The product of locally compact spaces is locally compact











up vote
2
down vote

favorite












My definition of local compactness is as follows:




A topological space $(X, mathcal{T})$ is locally compact if, for every $x in X$, there is a $U in mathcal{T}$ with $x in U$ such that $overline{U}$ is compact.




I am then asked to prove that, if $X$ and $Y$ are locally compact and Hausdorff, then $X times Y$ equipped with the product topology is also locally compact. Trouble is, I don't believe I use Hausdorff in my proof! And I don't see where it breaks down. Here it is:



Let $(X, mathcal{T}_X)$ and $(Y, mathcal{T}_Y)$ be locally compact. Consider a point $(x, y) in X times Y$. Then there exist $U in mathcal{T}_X$ with $x in U$ and $V in mathcal{T}_Y$ with $y in V$ such that $overline{U}$ is compact in $X$ and $overline{V}$ is compact in $Y$.



Firstly, $U times V$ is open in $X times Y$, and $(x,y) in U times V$. Furthermore, $overline{U}$ is compact as a subset of $X$, so considered as a topological space with the subspace topology, it is also compact. Similarly for $overline{V}$. Then the space $overline{U} times overline{V}$ equipped with the product topology is also compact. Now I believe that the product topology induced by the topologies on $overline{U}$ and $overline{V}$ is the same as the subspace topology induced by considering $overline{U} times overline{V}$ as a subspace of $X times Y$. So $overline{U} times overline{V}$ is also compact with respect to the subspace topology, and so it is a compact subset of $X times Y$. But $overline{U} times overline{V} = overline{U times V}$, so we're done.










share|cite|improve this question
























  • @mathworker21 (I wrote "I believe" since I haven't seen a proof of that fact, but I have tried proving it myself, and I think my proof is correct. So I didn't want to claim it as definitely true!) Well, the problem specifically asks me to assume Hausdorff, but other than that I don't have any reason to believe so. It also didn't make sense to me intuitive why I would need it, so I tried proving it without that assumption.
    – Danny Hansen
    Nov 24 at 14:36












  • ok, I guess if it said "locally compact and Hausdorff" that is weird. usually things will say "locally compact Hausdorff" or just assume everything is Hausdorff just for ease. sorry for my first comment.
    – mathworker21
    Nov 24 at 14:39












  • @mathworker21 Well, it's not in English, but directly translated it does say "locally compact Hausdorff" (it says "assume $X$ and $Y$ are locally compact Hausdorff spaces"). But unless I am mistaken that just means that the space is both locally compact and Hausdorff. Sorry if that makes a difference. Local compactness is only defined in the context of this problem, so I don't know any results about it.
    – Danny Hansen
    Nov 24 at 14:42












  • It doesn't make a real difference. What I was trying to say is that people usually assume things are Hausdorff, just to sometimes make their life easier. When they do this, they say "locally compact Hausdorff". But if someone said "locally compact and Hausdorff", I would think they were emphasizing the word "Hausdorff", which would be weird. If what I'm saying sounds nonsensical to you, it's because it mostly is. Goodnight
    – mathworker21
    Nov 24 at 14:46






  • 1




    A brief comment: to say that a space $X$ is locally foo generally means that given any point $x in X$ in the space and any neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ with $V subseteq U$ and having the property foo. In the presence of the Hausdorff condition, your definition is equivalent to what I would therefore call locally compact.
    – Stephen
    Nov 24 at 19:04

















up vote
2
down vote

favorite












My definition of local compactness is as follows:




A topological space $(X, mathcal{T})$ is locally compact if, for every $x in X$, there is a $U in mathcal{T}$ with $x in U$ such that $overline{U}$ is compact.




I am then asked to prove that, if $X$ and $Y$ are locally compact and Hausdorff, then $X times Y$ equipped with the product topology is also locally compact. Trouble is, I don't believe I use Hausdorff in my proof! And I don't see where it breaks down. Here it is:



Let $(X, mathcal{T}_X)$ and $(Y, mathcal{T}_Y)$ be locally compact. Consider a point $(x, y) in X times Y$. Then there exist $U in mathcal{T}_X$ with $x in U$ and $V in mathcal{T}_Y$ with $y in V$ such that $overline{U}$ is compact in $X$ and $overline{V}$ is compact in $Y$.



