Possible use of choice in proving “Compactness implies limit point compactness”
$begingroup$
A standard proof can be found here. Basically, the idea is to prove the contrapositive:
Let $Asubseteq X$. If $X$ is compact and $A$ doesn't have any limit
point, then A is finite.
Since A has no limit points, for each $ain A$, there exists a neighboorhood $U_a$ such that $U_acap A =
{a}$. ${X-A}cup{U_a:ain A}$ forms an open cover of $X$. By
compactness of $X$, only finitely many covers $X$. Because $X-A$ and $A$ are disjoint, A must be finite.
However, the above proof seems to rely on the axiom of choice, since we need to choose a neighborhood $U_a$ for each $a in A$, and there isn't any definite way to do so.
So can somebody clarify whether AC is necessary here? If not needed, please provide an alternative proof. Thanks in advance.
general-topology set-theory compactness axiom-of-choice
asked Jan 18 at 7:34
YuiTo ChengYuiTo Cheng
2,52341037
$endgroup$
add a comment |
$begingroup$
A standard proof can be found here. Basically, the idea is to prove the contrapositive:
Let $Asubseteq X$. If $X$ is compact and $A$ doesn't have any limit
point, then A is finite.
Since A has no limit points, for each $ain A$, there exists a neighboorhood $U_a$ such that $U_acap A =
{a}$. ${X-A}cup{U_a:ain A}$ forms an open cover of $X$. By
compactness of $X$, only finitely many covers $X$. Because $X-A$ and $A$ are disjoint, A must be finite.
However, the above proof seems to rely on the axiom of choice, since we need to choose a neighborhood $U_a$ for each $a in A$, and there isn't any definite way to do so.
So can somebody clarify whether AC is necessary here? If not needed, please provide an alternative proof. Thanks in advance.
general-topology set-theory compactness axiom-of-choice
asked Jan 18 at 7:34
YuiTo ChengYuiTo Cheng
2,52341037
$endgroup$
$begingroup$
Some authors who find using the AC a bit too much tend to present proofs only for separable and metrizable spaces. In those cases, you can avoid using the AC by an inductive argument. However, AC is absolutely necessary whenever you take infinitely many elements of a set and you do not provide a function explicitly. That is to say, in general topological settings, either you lose lots of results or you accept the AC imo.
$endgroup$
– Will M.
Jan 18 at 7:58
$begingroup$
Indeed it is. Topology is definitely pro-choice.
$endgroup$
– William Elliot
Jan 18 at 7:58
add a comment |
$begingroup$
A standard proof can be found here. Basically, the idea is to prove the contrapositive:
Let $Asubseteq X$. If $X$ is compact and $A$ doesn't have any limit
point, then A is finite.
Since A has no limit points, for each $ain A$, there exists a neighboorhood $U_a$ such that $U_acap A =
{a}$. ${X-A}cup{U_a:ain A}$ forms an open cover of $X$. By
compactness of $X$, only finitely many covers $X$. Because $X-A$ and $A$ are disjoint, A must be finite.
However, the above proof seems to rely on the axiom of choice, since we need to choose a neighborhood $U_a$ for each $a in A$, and there isn't any definite way to do so.
So can somebody clarify whether AC is necessary here? If not needed, please provide an alternative proof. Thanks in advance.
general-topology set-theory compactness axiom-of-choice
asked Jan 18 at 7:34
YuiTo ChengYuiTo Cheng
2,52341037
$endgroup$
A standard proof can be found here. Basically, the idea is to prove the contrapositive:
Let $Asubseteq X$. If $X$ is compact and $A$ doesn't have any limit
point, then A is finite.
Since A has no limit points, for each $ain A$, there exists a neighboorhood $U_a$ such that $U_acap A =
{a}$. ${X-A}cup{U_a:ain A}$ forms an open cover of $X$. By
compactness of $X$, only finitely many covers $X$. Because $X-A$ and $A$ are disjoint, A must be finite.
However, the above proof seems to rely on the axiom of choice, since we need to choose a neighborhood $U_a$ for each $a in A$, and there isn't any definite way to do so.
So can somebody clarify whether AC is necessary here? If not needed, please provide an alternative proof. Thanks in advance.
general-topology set-theory compactness axiom-of-choice
general-topology set-theory compactness axiom-of-choice
asked Jan 18 at 7:34
YuiTo ChengYuiTo Cheng
2,52341037
asked Jan 18 at 7:34
YuiTo ChengYuiTo Cheng
2,52341037
edited Jan 18 at 7:56
YuiTo Cheng
asked Jan 18 at 7:34
YuiTo ChengYuiTo Cheng
2,52341037
asked Jan 18 at 7:34
YuiTo ChengYuiTo Cheng
2,52341037
asked Jan 18 at 7:34
YuiTo ChengYuiTo Cheng
2,52341037
2,52341037
$begingroup$
Some authors who find using the AC a bit too much tend to present proofs only for separable and metrizable spaces. In those cases, you can avoid using the AC by an inductive argument. However, AC is absolutely necessary whenever you take infinitely many elements of a set and you do not provide a function explicitly. That is to say, in general topological settings, either you lose lots of results or you accept the AC imo.
