Does there exist a compact topological space that doesn't possess the Bolzano-Weierstrass property?












3












$begingroup$


Limit point $a$ of $A$: each neighbourhood of $a$ contains a point of $A$ unequal to $a$ itself.



Point of accumulation $a$ of a set $A$: each neighbourhood of $a$ contains infinitely many distinct points of $A$.



Topological space $X$ with the Bolzano Weierstrass property: each infinite subset of $X$ has at least one point of accumulation.



In my book it was proven that in Hausdorff spaces, a point $a$ is a limit point of the set $A$ if and only if it is an accumulation point of $A$.



Another relevant theorem is that in compact spaces, every infinite subset of a topological space has at least one limit point in the topological space.



I was wondering if there is an example of a topological space which is compact but doesn't possess the Bolzano-Weierstrass property. I couldn't come up with any, but we know that such a space must be non-Hausdorff by the aforementioned theorems and definitions.



Thanks in advance!






general-topology compactness







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$endgroup$

















    3












    $begingroup$


    Limit point $a$ of $A$: each neighbourhood of $a$ contains a point of $A$ unequal to $a$ itself.



    Point of accumulation $a$ of a set $A$: each neighbourhood of $a$ contains infinitely many distinct points of $A$.



    Topological space $X$ with the Bolzano Weierstrass property: each infinite subset of $X$ has at least one point of accumulation.



    In my book it was proven that in Hausdorff spaces, a point $a$ is a limit point of the set $A$ if and only if it is an accumulation point of $A$.



    Another relevant theorem is that in compact spaces, every infinite subset of a topological space has at least one limit point in the topological space.



    I was wondering if there is an example of a topological space which is compact but doesn't possess the Bolzano-Weierstrass property. I couldn't come up with any, but we know that such a space must be non-Hausdorff by the aforementioned theorems and definitions.



    Thanks in advance!






    general-topology compactness







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Limit point $a$ of $A$: each neighbourhood of $a$ contains a point of $A$ unequal to $a$ itself.



      Point of accumulation $a$ of a set $A$: each neighbourhood of $a$ contains infinitely many distinct points of $A$.



      Topological space $X$ with the Bolzano Weierstrass property: each infinite subset of $X$ has at least one point of accumulation.



      In my book it was proven that in Hausdorff spaces, a point $a$ is a limit point of the set $A$ if and only if it is an accumulation point of $A$.



      Another relevant theorem is that in compact spaces, every infinite subset of a topological space has at least one limit point in the topological space.



      I was wondering if there is an example of a topological space which is compact but doesn't possess the Bolzano-Weierstrass property. I couldn't come up with any, but we know that such a space must be non-Hausdorff by the aforementioned theorems and definitions.



      Thanks in advance!






      general-topology compactness







      share|cite|improve this question









      $endgroup$




      Limit point $a$ of $A$: each neighbourhood of $a$ contains a point of $A$ unequal to $a$ itself.



      Point of accumulation $a$ of a set $A$: each neighbourhood of $a$ contains infinitely many distinct points of $A$.



      Topological space $X$ with the Bolzano Weierstrass property: each infinite subset of $X$ has at least one point of accumulation.



      In my book it was proven that in Hausdorff spaces, a point $a$ is a limit point of the set $A$ if and only if it is an accumulation point of $A$.



      Another relevant theorem is that in compact spaces, every infinite subset of a topological space has at least one limit point in the topological space.



      I was wondering if there is an example of a topological space which is compact but doesn't possess the Bolzano-Weierstrass property. I couldn't come up with any, but we know that such a space must be non-Hausdorff by the aforementioned theorems and definitions.



      Thanks in advance!






      general-topology compactness




      general-topology compactness






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      asked Jan 4 at 17:18









      Steven WagterSteven Wagter

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          $begingroup$

          Every compact topological space has the Bolzano-Weierstrass property.



          Let $X$ be a compact space, and suppose that $A subseteq X$ has no accumulation points. That is, every $x in X$ has an open neighbourhood $U_x$ such that $U_x cap A$ is finite. Then ${ U_x : x in X }$ is an open cover of $X$, and so by compactness there are $x_1 , ldots , x_n in X$ such that $U_{x_1} cup cdots cup U_{x_n} = X$. But then $A = A cap ( U_{x_1} cup cdots cup U_{x_n} ) = ( U_{x_1} cap A ) cup cdots cup ( U_{x_n} cap A )$ is a finite union of finite sets, and so is finite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In fact $X$ is compact iff for every infinite subset $A$ there is a point $x in X$ such that for all open neighbourhoods $O$ of $x$ we have that $|O cap A| = |A|$. This is the countable case of the right hand property. @StevenWagter
            $endgroup$
            – Henno Brandsma
            Jan 4 at 23:40