Firstly, $U times V$ is open in $X times Y$, and $(x,y) in U times V$. Furthermore, $overline{U}$ is compact as a subset of $X$, so considered as a topological space with the subspace topology, it is also compact. Similarly for $overline{V}$. Then the space $overline{U} times overline{V}$ equipped with the product topology is also compact. Now I believe that the product topology induced by the topologies on $overline{U}$ and $overline{V}$ is the same as the subspace topology induced by considering $overline{U} times overline{V}$ as a subspace of $X times Y$. So $overline{U} times overline{V}$ is also compact with respect to the subspace topology, and so it is a compact subset of $X times Y$. But $overline{U} times overline{V} = overline{U times V}$, so we're done.










share|cite|improve this question
























  • @mathworker21 (I wrote "I believe" since I haven't seen a proof of that fact, but I have tried proving it myself, and I think my proof is correct. So I didn't want to claim it as definitely true!) Well, the problem specifically asks me to assume Hausdorff, but other than that I don't have any reason to believe so. It also didn't make sense to me intuitive why I would need it, so I tried proving it without that assumption.
    – Danny Hansen
    Nov 24 at 14:36












  • ok, I guess if it said "locally compact and Hausdorff" that is weird. usually things will say "locally compact Hausdorff" or just assume everything is Hausdorff just for ease. sorry for my first comment.
    – mathworker21
    Nov 24 at 14:39












  • @mathworker21 Well, it's not in English, but directly translated it does say "locally compact Hausdorff" (it says "assume $X$ and $Y$ are locally compact Hausdorff spaces"). But unless I am mistaken that just means that the space is both locally compact and Hausdorff. Sorry if that makes a difference. Local compactness is only defined in the context of this problem, so I don't know any results about it.
    – Danny Hansen
    Nov 24 at 14:42












  • It doesn't make a real difference. What I was trying to say is that people usually assume things are Hausdorff, just to sometimes make their life easier. When they do this, they say "locally compact Hausdorff". But if someone said "locally compact and Hausdorff", I would think they were emphasizing the word "Hausdorff", which would be weird. If what I'm saying sounds nonsensical to you, it's because it mostly is. Goodnight
    – mathworker21
    Nov 24 at 14:46






  • 1




    A brief comment: to say that a space $X$ is locally foo generally means that given any point $x in X$ in the space and any neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ with $V subseteq U$ and having the property foo. In the presence of the Hausdorff condition, your definition is equivalent to what I would therefore call locally compact.
    – Stephen
    Nov 24 at 19:04















up vote
2
down vote

favorite









up vote
2
down vote

favorite











My definition of local compactness is as follows:




A topological space $(X, mathcal{T})$ is locally compact if, for every $x in X$, there is a $U in mathcal{T}$ with $x in U$ such that $overline{U}$ is compact.




I am then asked to prove that, if $X$ and $Y$ are locally compact and Hausdorff, then $X times Y$ equipped with the product topology is also locally compact. Trouble is, I don't believe I use Hausdorff in my proof! And I don't see where it breaks down. Here it is:



Let $(X, mathcal{T}_X)$ and $(Y, mathcal{T}_Y)$ be locally compact. Consider a point $(x, y) in X times Y$. Then there exist $U in mathcal{T}_X$ with $x in U$ and $V in mathcal{T}_Y$ with $y in V$ such that $overline{U}$ is compact in $X$ and $overline{V}$ is compact in $Y$.



Firstly, $U times V$ is open in $X times Y$, and $(x,y) in U times V$. Furthermore, $overline{U}$ is compact as a subset of $X$, so considered as a topological space with the subspace topology, it is also compact. Similarly for $overline{V}$. Then the space $overline{U} times overline{V}$ equipped with the product topology is also compact. Now I believe that the product topology induced by the topologies on $overline{U}$ and $overline{V}$ is the same as the subspace topology induced by considering $overline{U} times overline{V}$ as a subspace of $X times Y$. So $overline{U} times overline{V}$ is also compact with respect to the subspace topology, and so it is a compact subset of $X times Y$. But $overline{U} times overline{V} = overline{U times V}$, so we're done.










share|cite|improve this question















My definition of local compactness is as follows:




A topological space $(X, mathcal{T})$ is locally compact if, for every $x in X$, there is a $U in mathcal{T}$ with $x in U$ such that $overline{U}$ is compact.




