for continuity's sake) natural numbers, would there be a unique function $Sort:mathbb{N}tomathbb{N}$ which assigns a unique position in $v$ to each $nin mathbb{N}$?
$endgroup$
– R. Burton
Dec 30 '18 at 1:14
1
$begingroup$
@R.Burton I'm not sure if I understand your comment, but I think what you're talking about is called the inverse function and is denoted as $v^{-1}$. Obviously, for $v$, $v^{-1}(n)=n$ since the position of each element $n$ in $v$ is simply $n$. However, the inverse function does not exist for all elements of $Bbb{N}^infty$. See $zeta(n)=0$ for all $ninBbb{N}$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:15
$begingroup$
@NobleMushtak Correction: for each totally ordered $Xsubseteqmathbb{N}$ there is a unique function $Sort:2^mathbb{N}tobigcup_{i=0}^inftymathbb{N}^i$ such that $Sort(X)=textbf{v}$, where $textbf{v}$ is the equivalent vector. The converse, that every vector is the image of a set under $Sort$ is only true if $Sort$ admits multisets as inputs, or is multivalued, otherwise, only vectors with unique entries can be described this way.
$endgroup$
– R. Burton
Jan 1 at 18:08
$begingroup$
@R.Burton I see what you mean now. Yes, for any subset of $X subset Bbb{N}$, there is a corresponding vector $mathbf{v}$ such that the set of all components in $mathbf{v}$ is $X$. However, this does not mean $mathbf{v}$ is the "equivalent" vector or sequence to $X$, because there are multiple vectors for which the set of all components is $X$.
$endgroup$
– Noble Mushtak
Jan 1 at 20:06
|
show 1 more comment
1
$begingroup$
Good answer. I would add that a vector is only a vector in virtue of the larger structure of a vector space.
$endgroup$
– Ben W
Dec 30 '18 at 1:13
$begingroup$
For each vector whose entries are (for continuity's sake) natural numbers, would there be a unique function $Sort:mathbb{N}tomathbb{N}$ which assigns a unique position in $v$ to each $nin mathbb{N}$?
$endgroup$
– R. Burton
Dec 30 '18 at 1:14
1
$begingroup$
@R.Burton I'm not sure if I understand your comment, but I think what you're talking about is called the inverse function and is denoted as $v^{-1}$. Obviously, for $v$, $v^{-1}(n)=n$ since the position of each element $n$ in $v$ is simply $n$. However, the inverse function does not exist for all elements of $Bbb{N}^infty$. See $zeta(n)=0$ for all $ninBbb{N}$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:15
$begingroup$
@NobleMushtak Correction: for each totally ordered $Xsubseteqmathbb{N}$ there is a unique function $Sort:2^mathbb{N}tobigcup_{i=0}^inftymathbb{N}^i$ such that $Sort(X)=textbf{v}$, where $textbf{v}$ is the equivalent vector. The converse, that every vector is the image of a set under $Sort$ is only true if $Sort$ admits multisets as inputs, or is multivalued, otherwise, only vectors with unique entries can be described this way.
$endgroup$
– R. Burton
Jan 1 at 18:08
$begingroup$
@R.Burton I see what you mean now. Yes, for any subset of $X subset Bbb{N}$, there is a corresponding vector $mathbf{v}$ such that the set of all components in $mathbf{v}$ is $X$. However, this does not mean $mathbf{v}$ is the "equivalent" vector or sequence to $X$, because there are multiple vectors for which the set of all components is $X$.
$endgroup$
– Noble Mushtak
Jan 1 at 20:06
1
1
$begingroup$
Good answer. I would add that a vector is only a vector in virtue of the larger structure of a vector space.
$endgroup$
– Ben W
Dec 30 '18 at 1:13
$begingroup$
Good answer. I would add that a vector is only a vector in virtue of the larger structure of a vector space.
$endgroup$
– Ben W
Dec 30 '18 at 1:13
$begingroup$
For each vector whose entries are (for continuity's sake) natural numbers, would there be a unique function $Sort:mathbb{N}tomathbb{N}$ which assigns a unique position in $v$ to each $nin mathbb{N}$?
$endgroup$
– R. Burton
Dec 30 '18 at 1:14
$begingroup$
For each vector whose entries are (for continuity's sake) natural numbers, would there be a unique function $Sort:mathbb{N}tomathbb{N}$ which assigns a unique position in $v$ to each $nin mathbb{N}$?