$endgroup$
– Will M.
Jan 18 at 7:58
$begingroup$
Indeed it is. Topology is definitely pro-choice.
$endgroup$
– William Elliot
Jan 18 at 7:58
add a comment |
$begingroup$
Some authors who find using the AC a bit too much tend to present proofs only for separable and metrizable spaces. In those cases, you can avoid using the AC by an inductive argument. However, AC is absolutely necessary whenever you take infinitely many elements of a set and you do not provide a function explicitly. That is to say, in general topological settings, either you lose lots of results or you accept the AC imo.
$endgroup$
– Will M.
Jan 18 at 7:58
$begingroup$
Indeed it is. Topology is definitely pro-choice.
$endgroup$
– William Elliot
Jan 18 at 7:58
$begingroup$
Some authors who find using the AC a bit too much tend to present proofs only for separable and metrizable spaces. In those cases, you can avoid using the AC by an inductive argument. However, AC is absolutely necessary whenever you take infinitely many elements of a set and you do not provide a function explicitly. That is to say, in general topological settings, either you lose lots of results or you accept the AC imo.
$endgroup$
– Will M.
Jan 18 at 7:58
$begingroup$
Some authors who find using the AC a bit too much tend to present proofs only for separable and metrizable spaces. In those cases, you can avoid using the AC by an inductive argument. However, AC is absolutely necessary whenever you take infinitely many elements of a set and you do not provide a function explicitly. That is to say, in general topological settings, either you lose lots of results or you accept the AC imo.
$endgroup$
– Will M.
Jan 18 at 7:58
$begingroup$
Indeed it is. Topology is definitely pro-choice.
$endgroup$
– William Elliot
Jan 18 at 7:58
$begingroup$
Indeed it is. Topology is definitely pro-choice.
$endgroup$
– William Elliot
Jan 18 at 7:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It does not require the Axiom of Choice.
For every $ain A$, set
$$
mathcal W_a=big{Usubset X: U ,text{open},,&,,Ucap A={a} big}
$$
clearly $mathcal W_anevarnothing$ and its definition DOES NOT require the Axiom of Choice.
Then set $U_a=bigcup mathcal W_a$.
Now, $mathcal W={U_a: ain A}$ is a cover without a finite subcover, and it is defined without appealing to the AC.
answered Jan 18 at 8:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
$begingroup$
"Its definition does require the Axiom of choice." You mean without AC, $mathcal W_a$ cannot be formed, right?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:09
$begingroup$
Or you are intended to mean "doesn't"?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:19
$begingroup$
@YuiToCheng Thanx! - I corrected my answer
$endgroup$
– Yiorgos S. Smyrlis
Jan 18 at 8:21
$begingroup$
Thanks, that's a perfect answer.
$endgroup$
– YuiTo Cheng
Jan 18 at 8:21
$begingroup$
You do need to remark that $A$ is closed and hence itself compact to complete the contradiction. Or like the OP add its complement to the cover.
$endgroup$
– Henno Brandsma
Jan 18 at 9:29
add a comment |
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
It does not require the Axiom of Choice.
For every $ain A$, set
$$
mathcal W_a=big{Usubset X: U ,text{open},,&,,Ucap A={a} big}
$$
clearly $mathcal W_anevarnothing$ and its definition DOES NOT require the Axiom of Choice.
Then set $U_a=bigcup mathcal W_a$.
Now, $mathcal W={U_a: ain A}$ is a cover without a finite subcover, and it is defined without appealing to the AC.
answered Jan 18 at 8:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
$begingroup$
"Its definition does require the Axiom of choice." You mean without AC, $mathcal W_a$ cannot be formed, right?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:09
$begingroup$
Or you are intended to mean "doesn't"?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:19
$begingroup$
@YuiToCheng Thanx! - I corrected my answer
$endgroup$
– Yiorgos S. Smyrlis
Jan 18 at 8:21
$begingroup$
Thanks, that's a perfect answer.
$endgroup$
– YuiTo Cheng
Jan 18 at 8:21
$begingroup$
You do need to remark that $A$ is closed and hence itself compact to complete the contradiction. Or like the OP add its complement to the cover.
$endgroup$
– Henno Brandsma
Jan 18 at 9:29
add a comment |
$begingroup$
It does not require the Axiom of Choice.
For every $ain A$, set
$$
mathcal W_a=big{Usubset X: U ,text{open},,&,,Ucap A={a} big}
$$
clearly $mathcal W_anevarnothing$ and its definition DOES NOT require the Axiom of Choice.
Then set $U_a=bigcup mathcal W_a$.
Now, $mathcal W={U_a: ain A}$ is a cover without a finite subcover, and it is defined without appealing to the AC.
answered Jan 18 at 8:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
$begingroup$
"Its definition does require the Axiom of choice." You mean without AC, $mathcal W_a$ cannot be formed, right?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:09
$begingroup$
Or you are intended to mean "doesn't"?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:19
$begingroup$
@YuiToCheng Thanx! - I corrected my answer
$endgroup$
– Yiorgos S. Smyrlis
Jan 18 at 8:21
$begingroup$
Thanks, that's a perfect answer.