I am then asked to prove that, if $X$ and $Y$ are locally compact and Hausdorff, then $X times Y$ equipped with the product topology is also locally compact. Trouble is, I don't believe I use Hausdorff in my proof! And I don't see where it breaks down. Here it is:



Let $(X, mathcal{T}_X)$ and $(Y, mathcal{T}_Y)$ be locally compact. Consider a point $(x, y) in X times Y$. Then there exist $U in mathcal{T}_X$ with $x in U$ and $V in mathcal{T}_Y$ with $y in V$ such that $overline{U}$ is compact in $X$ and $overline{V}$ is compact in $Y$.



Firstly, $U times V$ is open in $X times Y$, and $(x,y) in U times V$. Furthermore, $overline{U}$ is compact as a subset of $X$, so considered as a topological space with the subspace topology, it is also compact. Similarly for $overline{V}$. Then the space $overline{U} times overline{V}$ equipped with the product topology is also compact. Now I believe that the product topology induced by the topologies on $overline{U}$ and $overline{V}$ is the same as the subspace topology induced by considering $overline{U} times overline{V}$ as a subspace of $X times Y$. So $overline{U} times overline{V}$ is also compact with respect to the subspace topology, and so it is a compact subset of $X times Y$. But $overline{U} times overline{V} = overline{U times V}$, so we're done.







general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 18:28

























asked Nov 24 at 14:26









Danny Hansen

816




816












  • @mathworker21 (I wrote "I believe" since I haven't seen a proof of that fact, but I have tried proving it myself, and I think my proof is correct. So I didn't want to claim it as definitely true!) Well, the problem specifically asks me to assume Hausdorff, but other than that I don't have any reason to believe so. It also didn't make sense to me intuitive why I would need it, so I tried proving it without that assumption.
    – Danny Hansen
    Nov 24 at 14:36












  • ok, I guess if it said "locally compact and Hausdorff" that is weird. usually things will say "locally compact Hausdorff" or just assume everything is Hausdorff just for ease. sorry for my first comment.
    – mathworker21
    Nov 24 at 14:39












  • @mathworker21 Well, it's not in English, but directly translated it does say "locally compact Hausdorff" (it says "assume $X$ and $Y$ are locally compact Hausdorff spaces"). But unless I am mistaken that just means that the space is both locally compact and Hausdorff. Sorry if that makes a difference. Local compactness is only defined in the context of this problem, so I don't know any results about it.
    – Danny Hansen
    Nov 24 at 14:42












  • It doesn't make a real difference. What I was trying to say is that people usually assume things are Hausdorff, just to sometimes make their life easier. When they do this, they say "locally compact Hausdorff". But if someone said "locally compact and Hausdorff", I would think they were emphasizing the word "Hausdorff", which would be weird. If what I'm saying sounds nonsensical to you, it's because it mostly is. Goodnight
    – mathworker21
    Nov 24 at 14:46






  • 1




    A brief comment: to say that a space $X$ is locally foo generally means that given any point $x in X$ in the space and any neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ with $V subseteq U$ and having the property foo. In the presence of the Hausdorff condition, your definition is equivalent to what I would therefore call locally compact.
    – Stephen
    Nov 24 at 19:04




















  • @mathworker21 (I wrote "I believe" since I haven't seen a proof of that fact, but I have tried proving it myself, and I think my proof is correct. So I didn't want to claim it as definitely true!) Well, the problem specifically asks me to assume Hausdorff, but other than that I don't have any reason to believe so. It also didn't make sense to me intuitive why I would need it, so I tried proving it without that assumption.
    – Danny Hansen
    Nov 24 at 14:36












  • ok, I guess if it said "locally compact and Hausdorff" that is weird. usually things will say "locally compact Hausdorff" or just assume everything is Hausdorff just for ease. sorry for my first comment.
    – mathworker21
    Nov 24 at 14:39












  • @mathworker21 Well, it's not in English, but directly translated it does say "locally compact Hausdorff" (it says "assume $X$ and $Y$ are locally compact Hausdorff spaces"). But unless I am mistaken that just means that the space is both locally compact and Hausdorff. Sorry if that makes a difference. Local compactness is only defined in the context of this problem, so I don't know any results about it.
    – Danny Hansen
    Nov 24 at 14:42