$endgroup$
– R. Burton
Dec 30 '18 at 1:14
1
1
$begingroup$
@R.Burton I'm not sure if I understand your comment, but I think what you're talking about is called the inverse function and is denoted as $v^{-1}$. Obviously, for $v$, $v^{-1}(n)=n$ since the position of each element $n$ in $v$ is simply $n$. However, the inverse function does not exist for all elements of $Bbb{N}^infty$. See $zeta(n)=0$ for all $ninBbb{N}$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:15
$begingroup$
@R.Burton I'm not sure if I understand your comment, but I think what you're talking about is called the inverse function and is denoted as $v^{-1}$. Obviously, for $v$, $v^{-1}(n)=n$ since the position of each element $n$ in $v$ is simply $n$. However, the inverse function does not exist for all elements of $Bbb{N}^infty$. See $zeta(n)=0$ for all $ninBbb{N}$.
$endgroup$
– Noble Mushtak
Dec 30 '18 at 1:15
$begingroup$
@NobleMushtak Correction: for each totally ordered $Xsubseteqmathbb{N}$ there is a unique function $Sort:2^mathbb{N}tobigcup_{i=0}^inftymathbb{N}^i$ such that $Sort(X)=textbf{v}$, where $textbf{v}$ is the equivalent vector. The converse, that every vector is the image of a set under $Sort$ is only true if $Sort$ admits multisets as inputs, or is multivalued, otherwise, only vectors with unique entries can be described this way.
$endgroup$
– R. Burton
Jan 1 at 18:08
$begingroup$
@NobleMushtak Correction: for each totally ordered $Xsubseteqmathbb{N}$ there is a unique function $Sort:2^mathbb{N}tobigcup_{i=0}^inftymathbb{N}^i$ such that $Sort(X)=textbf{v}$, where $textbf{v}$ is the equivalent vector. The converse, that every vector is the image of a set under $Sort$ is only true if $Sort$ admits multisets as inputs, or is multivalued, otherwise, only vectors with unique entries can be described this way.
$endgroup$
– R. Burton
Jan 1 at 18:08
$begingroup$
@R.Burton I see what you mean now. Yes, for any subset of $X subset Bbb{N}$, there is a corresponding vector $mathbf{v}$ such that the set of all components in $mathbf{v}$ is $X$. However, this does not mean $mathbf{v}$ is the "equivalent" vector or sequence to $X$, because there are multiple vectors for which the set of all components is $X$.
$endgroup$
– Noble Mushtak
Jan 1 at 20:06
$begingroup$
@R.Burton I see what you mean now. Yes, for any subset of $X subset Bbb{N}$, there is a corresponding vector $mathbf{v}$ such that the set of all components in $mathbf{v}$ is $X$. However, this does not mean $mathbf{v}$ is the "equivalent" vector or sequence to $X$, because there are multiple vectors for which the set of all components is $X$.
$endgroup$
– Noble Mushtak
Jan 1 at 20:06
|
show 1 more comment
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056416%2fwhat-is-the-difference-between-a-vector-and-a-countable-nowhere-dense-totally%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I don't understand. Do you know what a vector is?
$endgroup$
– lulu
Dec 30 '18 at 1:06
6
$begingroup$
Reminds me of the old riddle: "What is the difference between Marilyn Monroe and a submarine?" "Marilyn Monroe is a movie actress, and a submarine is a boat that goes under water."
$endgroup$
– bof
Dec 30 '18 at 1:44
$begingroup$
What does "nowhere dense" have to do with it? Why is a countable dense set like the rational numbers any less like a vector than a countable nowhere dense set?
$endgroup$
– bof
Jan 1 at 3:32
$begingroup$
@bof Some orderings of a dense set cannot be encoded by a vector. Take, for example, $mathbb{Q}cap[0,1]$, ordered by $i<jiff a_i<a_j$. The first element of the vector $(a_1,ldots,a_infty)$, would obviously be $0$, but since the next element can be arbitrarily close to $0$, there cannot be a unique $a_2$. Thus, there is no unambiguous vector which encodes both the specified, order and all elements of the rationals in $[0,1]$. Granted, it is possible to create a vector to represent any countable set. But only non-dense sets can be represented this way for any order.
$endgroup$
– R. Burton
Jan 1 at 17:59
$begingroup$
How do you represent $mathbb Z$ with the natural order as a vector? Better yet, how about the nowhere dense countable set consisting of the midpoints of the "deleted intervals" of the Cantor set? This can be described as the set of numbers whose base $3$ representation has a $1$ in it, and all digits after the first $1$ are $1$'s, i.e., $0.overline1$, $0.0overline1$, $0.2overline1$, $0.00overline1$, $0.02overline1$, $0.20overline1$, $0.22overline1$, $0.000overline1$, $0.002overline1$, $dots$, This is a countable nowhere dense set, and it is order-isomorphic to $mathbb Q$.
$endgroup$
– bof
Jan 2 at 3:16