Mixing
$Y=prod_{ninmathbb{N}}X$.
$begingroup$
Let X be a compact CW-complex. The infinite cartesian product $Y=prod_{ninmathbb{N}}X$ is a compact topological space, and as CW-complex It should have a finite number of cells. But in the CW-decomposition of Y given in Cartesian product of two CW-complexes infinite cells appear.
What is happening?
cw-complexes
asked Jan 23 at 19:25
Rodrigo Moron SanzRodrigo Moron Sanz
415
$endgroup$
add a comment |
$begingroup$
Let X be a compact CW-complex. The infinite cartesian product $Y=prod_{ninmathbb{N}}X$ is a compact topological space, and as CW-complex It should have a finite number of cells. But in the CW-decomposition of Y given in Cartesian product of two CW-complexes infinite cells appear.
What is happening?
cw-complexes
asked Jan 23 at 19:25
Rodrigo Moron SanzRodrigo Moron Sanz
415
$endgroup$
$begingroup$
Is that a CW-complex?
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:27
1
$begingroup$
good question, maybe it does not admits a Cw structure.
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 19:32
add a comment |
$begingroup$
Let X be a compact CW-complex. The infinite cartesian product $Y=prod_{ninmathbb{N}}X$ is a compact topological space, and as CW-complex It should have a finite number of cells. But in the CW-decomposition of Y given in Cartesian product of two CW-complexes infinite cells appear.
What is happening?
cw-complexes
asked Jan 23 at 19:25
Rodrigo Moron SanzRodrigo Moron Sanz
415
$endgroup$
Let X be a compact CW-complex. The infinite cartesian product $Y=prod_{ninmathbb{N}}X$ is a compact topological space, and as CW-complex It should have a finite number of cells. But in the CW-decomposition of Y given in Cartesian product of two CW-complexes infinite cells appear.
What is happening?
cw-complexes
cw-complexes
asked Jan 23 at 19:25
Rodrigo Moron SanzRodrigo Moron Sanz
415
asked Jan 23 at 19:25
Rodrigo Moron SanzRodrigo Moron Sanz
415
edited Jan 30 at 19:35
Rodrigo Moron Sanz
asked Jan 23 at 19:25
Rodrigo Moron SanzRodrigo Moron Sanz
415
asked Jan 23 at 19:25
Rodrigo Moron SanzRodrigo Moron Sanz
415
asked Jan 23 at 19:25
Rodrigo Moron SanzRodrigo Moron Sanz
415
415
$begingroup$
Is that a CW-complex?
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:27
1
$begingroup$
good question, maybe it does not admits a Cw structure.
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 19:32
add a comment |
$begingroup$
Is that a CW-complex?
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:27
1
$begingroup$
good question, maybe it does not admits a Cw structure.
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 19:32
$begingroup$
Is that a CW-complex?
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:27
$begingroup$
Is that a CW-complex?
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:27
1
1
$begingroup$
good question, maybe it does not admits a Cw structure.
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 19:32
$begingroup$
good question, maybe it does not admits a Cw structure.
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 19:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is not a CW-complex unless $X$ has only a single point. To see this, let $X$ be a CW-complex which has more than one point.
Case 1. $dim(X) = 0$. Then $X$ is a finite discrete space and $Y$ is an infinite space. If it were a compact CW-complex, then it could have only finitely many $0$-cells. Hence it must have at least one (open) cell $e$ of dimension $n > 0$ which is a homeomorphic copy of $mathring{D}^n$. We conclude that $Y$ has a connected component with more than one point. On the other hand each component of $Y$ is the product of components of $X$, i.e. has only a single point. This is a contradiction.
Case 2. $dim(X) > 0$. Then $X$ has an open cell $e$ of dimension $n > 0$ and $Y$ contains a subspace homeomorphic to $E = prod_{n in mathbb{N}} e$. If $Y$ were a CW-complex (which is compact!), it would embed into some $mathbb{R}^m$. See Hatcher "Algebraic Topology" Corollary A.10. Hence also $E$ would embed into $mathbb{R}^m$. But now $E$ contains a copy of $mathbb{R}^{m+1}$, hence $mathbb{R}^{m+1}$ would embed into $mathbb{R}^m$. But this is impossible which is the consequence of topological dimension theory. See for example Chapter 7 of Engelking "General Topology" (in particular Theorems 7.1.1, 7.3.3, 7.3.19).
answered Jan 23 at 21:18
Paul FrostPaul Frost
12.2k3935
$endgroup$
$begingroup$
To nitpick, it is also a CW complex if $X$ is empty.
