The product of locally compact spaces is locally compact
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
asked Nov 24 at 14:26
Danny Hansen
816
add a comment |
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
asked Nov 24 at 14:26
Danny Hansen
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
add a comment |
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
asked Nov 24 at 14:26
Danny Hansen
816
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
general-topology
asked Nov 24 at 14:26
Danny Hansen
816
asked Nov 24 at 14:26
Danny Hansen
816
edited Nov 24 at 18:28
asked Nov 24 at 14:26
Danny Hansen
816
asked Nov 24 at 14:26
Danny Hansen
816
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
add a comment |
@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
add a comment |
1 Answer
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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.
answered Nov 24 at 18:16
Henno Brandsma
105k346113
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
add a comment |
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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.
answered Nov 24 at 18:16
Henno Brandsma
105k346113
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
add a comment |
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.
answered Nov 24 at 18:16
Henno Brandsma
105k346113
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
add a comment |
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.
answered Nov 24 at 18:16
Henno Brandsma
105k346113
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.
answered Nov 24 at 18:16
Henno Brandsma
105k346113
answered Nov 24 at 18:16
Henno Brandsma
105k346113
answered Nov 24 at 18:16
Henno Brandsma
105k346113
answered Nov 24 at 18:16
Henno Brandsma
105k346113
105k346113
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
add a comment |
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
add a comment |
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@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