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Every extremally disconnected collectionwise Hausdorff space is discrete.
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How to prove that every extremally disconnected collectionwise Hausdorff space is discrete?
Can someone at least give me a hint as to how to construct the proof?
general-topology
asked Jul 15 at 1:52
zhongyuan chen
13110
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up vote
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favorite
How to prove that every extremally disconnected collectionwise Hausdorff space is discrete?
Can someone at least give me a hint as to how to construct the proof?
general-topology
asked Jul 15 at 1:52
zhongyuan chen
13110
According to Wikipedia (no reference given there) this holds for first countable spaces, not general ones?
– Henno Brandsma
Jul 15 at 5:24
This is false as stated: $beta omega$ is not discrete, but it is e.d. and cwH.
– Henno Brandsma
Jul 15 at 9:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to prove that every extremally disconnected collectionwise Hausdorff space is discrete?
Can someone at least give me a hint as to how to construct the proof?
general-topology
asked Jul 15 at 1:52
zhongyuan chen
13110
How to prove that every extremally disconnected collectionwise Hausdorff space is discrete?
Can someone at least give me a hint as to how to construct the proof?
general-topology
asked Jul 15 at 1:52
zhongyuan chen
13110
asked Jul 15 at 1:52
zhongyuan chen
13110
asked Jul 15 at 1:52
zhongyuan chen
13110
asked Jul 15 at 1:52
zhongyuan chen
13110
13110
According to Wikipedia (no reference given there) this holds for first countable spaces, not general ones?
– Henno Brandsma
Jul 15 at 5:24
This is false as stated: $beta omega$ is not discrete, but it is e.d. and cwH.
– Henno Brandsma
Jul 15 at 9:42
add a comment |Â
According to Wikipedia (no reference given there) this holds for first countable spaces, not general ones?
– Henno Brandsma
Jul 15 at 5:24
This is false as stated: $beta omega$ is not discrete, but it is e.d. and cwH.
– Henno Brandsma
Jul 15 at 9:42
According to Wikipedia (no reference given there) this holds for first countable spaces, not general ones?
– Henno Brandsma
Jul 15 at 5:24
According to Wikipedia (no reference given there) this holds for first countable spaces, not general ones?
– Henno Brandsma
Jul 15 at 5:24
This is false as stated: $beta omega$ is not discrete, but it is e.d. and cwH.
– Henno Brandsma
Jul 15 at 9:42
This is false as stated: $beta omega$ is not discrete, but it is e.d. and cwH.
– Henno Brandsma
Jul 15 at 9:42
add a comment |Â
1 Answer
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Suppose $X$ is extremally disconnected (this implies the following property: if $U$ and $V$ are disjoint open sets then $overlineU$ and $overlineV$ are also disjoint (and open)), Hausdorff and cwH as well.
Suppose that $(x_n)$ is a sequence in $X$ of all distinct points, such that $x_n to x in X$, where $x notin x_n: n in omega$. Then as $X$ is cwH, and the set $x_n: n in omega$ is discrete (this uses that $X$ is a Hausdorff space), we can find open sets $U_n$ such that $x_n in U_n$ and such that the $U_n$ are pairwise disjoint.
But then $U = bigcup_n U_2n$ and $V = bigcup_n U_2n+1$ are open and disjoint and $x in overlineU cap overlineV neq emptyset$ and this contradicts $X$ being extremally disconnected.
It follows that $X$ can only have trivial convergent sequences: ones that are eventually constant, in essence, or we could extract a sequence as above from it.
So if $X$ were also first countable or a sequential space (more generally), if $X$ were non-discrete, there would be a non-sequentially closed set $A$, and we'd have $x_n$, all from $A$, converging to some $x notin A$, and we could apply the above argument to get a contradiction. But I don't think "$X$ cwH and e.d. implies $X$ discrete" holds in full generality. A counterexample is $beta omega$, the ÄŒech-Stone compactification of the countable discrete $omega$, which is even collectionwise normal.
answered Jul 15 at 6:55
Henno Brandsma
91.6k342100
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Suppose $X$ is extremally disconnected (this implies the following property: if $U$ and $V$ are disjoint open sets then $overlineU$ and $overlineV$ are also disjoint (and open)), Hausdorff and cwH as well.
Suppose that $(x_n)$ is a sequence in $X$ of all distinct points, such that $x_n to x in X$, where $x notin x_n: n in omega$. Then as $X$ is cwH, and the set $x_n: n in omega$ is discrete (this uses that $X$ is a Hausdorff space), we can find open sets $U_n$ such that $x_n in U_n$ and such that the $U_n$ are pairwise disjoint.
But then $U = bigcup_n U_2n$ and $V = bigcup_n U_2n+1$ are open and disjoint and $x in overlineU cap overlineV neq emptyset$ and this contradicts $X$ being extremally disconnected.
It follows that $X$ can only have trivial convergent sequences: ones that are eventually constant, in essence, or we could extract a sequence as above from it.