$endgroup$
– YuiTo Cheng
Jan 18 at 8:21
$begingroup$
You do need to remark that $A$ is closed and hence itself compact to complete the contradiction. Or like the OP add its complement to the cover.
$endgroup$
– Henno Brandsma
Jan 18 at 9:29
add a comment |
$begingroup$
It does not require the Axiom of Choice.
For every $ain A$, set
$$
mathcal W_a=big{Usubset X: U ,text{open},,&,,Ucap A={a} big}
$$
clearly $mathcal W_anevarnothing$ and its definition DOES NOT require the Axiom of Choice.
Then set $U_a=bigcup mathcal W_a$.
Now, $mathcal W={U_a: ain A}$ is a cover without a finite subcover, and it is defined without appealing to the AC.
answered Jan 18 at 8:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
It does not require the Axiom of Choice.
For every $ain A$, set
$$
mathcal W_a=big{Usubset X: U ,text{open},,&,,Ucap A={a} big}
$$
clearly $mathcal W_anevarnothing$ and its definition DOES NOT require the Axiom of Choice.
Then set $U_a=bigcup mathcal W_a$.
Now, $mathcal W={U_a: ain A}$ is a cover without a finite subcover, and it is defined without appealing to the AC.
answered Jan 18 at 8:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
edited Jan 18 at 8:21
answered Jan 18 at 8:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
answered Jan 18 at 8:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
answered Jan 18 at 8:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
63.7k1385165
$begingroup$
"Its definition does require the Axiom of choice." You mean without AC, $mathcal W_a$ cannot be formed, right?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:09
$begingroup$
Or you are intended to mean "doesn't"?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:19
$begingroup$
@YuiToCheng Thanx! - I corrected my answer
$endgroup$
– Yiorgos S. Smyrlis
Jan 18 at 8:21
$begingroup$
Thanks, that's a perfect answer.
$endgroup$
– YuiTo Cheng
Jan 18 at 8:21
$begingroup$
You do need to remark that $A$ is closed and hence itself compact to complete the contradiction. Or like the OP add its complement to the cover.
$endgroup$
– Henno Brandsma
Jan 18 at 9:29
add a comment |
$begingroup$
"Its definition does require the Axiom of choice." You mean without AC, $mathcal W_a$ cannot be formed, right?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:09
$begingroup$
Or you are intended to mean "doesn't"?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:19
$begingroup$
@YuiToCheng Thanx! - I corrected my answer
$endgroup$
– Yiorgos S. Smyrlis
Jan 18 at 8:21
$begingroup$
Thanks, that's a perfect answer.
$endgroup$
– YuiTo Cheng
Jan 18 at 8:21
$begingroup$
You do need to remark that $A$ is closed and hence itself compact to complete the contradiction. Or like the OP add its complement to the cover.
$endgroup$
– Henno Brandsma
Jan 18 at 9:29
$begingroup$
"Its definition does require the Axiom of choice." You mean without AC, $mathcal W_a$ cannot be formed, right?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:09
$begingroup$
"Its definition does require the Axiom of choice." You mean without AC, $mathcal W_a$ cannot be formed, right?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:09
$begingroup$
Or you are intended to mean "doesn't"?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:19
$begingroup$
Or you are intended to mean "doesn't"?
$endgroup$
– YuiTo Cheng
Jan 18 at 8:19
$begingroup$
@YuiToCheng Thanx! - I corrected my answer
$endgroup$
– Yiorgos S. Smyrlis
Jan 18 at 8:21
$begingroup$
@YuiToCheng Thanx! - I corrected my answer
$endgroup$
– Yiorgos S. Smyrlis
Jan 18 at 8:21
$begingroup$
Thanks, that's a perfect answer.
$endgroup$
– YuiTo Cheng
Jan 18 at 8:21
$begingroup$
Thanks, that's a perfect answer.
$endgroup$
– YuiTo Cheng
Jan 18 at 8:21
$begingroup$
You do need to remark that $A$ is closed and hence itself compact to complete the contradiction. Or like the OP add its complement to the cover.
$endgroup$
– Henno Brandsma
Jan 18 at 9:29
$begingroup$
You do need to remark that $A$ is closed and hence itself compact to complete the contradiction. Or like the OP add its complement to the cover.
$endgroup$
– Henno Brandsma
Jan 18 at 9:29
add a comment |
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$begingroup$
Some authors who find using the AC a bit too much tend to present proofs only for separable and metrizable spaces. In those cases, you can avoid using the AC by an inductive argument. However, AC is absolutely necessary whenever you take infinitely many elements of a set and you do not provide a function explicitly. That is to say, in general topological settings, either you lose lots of results or you accept the AC imo.
$endgroup$
– Will M.
Jan 18 at 7:58
$begingroup$
Indeed it is. Topology is definitely pro-choice.
$endgroup$
– William Elliot
Jan 18 at 7:58