$endgroup$
– Eric Wofsey
Jan 23 at 21:22
$begingroup$
@EricWofsey :-)
$endgroup$
– Paul Frost
Jan 23 at 21:24
$begingroup$
So interesenting. Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 21:54
add a 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%2f3084944%2fy-prod-n-in-mathbbnx%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
$begingroup$
It is not a CW-complex unless $X$ has only a single point. To see this, let $X$ be a CW-complex which has more than one point.
Case 1. $dim(X) = 0$. Then $X$ is a finite discrete space and $Y$ is an infinite space. If it were a compact CW-complex, then it could have only finitely many $0$-cells. Hence it must have at least one (open) cell $e$ of dimension $n > 0$ which is a homeomorphic copy of $mathring{D}^n$. We conclude that $Y$ has a connected component with more than one point. On the other hand each component of $Y$ is the product of components of $X$, i.e. has only a single point. This is a contradiction.
Case 2. $dim(X) > 0$. Then $X$ has an open cell $e$ of dimension $n > 0$ and $Y$ contains a subspace homeomorphic to $E = prod_{n in mathbb{N}} e$. If $Y$ were a CW-complex (which is compact!), it would embed into some $mathbb{R}^m$. See Hatcher "Algebraic Topology" Corollary A.10. Hence also $E$ would embed into $mathbb{R}^m$. But now $E$ contains a copy of $mathbb{R}^{m+1}$, hence $mathbb{R}^{m+1}$ would embed into $mathbb{R}^m$. But this is impossible which is the consequence of topological dimension theory. See for example Chapter 7 of Engelking "General Topology" (in particular Theorems 7.1.1, 7.3.3, 7.3.19).
answered Jan 23 at 21:18
Paul FrostPaul Frost
12.2k3935
$endgroup$
$begingroup$
To nitpick, it is also a CW complex if $X$ is empty.
$endgroup$
– Eric Wofsey
Jan 23 at 21:22
$begingroup$
@EricWofsey :-)
$endgroup$
– Paul Frost
Jan 23 at 21:24
$begingroup$
So interesenting. Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 21:54
add a comment |
$begingroup$
It is not a CW-complex unless $X$ has only a single point. To see this, let $X$ be a CW-complex which has more than one point.
Case 1. $dim(X) = 0$. Then $X$ is a finite discrete space and $Y$ is an infinite space. If it were a compact CW-complex, then it could have only finitely many $0$-cells. Hence it must have at least one (open) cell $e$ of dimension $n > 0$ which is a homeomorphic copy of $mathring{D}^n$. We conclude that $Y$ has a connected component with more than one point. On the other hand each component of $Y$ is the product of components of $X$, i.e. has only a single point. This is a contradiction.
Case 2. $dim(X) > 0$. Then $X$ has an open cell $e$ of dimension $n > 0$ and $Y$ contains a subspace homeomorphic to $E = prod_{n in mathbb{N}} e$. If $Y$ were a CW-complex (which is compact!), it would embed into some $mathbb{R}^m$. See Hatcher "Algebraic Topology" Corollary A.10. Hence also $E$ would embed into $mathbb{R}^m$. But now $E$ contains a copy of $mathbb{R}^{m+1}$, hence $mathbb{R}^{m+1}$ would embed into $mathbb{R}^m$. But this is impossible which is the consequence of topological dimension theory. See for example Chapter 7 of Engelking "General Topology" (in particular Theorems 7.1.1, 7.3.3, 7.3.19).
answered Jan 23 at 21:18
Paul FrostPaul Frost
12.2k3935
$endgroup$
$begingroup$
To nitpick, it is also a CW complex if $X$ is empty.
$endgroup$
– Eric Wofsey
Jan 23 at 21:22
$begingroup$
@EricWofsey :-)
$endgroup$
– Paul Frost
Jan 23 at 21:24
$begingroup$
So interesenting. Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 21:54
add a comment |
$begingroup$
It is not a CW-complex unless $X$ has only a single point. To see this, let $X$ be a CW-complex which has more than one point.