          • $begingroup$
            @HennoBrandsma so if I understand you correctly, if $A$ is uncountable, then it doesn't suffice to show that there is an $x$ such that each nbhd contains infinite points of $A$; it would need to contain uncountably infinite points of $A$?
            $endgroup$
            – Steven Wagter
            Jan 5 at 16:32










          • $begingroup$
            @StevenWagter as many points as $A$ has, yes, to show compactness. And for all cardinalities. It shows how much stronger compactness is than countable compactness.
            $endgroup$
            – Henno Brandsma
            Jan 5 at 17:19











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          1 Answer
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          1 Answer
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          $begingroup$

          Every compact topological space has the Bolzano-Weierstrass property.



          Let $X$ be a compact space, and suppose that $A subseteq X$ has no accumulation points. That is, every $x in X$ has an open neighbourhood $U_x$ such that $U_x cap A$ is finite. Then ${ U_x : x in X }$ is an open cover of $X$, and so by compactness there are $x_1 , ldots , x_n in X$ such that $U_{x_1} cup cdots cup U_{x_n} = X$. But then $A = A cap ( U_{x_1} cup cdots cup U_{x_n} ) = ( U_{x_1} cap A ) cup cdots cup ( U_{x_n} cap A )$ is a finite union of finite sets, and so is finite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In fact $X$ is compact iff for every infinite subset $A$ there is a point $x in X$ such that for all open neighbourhoods $O$ of $x$ we have that $|O cap A| = |A|$. This is the countable case of the right hand property. @StevenWagter
            $endgroup$
            – Henno Brandsma
            Jan 4 at 23:40












          • $begingroup$
            @HennoBrandsma so if I understand you correctly, if $A$ is uncountable, then it doesn't suffice to show that there is an $x$ such that each nbhd contains infinite points of $A$; it would need to contain uncountably infinite points of $A$?
            $endgroup$
            – Steven Wagter
            Jan 5 at 16:32










          • $begingroup$
            @StevenWagter as many points as $A$ has, yes, to show compactness. And for all cardinalities. It shows how much stronger compactness is than countable compactness.
            $endgroup$
            – Henno Brandsma
            Jan 5 at 17:19
















          5












          $begingroup$

          Every compact topological space has the Bolzano-Weierstrass property.



          Let $X$ be a compact space, and suppose that $A subseteq X$ has no accumulation points. That is, every $x in X$ has an open neighbourhood $U_x$ such that $U_x cap A$ is finite. Then ${ U_x : x in X }$ is an open cover of $X$, and so by compactness there are $x_1 , ldots , x_n in X$ such that $U_{x_1} cup cdots cup U_{x_n} = X$. But then $A = A cap ( U_{x_1} cup cdots cup U_{x_n} ) = ( U_{x_1} cap A ) cup cdots cup ( U_{x_n} cap A )$ is a finite union of finite sets, and so is finite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            In fact $X$ is compact iff for every infinite subset $A$ there is a point $x in X$ such that for all open neighbourhoods $O$ of $x$ we have that $|O cap A| = |A|$. This is the countable case of the right hand property. @StevenWagter
            $endgroup$
            – Henno Brandsma
            Jan 4 at 23:40












          • $begingroup$
            @HennoBrandsma so if I understand you correctly, if $A$ is uncountable, then it doesn't suffice to show that there is an $x$ such that each nbhd contains infinite points of $A$; it would need to contain uncountably infinite points of $A$?
            $endgroup$
            – Steven Wagter
            Jan 5 at 16:32










          • $begingroup$
            @StevenWagter as many points as $A$ has, yes, to show compactness. And for all cardinalities. It shows how much stronger compactness is than countable compactness.
            $endgroup$
            – Henno Brandsma
            Jan 5 at 17:19














          5












          5








          5





          $begingroup$

          Every compact topological space has the Bolzano-Weierstrass property.



          Let $X$ be a compact space, and suppose that $A subseteq X$ has no accumulation points. That is, every $x in X$ has an open neighbourhood $U_x$ such that $U_x cap A$ is finite. Then ${ U_x : x in X }$ is an open cover of $X$, and so by compactness there are $x_1 , ldots , x_n in X$ such that $U_{x_1} cup cdots cup U_{x_n} = X$. But then $A = A cap ( U_{x_1} cup cdots cup U_{x_n} ) = ( U_{x_1} cap A ) cup cdots cup ( U_{x_n} cap A )$ is a finite union of finite sets, and so is finite.






          share|cite|improve this answer









          $endgroup$



          Every compact topological space has the Bolzano-Weierstrass property.



