Are open sets in locally compact spaces relatively compact?

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As the title suggests, I am investigating whether all open sets in locally compact
metric spaces are relatively compact. To me, this seems intuitive but I appear to be unable to produce a formal proof, that is, for an open set $U$ I cannot show that $barU$ is compact. I would therefore appreciate some help.



EDIT: The exact phrasing of the book from which this claim comes is : "If (X,d) is locally compact, we can even assume that $U$ is relatively compact.". Does this make sense?










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  • No. Pick any topological space that is locally compact, but not compact (e.g. $mathbbR$ under the usual topology). Then the space itself is open, but its closure is not compact.
    – Theo Bendit
    Sep 5 at 8:28










  • @TheoBendit Great example, thank you. Do you perhaps know of any conditions that will make this true?
    – JohnK
    Sep 5 at 8:37










  • Unfortunately, no I don't. Even in metric spaces, it's not that easy; it's possible for bounded sets in a locally compact space to not be relatively compact.
    – Theo Bendit
    Sep 5 at 8:40










  • @TheoBendit Please see the edit for the exact statement.
    – JohnK
    Sep 5 at 9:07






  • 2




    The phrase "we can even assume" makes it context-dependent. It depends on what you're using $U$ for. If you're using it in a situation where another smaller neighbourhood would suffice (i.e. one where we can always use some neighbourhood $V subseteq U$ instead of $U$), then yes this make sense. If your neighbourhood $U$ is not relatively compact, then replace it with $U cap V$, where $V$ is a relatively compact neighbourhood (which must exist by local compactness).
    – Theo Bendit
    Sep 5 at 9:11














up vote
0
down vote

favorite
1












As the title suggests, I am investigating whether all open sets in locally compact
metric spaces are relatively compact. To me, this seems intuitive but I appear to be unable to produce a formal proof, that is, for an open set $U$ I cannot show that $barU$ is compact. I would therefore appreciate some help.



EDIT: The exact phrasing of the book from which this claim comes is : "If (X,d) is locally compact, we can even assume that $U$ is relatively compact.". Does this make sense?










share|cite|improve this question























  • No. Pick any topological space that is locally compact, but not compact (e.g. $mathbbR$ under the usual topology). Then the space itself is open, but its closure is not compact.
    – Theo Bendit
    Sep 5 at 8:28










  • @TheoBendit Great example, thank you. Do you perhaps know of any conditions that will make this true?
    – JohnK
    Sep 5 at 8:37










  • Unfortunately, no I don't. Even in metric spaces, it's not that easy; it's possible for bounded sets in a locally compact space to not be relatively compact.
    – Theo Bendit
    Sep 5 at 8:40










  • @TheoBendit Please see the edit for the exact statement.
    – JohnK
    Sep 5 at 9:07






  • 2




    The phrase "we can even assume" makes it context-dependent. It depends on what you're using $U$ for. If you're using it in a situation where another smaller neighbourhood would suffice (i.e. one where we can always use some neighbourhood $V subseteq U$ instead of $U$), then yes this make sense. If your neighbourhood $U$ is not relatively compact, then replace it with $U cap V$, where $V$ is a relatively compact neighbourhood (which must exist by local compactness).
    – Theo Bendit
    Sep 5 at 9:11












up vote
0
down vote

favorite
1









up vote
0
down vote

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1






1





As the title suggests, I am investigating whether all open sets in locally compact
metric spaces are relatively compact. To me, this seems intuitive but I appear to be unable to produce a formal proof, that is, for an open set $U$ I cannot show that $barU$ is compact. I would therefore appreciate some help.



EDIT: The exact phrasing of the book from which this claim comes is : "If (X,d) is locally compact, we can even assume that $U$ is relatively compact.". Does this make sense?










share|cite|improve this question















As the title suggests, I am investigating whether all open sets in locally compact
metric spaces are relatively compact. To me, this seems intuitive but I appear to be unable to produce a formal proof, that is, for an open set $U$ I cannot show that $barU$ is compact. I would therefore appreciate some help.