So if $X$ were also first countable or a sequential space (more generally), if $X$ were non-discrete, there would be a non-sequentially closed set $A$, and we'd have $x_n$, all from $A$, converging to some $x notin A$, and we could apply the above argument to get a contradiction. But I don't think "$X$ cwH and e.d. implies $X$ discrete" holds in full generality. A counterexample is $beta omega$, the ÄŒech-Stone compactification of the countable discrete $omega$, which is even collectionwise normal.
answered Jul 15 at 6:55
Henno Brandsma
91.6k342100
add a comment |Â
up vote
1
down vote
accepted
Suppose $X$ is extremally disconnected (this implies the following property: if $U$ and $V$ are disjoint open sets then $overlineU$ and $overlineV$ are also disjoint (and open)), Hausdorff and cwH as well.
Suppose that $(x_n)$ is a sequence in $X$ of all distinct points, such that $x_n to x in X$, where $x notin x_n: n in omega$. Then as $X$ is cwH, and the set $x_n: n in omega$ is discrete (this uses that $X$ is a Hausdorff space), we can find open sets $U_n$ such that $x_n in U_n$ and such that the $U_n$ are pairwise disjoint.
But then $U = bigcup_n U_2n$ and $V = bigcup_n U_2n+1$ are open and disjoint and $x in overlineU cap overlineV neq emptyset$ and this contradicts $X$ being extremally disconnected.
It follows that $X$ can only have trivial convergent sequences: ones that are eventually constant, in essence, or we could extract a sequence as above from it.
So if $X$ were also first countable or a sequential space (more generally), if $X$ were non-discrete, there would be a non-sequentially closed set $A$, and we'd have $x_n$, all from $A$, converging to some $x notin A$, and we could apply the above argument to get a contradiction. But I don't think "$X$ cwH and e.d. implies $X$ discrete" holds in full generality. A counterexample is $beta omega$, the ÄŒech-Stone compactification of the countable discrete $omega$, which is even collectionwise normal.
answered Jul 15 at 6:55
Henno Brandsma
91.6k342100
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Suppose $X$ is extremally disconnected (this implies the following property: if $U$ and $V$ are disjoint open sets then $overlineU$ and $overlineV$ are also disjoint (and open)), Hausdorff and cwH as well.
Suppose that $(x_n)$ is a sequence in $X$ of all distinct points, such that $x_n to x in X$, where $x notin x_n: n in omega$. Then as $X$ is cwH, and the set $x_n: n in omega$ is discrete (this uses that $X$ is a Hausdorff space), we can find open sets $U_n$ such that $x_n in U_n$ and such that the $U_n$ are pairwise disjoint.
But then $U = bigcup_n U_2n$ and $V = bigcup_n U_2n+1$ are open and disjoint and $x in overlineU cap overlineV neq emptyset$ and this contradicts $X$ being extremally disconnected.
It follows that $X$ can only have trivial convergent sequences: ones that are eventually constant, in essence, or we could extract a sequence as above from it.
So if $X$ were also first countable or a sequential space (more generally), if $X$ were non-discrete, there would be a non-sequentially closed set $A$, and we'd have $x_n$, all from $A$, converging to some $x notin A$, and we could apply the above argument to get a contradiction. But I don't think "$X$ cwH and e.d. implies $X$ discrete" holds in full generality. A counterexample is $beta omega$, the ÄŒech-Stone compactification of the countable discrete $omega$, which is even collectionwise normal.
answered Jul 15 at 6:55
Henno Brandsma
91.6k342100
Suppose $X$ is extremally disconnected (this implies the following property: if $U$ and $V$ are disjoint open sets then $overlineU$ and $overlineV$ are also disjoint (and open)), Hausdorff and cwH as well.
Suppose that $(x_n)$ is a sequence in $X$ of all distinct points, such that $x_n to x in X$, where $x notin x_n: n in omega$. Then as $X$ is cwH, and the set $x_n: n in omega$ is discrete (this uses that $X$ is a Hausdorff space), we can find open sets $U_n$ such that $x_n in U_n$ and such that the $U_n$ are pairwise disjoint.
But then $U = bigcup_n U_2n$ and $V = bigcup_n U_2n+1$ are open and disjoint and $x in overlineU cap overlineV neq emptyset$ and this contradicts $X$ being extremally disconnected.
It follows that $X$ can only have trivial convergent sequences: ones that are eventually constant, in essence, or we could extract a sequence as above from it.
So if $X$ were also first countable or a sequential space (more generally), if $X$ were non-discrete, there would be a non-sequentially closed set $A$, and we'd have $x_n$, all from $A$, converging to some $x notin A$, and we could apply the above argument to get a contradiction. But I don't think "$X$ cwH and e.d. implies $X$ discrete" holds in full generality. A counterexample is $beta omega$, the ÄŒech-Stone compactification of the countable discrete $omega$, which is even collectionwise normal.
answered Jul 15 at 6:55
Henno Brandsma
91.6k342100
answered Jul 15 at 6:55
Henno Brandsma
91.6k342100
answered Jul 15 at 6:55
Henno Brandsma
91.6k342100
answered Jul 15 at 6:55
Henno Brandsma
91.6k342100
91.6k342100
add a comment |Â
add a comment |Â
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According to Wikipedia (no reference given there) this holds for first countable spaces, not general ones?
– Henno Brandsma
Jul 15 at 5:24
This is false as stated: $beta omega$ is not discrete, but it is e.d. and cwH.
– Henno Brandsma
Jul 15 at 9:42