Case 1. $dim(X) = 0$. Then $X$ is a finite discrete space and $Y$ is an infinite space. If it were a compact CW-complex, then it could have only finitely many $0$-cells. Hence it must have at least one (open) cell $e$ of dimension $n > 0$ which is a homeomorphic copy of $mathring{D}^n$. We conclude that $Y$ has a connected component with more than one point. On the other hand each component of $Y$ is the product of components of $X$, i.e. has only a single point. This is a contradiction.
Case 2. $dim(X) > 0$. Then $X$ has an open cell $e$ of dimension $n > 0$ and $Y$ contains a subspace homeomorphic to $E = prod_{n in mathbb{N}} e$. If $Y$ were a CW-complex (which is compact!), it would embed into some $mathbb{R}^m$. See Hatcher "Algebraic Topology" Corollary A.10. Hence also $E$ would embed into $mathbb{R}^m$. But now $E$ contains a copy of $mathbb{R}^{m+1}$, hence $mathbb{R}^{m+1}$ would embed into $mathbb{R}^m$. But this is impossible which is the consequence of topological dimension theory. See for example Chapter 7 of Engelking "General Topology" (in particular Theorems 7.1.1, 7.3.3, 7.3.19).
answered Jan 23 at 21:18
Paul FrostPaul Frost
12.2k3935
$endgroup$
It is not a CW-complex unless $X$ has only a single point. To see this, let $X$ be a CW-complex which has more than one point.
Case 1. $dim(X) = 0$. Then $X$ is a finite discrete space and $Y$ is an infinite space. If it were a compact CW-complex, then it could have only finitely many $0$-cells. Hence it must have at least one (open) cell $e$ of dimension $n > 0$ which is a homeomorphic copy of $mathring{D}^n$. We conclude that $Y$ has a connected component with more than one point. On the other hand each component of $Y$ is the product of components of $X$, i.e. has only a single point. This is a contradiction.
Case 2. $dim(X) > 0$. Then $X$ has an open cell $e$ of dimension $n > 0$ and $Y$ contains a subspace homeomorphic to $E = prod_{n in mathbb{N}} e$. If $Y$ were a CW-complex (which is compact!), it would embed into some $mathbb{R}^m$. See Hatcher "Algebraic Topology" Corollary A.10. Hence also $E$ would embed into $mathbb{R}^m$. But now $E$ contains a copy of $mathbb{R}^{m+1}$, hence $mathbb{R}^{m+1}$ would embed into $mathbb{R}^m$. But this is impossible which is the consequence of topological dimension theory. See for example Chapter 7 of Engelking "General Topology" (in particular Theorems 7.1.1, 7.3.3, 7.3.19).
answered Jan 23 at 21:18
Paul FrostPaul Frost
12.2k3935
answered Jan 23 at 21:18
Paul FrostPaul Frost
12.2k3935
answered Jan 23 at 21:18
Paul FrostPaul Frost
12.2k3935
answered Jan 23 at 21:18
Paul FrostPaul Frost
12.2k3935
12.2k3935
$begingroup$
To nitpick, it is also a CW complex if $X$ is empty.
$endgroup$
– Eric Wofsey
Jan 23 at 21:22
$begingroup$
@EricWofsey :-)
$endgroup$
– Paul Frost
Jan 23 at 21:24
$begingroup$
So interesenting. Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 21:54
add a comment |
$begingroup$
To nitpick, it is also a CW complex if $X$ is empty.
$endgroup$
– Eric Wofsey
Jan 23 at 21:22
$begingroup$
@EricWofsey :-)
$endgroup$
– Paul Frost
Jan 23 at 21:24
$begingroup$
So interesenting. Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 21:54
$begingroup$
To nitpick, it is also a CW complex if $X$ is empty.
$endgroup$
– Eric Wofsey
Jan 23 at 21:22
$begingroup$
To nitpick, it is also a CW complex if $X$ is empty.
$endgroup$
– Eric Wofsey
Jan 23 at 21:22
$begingroup$
@EricWofsey :-)
$endgroup$
– Paul Frost
Jan 23 at 21:24
$begingroup$
@EricWofsey :-)
$endgroup$
– Paul Frost
Jan 23 at 21:24
$begingroup$
So interesenting. Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 21:54
$begingroup$
So interesenting. Thanks a lot!
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 21:54
add a 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%2f3084944%2fy-prod-n-in-mathbbnx%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
$begingroup$
Is that a CW-complex?
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:27
1
$begingroup$
good question, maybe it does not admits a Cw structure.
$endgroup$
– Rodrigo Moron Sanz
Jan 23 at 19:32