EDIT: The exact phrasing of the book from which this claim comes is : "If (X,d) is locally compact, we can even assume that $U$ is relatively compact.". Does this make sense?







real-analysis general-topology






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edited Sep 5 at 9:04

























asked Sep 5 at 8:18









JohnK

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  • No. Pick any topological space that is locally compact, but not compact (e.g. $mathbbR$ under the usual topology). Then the space itself is open, but its closure is not compact.
    – Theo Bendit
    Sep 5 at 8:28










  • @TheoBendit Great example, thank you. Do you perhaps know of any conditions that will make this true?
    – JohnK
    Sep 5 at 8:37










  • Unfortunately, no I don't. Even in metric spaces, it's not that easy; it's possible for bounded sets in a locally compact space to not be relatively compact.
    – Theo Bendit
    Sep 5 at 8:40










  • @TheoBendit Please see the edit for the exact statement.
    – JohnK
    Sep 5 at 9:07






  • 2




    The phrase "we can even assume" makes it context-dependent. It depends on what you're using $U$ for. If you're using it in a situation where another smaller neighbourhood would suffice (i.e. one where we can always use some neighbourhood $V subseteq U$ instead of $U$), then yes this make sense. If your neighbourhood $U$ is not relatively compact, then replace it with $U cap V$, where $V$ is a relatively compact neighbourhood (which must exist by local compactness).
    – Theo Bendit
    Sep 5 at 9:11
















  • No. Pick any topological space that is locally compact, but not compact (e.g. $mathbbR$ under the usual topology). Then the space itself is open, but its closure is not compact.
    – Theo Bendit
    Sep 5 at 8:28










  • @TheoBendit Great example, thank you. Do you perhaps know of any conditions that will make this true?
    – JohnK
    Sep 5 at 8:37










  • Unfortunately, no I don't. Even in metric spaces, it's not that easy; it's possible for bounded sets in a locally compact space to not be relatively compact.
    – Theo Bendit
    Sep 5 at 8:40










  • @TheoBendit Please see the edit for the exact statement.
    – JohnK
    Sep 5 at 9:07






  • 2




    The phrase "we can even assume" makes it context-dependent. It depends on what you're using $U$ for. If you're using it in a situation where another smaller neighbourhood would suffice (i.e. one where we can always use some neighbourhood $V subseteq U$ instead of $U$), then yes this make sense. If your neighbourhood $U$ is not relatively compact, then replace it with $U cap V$, where $V$ is a relatively compact neighbourhood (which must exist by local compactness).
    – Theo Bendit
    Sep 5 at 9:11















No. Pick any topological space that is locally compact, but not compact (e.g. $mathbbR$ under the usual topology). Then the space itself is open, but its closure is not compact.
– Theo Bendit
Sep 5 at 8:28




No. Pick any topological space that is locally compact, but not compact (e.g. $mathbbR$ under the usual topology). Then the space itself is open, but its closure is not compact.
– Theo Bendit
Sep 5 at 8:28












@TheoBendit Great example, thank you. Do you perhaps know of any conditions that will make this true?
– JohnK
Sep 5 at 8:37




@TheoBendit Great example, thank you. Do you perhaps know of any conditions that will make this true?
– JohnK
Sep 5 at 8:37












Unfortunately, no I don't. Even in metric spaces, it's not that easy; it's possible for bounded sets in a locally compact space to not be relatively compact.
– Theo Bendit
Sep 5 at 8:40




Unfortunately, no I don't. Even in metric spaces, it's not that easy; it's possible for bounded sets in a locally compact space to not be relatively compact.
– Theo Bendit
Sep 5 at 8:40












@TheoBendit Please see the edit for the exact statement.
– JohnK
Sep 5 at 9:07




@TheoBendit Please see the edit for the exact statement.
– JohnK
Sep 5 at 9:07




2




2




The phrase "we can even assume" makes it context-dependent. It depends on what you're using $U$ for. If you're using it in a situation where another smaller neighbourhood would suffice (i.e. one where we can always use some neighbourhood $V subseteq U$ instead of $U$), then yes this make sense. If your neighbourhood $U$ is not relatively compact, then replace it with $U cap V$, where $V$ is a relatively compact neighbourhood (which must exist by local compactness).
– Theo Bendit
Sep 5 at 9:11




The phrase "we can even assume" makes it context-dependent. It depends on what you're using $U$ for. If you're using it in a situation where another smaller neighbourhood would suffice (i.e. one where we can always use some neighbourhood $V subseteq U$ instead of $U$), then yes this make sense. If your neighbourhood $U$ is not relatively compact, then replace it with $U cap V$, where $V$ is a relatively compact neighbourhood (which must exist by local compactness).
– Theo Bendit
Sep 5 at 9:11










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For your first question, no, the open sets in a locally compact space are not relatively compact. The entire topological space is a clopen set, and is relatively compact if and only if the entire space is compact.