  • It doesn't make a real difference. What I was trying to say is that people usually assume things are Hausdorff, just to sometimes make their life easier. When they do this, they say "locally compact Hausdorff". But if someone said "locally compact and Hausdorff", I would think they were emphasizing the word "Hausdorff", which would be weird. If what I'm saying sounds nonsensical to you, it's because it mostly is. Goodnight
    – mathworker21
    Nov 24 at 14:46






  • 1




    A brief comment: to say that a space $X$ is locally foo generally means that given any point $x in X$ in the space and any neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ with $V subseteq U$ and having the property foo. In the presence of the Hausdorff condition, your definition is equivalent to what I would therefore call locally compact.
    – Stephen
    Nov 24 at 19:04


















@mathworker21 (I wrote "I believe" since I haven't seen a proof of that fact, but I have tried proving it myself, and I think my proof is correct. So I didn't want to claim it as definitely true!) Well, the problem specifically asks me to assume Hausdorff, but other than that I don't have any reason to believe so. It also didn't make sense to me intuitive why I would need it, so I tried proving it without that assumption.
– Danny Hansen
Nov 24 at 14:36






@mathworker21 (I wrote "I believe" since I haven't seen a proof of that fact, but I have tried proving it myself, and I think my proof is correct. So I didn't want to claim it as definitely true!) Well, the problem specifically asks me to assume Hausdorff, but other than that I don't have any reason to believe so. It also didn't make sense to me intuitive why I would need it, so I tried proving it without that assumption.
– Danny Hansen
Nov 24 at 14:36














ok, I guess if it said "locally compact and Hausdorff" that is weird. usually things will say "locally compact Hausdorff" or just assume everything is Hausdorff just for ease. sorry for my first comment.
– mathworker21
Nov 24 at 14:39






ok, I guess if it said "locally compact and Hausdorff" that is weird. usually things will say "locally compact Hausdorff" or just assume everything is Hausdorff just for ease. sorry for my first comment.
– mathworker21
Nov 24 at 14:39














@mathworker21 Well, it's not in English, but directly translated it does say "locally compact Hausdorff" (it says "assume $X$ and $Y$ are locally compact Hausdorff spaces"). But unless I am mistaken that just means that the space is both locally compact and Hausdorff. Sorry if that makes a difference. Local compactness is only defined in the context of this problem, so I don't know any results about it.
– Danny Hansen
Nov 24 at 14:42






@mathworker21 Well, it's not in English, but directly translated it does say "locally compact Hausdorff" (it says "assume $X$ and $Y$ are locally compact Hausdorff spaces"). But unless I am mistaken that just means that the space is both locally compact and Hausdorff. Sorry if that makes a difference. Local compactness is only defined in the context of this problem, so I don't know any results about it.
– Danny Hansen
Nov 24 at 14:42














It doesn't make a real difference. What I was trying to say is that people usually assume things are Hausdorff, just to sometimes make their life easier. When they do this, they say "locally compact Hausdorff". But if someone said "locally compact and Hausdorff", I would think they were emphasizing the word "Hausdorff", which would be weird. If what I'm saying sounds nonsensical to you, it's because it mostly is. Goodnight
– mathworker21
Nov 24 at 14:46




It doesn't make a real difference. What I was trying to say is that people usually assume things are Hausdorff, just to sometimes make their life easier. When they do this, they say "locally compact Hausdorff". But if someone said "locally compact and Hausdorff", I would think they were emphasizing the word "Hausdorff", which would be weird. If what I'm saying sounds nonsensical to you, it's because it mostly is. Goodnight
– mathworker21
Nov 24 at 14:46




1




1




A brief comment: to say that a space $X$ is locally foo generally means that given any point $x in X$ in the space and any neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ with $V subseteq U$ and having the property foo. In the presence of the Hausdorff condition, your definition is equivalent to what I would therefore call locally compact.
– Stephen
Nov 24 at 19:04






A brief comment: to say that a space $X$ is locally foo generally means that given any point $x in X$ in the space and any neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ with $V subseteq U$ and having the property foo. In the presence of the Hausdorff condition, your definition is equivalent to what I would therefore call locally compact.
– Stephen
Nov 24 at 19:04












1 Answer
1






active

oldest

votes

















up vote
1
down vote













With that definition of local compactness (there are several, so I'm glad you included it) your proof is indeed correct.



$overline{Utimes V} = overline{U} times overline{V}$ in the product topology and the product of compact sets is compact, in short.



