Does there exist a compact topological space that doesn't possess the Bolzano-Weierstrass property?
$begingroup$
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
asked Jan 4 at 17:18
Steven WagterSteven Wagter
1639
$endgroup$
add a comment |
$begingroup$
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
asked Jan 4 at 17:18
Steven WagterSteven Wagter
1639
$endgroup$
add a comment |
$begingroup$
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
asked Jan 4 at 17:18
Steven WagterSteven Wagter
1639
$endgroup$
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
asked Jan 4 at 17:18
Steven WagterSteven Wagter
1639
asked Jan 4 at 17:18
Steven WagterSteven Wagter
1639
asked Jan 4 at 17:18
Steven WagterSteven Wagter
1639
asked Jan 4 at 17:18
Steven WagterSteven Wagter
1639
asked Jan 4 at 17:18
Steven WagterSteven Wagter
1639
1639
add a comment |
add a comment |
1 Answer
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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.
answered Jan 4 at 17:53
Meta-мета-μετα-meta-мета-μεταMeta-мета-μετα-meta-мета-μετα
4,724927
$endgroup$
$begingroup$
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
add a comment |
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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.
answered Jan 4 at 17:53
Meta-мета-μετα-meta-мета-μεταMeta-мета-μετα-meta-мета-μετα
4,724927
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 4 at 17:53
Meta-мета-μετα-meta-мета-μεταMeta-мета-μετα-meta-мета-μετα
4,724927
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 4 at 17:53
Meta-мета-μετα-meta-мета-μεταMeta-мета-μετα-meta-мета-μετα
4,724927
$endgroup$
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.
answered Jan 4 at 17:53
Meta-мета-μετα-meta-мета-μεταMeta-мета-μετα-meta-мета-μετα
4,724927
answered Jan 4 at 17:53
Meta-мета-μετα-meta-мета-μεταMeta-мета-μετα-meta-мета-μετα
4,724927
answered Jan 4 at 17:53
Meta-мета-μετα-meta-мета-μεταMeta-мета-μετα-meta-мета-μετα
4,724927
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
add a comment |
$begingroup$
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
add a comment |
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