For example, $mathbbR$ with the usual topology, is locally compact, but not compact. Therefore, the subset $mathbbR$ is open, but not relatively compact.



Another example: any infinite set equipped with the discrete metric. The space is locally compact, since for any point $x$ in the space, the open ball $B(x; 1/2) = lbrace x rbrace$ is open and compact. But, the open ball $B(x; 2)$ is the entire space, which is non-compact (all the singletons form an infinite open cover with no proper subcover). This illustrates how, in a metric space (unlike in a normed linear space), it's possible to have certain balls be relatively compact, and some not. It also shows an example of a bounded set in a locally compact space that isn't relatively compact.




As for your second question, in the edit, it may make sense depending on context. While not every open neighbourhood $mathcalU$ of a point $x$ in a locally compact space is relatively compact, it is true that we can always find a smaller relatively compact neighbourhood $mathcalV$ of $x$ contained in $mathcalU$. All we do is use the definition of local compactness to find an open neighbourhood $mathcalW$ of $x$ with compact closure. Then $mathcalU cap mathcalW$ is an open set containing $x$. It has compact closure, because
$$mathcalU cap mathcalW subseteq mathcalW implies overlinemathcalU cap mathcalW subseteq overlinemathcalW,$$
and closed subsets of compact sets are compact.



So, if the text uses $mathcalU$ in such a way that we may always substitute $mathcalU$ for a sub-neighbourhood $mathcalV$, then we can safely replace $mathcalU$ with a relatively compact neighbourhood $mathcalV$. In most usages of neighbourhoods in topology, restricting neighbourhoods like this is acceptable (for example, showing convergence of a sequence/net).






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    For your first question, no, the open sets in a locally compact space are not relatively compact. The entire topological space is a clopen set, and is relatively compact if and only if the entire space is compact.



    For example, $mathbbR$ with the usual topology, is locally compact, but not compact. Therefore, the subset $mathbbR$ is open, but not relatively compact.



    Another example: any infinite set equipped with the discrete metric. The space is locally compact, since for any point $x$ in the space, the open ball $B(x; 1/2) = lbrace x rbrace$ is open and compact. But, the open ball $B(x; 2)$ is the entire space, which is non-compact (all the singletons form an infinite open cover with no proper subcover). This illustrates how, in a metric space (unlike in a normed linear space), it's possible to have certain balls be relatively compact, and some not. It also shows an example of a bounded set in a locally compact space that isn't relatively compact.




    As for your second question, in the edit, it may make sense depending on context. While not every open neighbourhood $mathcalU$ of a point $x$ in a locally compact space is relatively compact, it is true that we can always find a smaller relatively compact neighbourhood $mathcalV$ of $x$ contained in $mathcalU$. All we do is use the definition of local compactness to find an open neighbourhood $mathcalW$ of $x$ with compact closure. Then $mathcalU cap mathcalW$ is an open set containing $x$. It has compact closure, because
    $$mathcalU cap mathcalW subseteq mathcalW implies overlinemathcalU cap mathcalW subseteq overlinemathcalW,$$
    and closed subsets of compact sets are compact.



    So, if the text uses $mathcalU$ in such a way that we may always substitute $mathcalU$ for a sub-neighbourhood $mathcalV$, then we can safely replace $mathcalU$ with a relatively compact neighbourhood $mathcalV$. In most usages of neighbourhoods in topology, restricting neighbourhoods like this is acceptable (for example, showing convergence of a sequence/net).






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      For your first question, no, the open sets in a locally compact space are not relatively compact. The entire topological space is a clopen set, and is relatively compact if and only if the entire space is compact.



      For example, $mathbbR$ with the usual topology, is locally compact, but not compact. Therefore, the subset $mathbbR$ is open, but not relatively compact.



      Another example: any infinite set equipped with the discrete metric. The space is locally compact, since for any point $x$ in the space, the open ball $B(x; 1/2) = lbrace x rbrace$ is open and compact. But, the open ball $B(x; 2)$ is the entire space, which is non-compact (all the singletons form an infinite open cover with no proper subcover). This illustrates how, in a metric space (unlike in a normed linear space), it's possible to have certain balls be relatively compact, and some not. It also shows an example of a bounded set in a locally compact space that isn't relatively compact.