It's very common for people to consider local compactness in the context of Hausdorff spaces only, as then all the different usual definitions are all equivalent to each other, and we can prove many more theorems (and we have a nicely behaved one-point compactification as well). The exercise poser could just have been lazy and didn't realise that for this one result he could do away with the Hausdorffness altogether.






share|cite|improve this answer





















  • Thanks for that clarification. I did indeed find several different definitions before asking this question, so I thought they might not be equivalent in general. But interesting that they are in a Hausdorff space.
    – Danny Hansen
    Nov 24 at 18:30











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',
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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011614%2fthe-product-of-locally-compact-spaces-is-locally-compact%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








up vote
1
down vote













With that definition of local compactness (there are several, so I'm glad you included it) your proof is indeed correct.



$overline{Utimes V} = overline{U} times overline{V}$ in the product topology and the product of compact sets is compact, in short.



It's very common for people to consider local compactness in the context of Hausdorff spaces only, as then all the different usual definitions are all equivalent to each other, and we can prove many more theorems (and we have a nicely behaved one-point compactification as well). The exercise poser could just have been lazy and didn't realise that for this one result he could do away with the Hausdorffness altogether.






share|cite|improve this answer





















  • Thanks for that clarification. I did indeed find several different definitions before asking this question, so I thought they might not be equivalent in general. But interesting that they are in a Hausdorff space.
    – Danny Hansen
    Nov 24 at 18:30















up vote
1
down vote













With that definition of local compactness (there are several, so I'm glad you included it) your proof is indeed correct.



$overline{Utimes V} = overline{U} times overline{V}$ in the product topology and the product of compact sets is compact, in short.



It's very common for people to consider local compactness in the context of Hausdorff spaces only, as then all the different usual definitions are all equivalent to each other, and we can prove many more theorems (and we have a nicely behaved one-point compactification as well). The exercise poser could just have been lazy and didn't realise that for this one result he could do away with the Hausdorffness altogether.






share|cite|improve this answer





















  • Thanks for that clarification. I did indeed find several different definitions before asking this question, so I thought they might not be equivalent in general. But interesting that they are in a Hausdorff space.
    – Danny Hansen
    Nov 24 at 18:30













up vote
1
down vote










up vote
1
down vote









With that definition of local compactness (there are several, so I'm glad you included it) your proof is indeed correct.



$overline{Utimes V} = overline{U} times overline{V}$ in the product topology and the product of compact sets is compact, in short.



It's very common for people to consider local compactness in the context of Hausdorff spaces only, as then all the different usual definitions are all equivalent to each other, and we can prove many more theorems (and we have a nicely behaved one-point compactification as well). The exercise poser could just have been lazy and didn't realise that for this one result he could do away with the Hausdorffness altogether.






share|cite|improve this answer












With that definition of local compactness (there are several, so I'm glad you included it) your proof is indeed correct.



$overline{Utimes V} = overline{U} times overline{V}$ in the product topology and the product of compact sets is compact, in short.



It's very common for people to consider local compactness in the context of Hausdorff spaces only, as then all the different usual definitions are all equivalent to each other, and we can prove many more theorems (and we have a nicely behaved one-point compactification as well). The exercise poser could just have been lazy and didn't realise that for this one result he could do away with the Hausdorffness altogether.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 18:16









Henno Brandsma

102k345109




102k345109












  • Thanks for that clarification. I did indeed find several different definitions before asking this question, so I thought they might not be equivalent in general. But interesting that they are in a Hausdorff space.
    – Danny Hansen
    Nov 24 at 18:30


















  • Thanks for that clarification. I did indeed find several different definitions before asking this question, so I thought they might not be equivalent in general. But interesting that they are in a Hausdorff space.
    – Danny Hansen
    Nov 24 at 18:30
















Thanks for that clarification. I did indeed find several different definitions before asking this question, so I thought they might not be equivalent in general. But interesting that they are in a Hausdorff space.
– Danny Hansen
Nov 24 at 18:30




Thanks for that clarification. I did indeed find several different definitions before asking this question, so I thought they might not be equivalent in general. But interesting that they are in a Hausdorff space.
– Danny Hansen
Nov 24 at 18:30


















draft saved

draft discarded




















































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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011614%2fthe-product-of-locally-compact-spaces-is-locally-compact%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Berounka

Sphinx de Gizeh

Different font size/position of beamer's navigation symbols template's content depending on regular/plain...