          Let $X$ be a compact space, and suppose that $A subseteq X$ has no accumulation points. That is, every $x in X$ has an open neighbourhood $U_x$ such that $U_x cap A$ is finite. Then ${ U_x : x in X }$ is an open cover of $X$, and so by compactness there are $x_1 , ldots , x_n in X$ such that $U_{x_1} cup cdots cup U_{x_n} = X$. But then $A = A cap ( U_{x_1} cup cdots cup U_{x_n} ) = ( U_{x_1} cap A ) cup cdots cup ( U_{x_n} cap A )$ is a finite union of finite sets, and so is finite.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 17:53









          Meta-мета-μετα-meta-мета-μεταMeta-мета-μετα-meta-мета-μετα

          4,724927




          4,724927












          • $begingroup$
            In fact $X$ is compact iff for every infinite subset $A$ there is a point $x in X$ such that for all open neighbourhoods $O$ of $x$ we have that $|O cap A| = |A|$. This is the countable case of the right hand property. @StevenWagter
            $endgroup$
            – Henno Brandsma
            Jan 4 at 23:40












          • $begingroup$
            @HennoBrandsma so if I understand you correctly, if $A$ is uncountable, then it doesn't suffice to show that there is an $x$ such that each nbhd contains infinite points of $A$; it would need to contain uncountably infinite points of $A$?
            $endgroup$
            – Steven Wagter
            Jan 5 at 16:32










          • $begingroup$
            @StevenWagter as many points as $A$ has, yes, to show compactness. And for all cardinalities. It shows how much stronger compactness is than countable compactness.
            $endgroup$
            – Henno Brandsma
            Jan 5 at 17:19


















          • $begingroup$
            In fact $X$ is compact iff for every infinite subset $A$ there is a point $x in X$ such that for all open neighbourhoods $O$ of $x$ we have that $|O cap A| = |A|$. This is the countable case of the right hand property. @StevenWagter
            $endgroup$
            – Henno Brandsma
            Jan 4 at 23:40












          • $begingroup$
            @HennoBrandsma so if I understand you correctly, if $A$ is uncountable, then it doesn't suffice to show that there is an $x$ such that each nbhd contains infinite points of $A$; it would need to contain uncountably infinite points of $A$?
            $endgroup$
            – Steven Wagter
            Jan 5 at 16:32










          • $begingroup$
            @StevenWagter as many points as $A$ has, yes, to show compactness. And for all cardinalities. It shows how much stronger compactness is than countable compactness.
            $endgroup$
            – Henno Brandsma
            Jan 5 at 17:19
















          $begingroup$
          In fact $X$ is compact iff for every infinite subset $A$ there is a point $x in X$ such that for all open neighbourhoods $O$ of $x$ we have that $|O cap A| = |A|$. This is the countable case of the right hand property. @StevenWagter
          $endgroup$
          – Henno Brandsma
          Jan 4 at 23:40






          $begingroup$
          In fact $X$ is compact iff for every infinite subset $A$ there is a point $x in X$ such that for all open neighbourhoods $O$ of $x$ we have that $|O cap A| = |A|$. This is the countable case of the right hand property. @StevenWagter
          $endgroup$
          – Henno Brandsma
          Jan 4 at 23:40














          $begingroup$
          @HennoBrandsma so if I understand you correctly, if $A$ is uncountable, then it doesn't suffice to show that there is an $x$ such that each nbhd contains infinite points of $A$; it would need to contain uncountably infinite points of $A$?
          $endgroup$
          – Steven Wagter
          Jan 5 at 16:32




          $begingroup$
          @HennoBrandsma so if I understand you correctly, if $A$ is uncountable, then it doesn't suffice to show that there is an $x$ such that each nbhd contains infinite points of $A$; it would need to contain uncountably infinite points of $A$?
          $endgroup$
          – Steven Wagter
          Jan 5 at 16:32












          $begingroup$
          @StevenWagter as many points as $A$ has, yes, to show compactness. And for all cardinalities. It shows how much stronger compactness is than countable compactness.
          $endgroup$
          – Henno Brandsma
          Jan 5 at 17:19




          $begingroup$
          @StevenWagter as many points as $A$ has, yes, to show compactness. And for all cardinalities. It shows how much stronger compactness is than countable compactness.
          $endgroup$
          – Henno Brandsma
          Jan 5 at 17:19


















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