      As for your second question, in the edit, it may make sense depending on context. While not every open neighbourhood $mathcalU$ of a point $x$ in a locally compact space is relatively compact, it is true that we can always find a smaller relatively compact neighbourhood $mathcalV$ of $x$ contained in $mathcalU$. All we do is use the definition of local compactness to find an open neighbourhood $mathcalW$ of $x$ with compact closure. Then $mathcalU cap mathcalW$ is an open set containing $x$. It has compact closure, because
      $$mathcalU cap mathcalW subseteq mathcalW implies overlinemathcalU cap mathcalW subseteq overlinemathcalW,$$
      and closed subsets of compact sets are compact.



      So, if the text uses $mathcalU$ in such a way that we may always substitute $mathcalU$ for a sub-neighbourhood $mathcalV$, then we can safely replace $mathcalU$ with a relatively compact neighbourhood $mathcalV$. In most usages of neighbourhoods in topology, restricting neighbourhoods like this is acceptable (for example, showing convergence of a sequence/net).






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        For your first question, no, the open sets in a locally compact space are not relatively compact. The entire topological space is a clopen set, and is relatively compact if and only if the entire space is compact.



        For example, $mathbbR$ with the usual topology, is locally compact, but not compact. Therefore, the subset $mathbbR$ is open, but not relatively compact.



        Another example: any infinite set equipped with the discrete metric. The space is locally compact, since for any point $x$ in the space, the open ball $B(x; 1/2) = lbrace x rbrace$ is open and compact. But, the open ball $B(x; 2)$ is the entire space, which is non-compact (all the singletons form an infinite open cover with no proper subcover). This illustrates how, in a metric space (unlike in a normed linear space), it's possible to have certain balls be relatively compact, and some not. It also shows an example of a bounded set in a locally compact space that isn't relatively compact.




        As for your second question, in the edit, it may make sense depending on context. While not every open neighbourhood $mathcalU$ of a point $x$ in a locally compact space is relatively compact, it is true that we can always find a smaller relatively compact neighbourhood $mathcalV$ of $x$ contained in $mathcalU$. All we do is use the definition of local compactness to find an open neighbourhood $mathcalW$ of $x$ with compact closure. Then $mathcalU cap mathcalW$ is an open set containing $x$. It has compact closure, because
        $$mathcalU cap mathcalW subseteq mathcalW implies overlinemathcalU cap mathcalW subseteq overlinemathcalW,$$
        and closed subsets of compact sets are compact.



        So, if the text uses $mathcalU$ in such a way that we may always substitute $mathcalU$ for a sub-neighbourhood $mathcalV$, then we can safely replace $mathcalU$ with a relatively compact neighbourhood $mathcalV$. In most usages of neighbourhoods in topology, restricting neighbourhoods like this is acceptable (for example, showing convergence of a sequence/net).






        share|cite|improve this answer












        For your first question, no, the open sets in a locally compact space are not relatively compact. The entire topological space is a clopen set, and is relatively compact if and only if the entire space is compact.



        For example, $mathbbR$ with the usual topology, is locally compact, but not compact. Therefore, the subset $mathbbR$ is open, but not relatively compact.



        Another example: any infinite set equipped with the discrete metric. The space is locally compact, since for any point $x$ in the space, the open ball $B(x; 1/2) = lbrace x rbrace$ is open and compact. But, the open ball $B(x; 2)$ is the entire space, which is non-compact (all the singletons form an infinite open cover with no proper subcover). This illustrates how, in a metric space (unlike in a normed linear space), it's possible to have certain balls be relatively compact, and some not. It also shows an example of a bounded set in a locally compact space that isn't relatively compact.




        As for your second question, in the edit, it may make sense depending on context. While not every open neighbourhood $mathcalU$ of a point $x$ in a locally compact space is relatively compact, it is true that we can always find a smaller relatively compact neighbourhood $mathcalV$ of $x$ contained in $mathcalU$. All we do is use the definition of local compactness to find an open neighbourhood $mathcalW$ of $x$ with compact closure. Then $mathcalU cap mathcalW$ is an open set containing $x$. It has compact closure, because
        $$mathcalU cap mathcalW subseteq mathcalW implies overlinemathcalU cap mathcalW subseteq overlinemathcalW,$$
        and closed subsets of compact sets are compact.



        So, if the text uses $mathcalU$ in such a way that we may always substitute $mathcalU$ for a sub-neighbourhood $mathcalV$, then we can safely replace $mathcalU$ with a relatively compact neighbourhood $mathcalV$. In most usages of neighbourhoods in topology, restricting neighbourhoods like this is acceptable (for example, showing convergence of a sequence/net).







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        answered Sep 5 at 12:16









        Theo Bendit

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