All Questions
Tagged with axiom-of-choice compactness
45
questions
1
vote
1
answer
49
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Point-separating fields of clopen sets on compact spaces without Choice
In Matthew Dirk's Paper on Stone's representation theorem there is a proof of
Lemma 3.8. If X is a Stone space and F is a separating field of clopen subsets of
X, then F is the dual algebra of X; that ...
4
votes
1
answer
97
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Where do I use finiteness in this proof of: In ZF, the compactness theorem implies the Axiom of Choice for collections of finite sets?
Work in ZF, and assume the compactness theorem. Let $\mathsf{AC}^\text{fin}$ be the sentence "every collection of finite non-empty sets has a choice function".
UPDATE: Thank you to the ...
- 43
1
vote
1
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111
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If $Y$ is a regular, then $\delta(X,Y)$ with compact-open topology is regular
Suppose $Y$ is regular.
Let $f^*$ be point in $δ(X,Y)$ and $\{f:f(K)∈G\}$ open set such that
$f^*∈\{f:f(K)⊆G\}$.
Since $f^* (K)⊆G$, and ($Y$ regular) then $∀y∈f^* (K), ∃v_y$ open set such that $y∈v_y⊆(...
- 2,013
2
votes
1
answer
144
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Avoid AC in the proof that compact subspaces of a Hausdorff space are closed.
In Munkres Topology Theorem 26.3 there is a proof that a compact subspace of a Hausdorff space is closed. The proof uses the axiom of choice (AC) at some point. Here we can find a proof by ...
- 4,827
24
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1
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502
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Does Tychonoff's Theorem imply Excluded Middle?
It is well-known that using excluded middle, we can prove that Tychonoff's Theorem implies the axiom of choice. This was proved by Kelley in 1950.
However, the standard proof requires excluded middle ...
- 32.2k
2
votes
1
answer
122
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Heine's theorem and axiom of choice
I recently read a proof on Wikipedia of Heine's theorem here. I am using their notations.
The proof aims at showing that given two metric spaces $M$ and $N$ where $M$ is compact, if $f$ is a ...
- 143
3
votes
1
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251
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Every compact subset of a CW-complex is contained in a finite subcomplex: does this statement require choice?
In Introduction to Topological Manifolds by J. M. Lee this result proved with a use of an axiom of choice. In my words:
If $K$ is a compact subset of CW-complex $(X,\mathcal{E},\varphi)$ in question, ...
- 1,879
2
votes
1
answer
109
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How much choice is needed to prove that every compact metric space is a continuous image of the Cantor set?
Having inspected the proof of the fact that for every compact metric space $X$ there exists a continuous surjection from $\{0,1\}^{\mathbb N}$ onto $X$, it looks to me it is a theorem of ZF + DC (...
- 16.5k
2
votes
1
answer
190
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Separating a point and a compact subset in a Hausdorff space without Choice
Let $K$ be a compact subset of a Hausdorff space $X$, and $x \not \in K$. Then there are disjoint open sets $U,V$ with $K \subseteq U$ and $x \in V$.
The proof of this result that I have seen uses ...
- 97
1
vote
0
answers
108
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compact $\Leftrightarrow$ sequentially compact $\Leftrightarrow$ totally bounded and complete
Let $X$ be a metric space. Assuming the axiom of countable choice, the following are equivalent:
$X$ is compact.
$X$ is sequentially compact.
$X$ is totally bounded and complete.
What if we don't ...
- 318
6
votes
1
answer
203
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Is the axiom of choice needed in proving that metric spaces in which every infinite subset has a limit point are compact?
The following proof is based on Rudin's Principles, chapter 2, exercise 26. I wanted to confirm that the axiom of (countable) choice is used in the line in bold, and was curious to know whether this ...
- 2,306
4
votes
2
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223
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"Indexed" version of compactness and Axiom of Choice
When thinking about some problems related to compactness, I thought of this notion which, at the first glance, seems similar to the usual definition of compactness.
A topological space is compact iff ...
- 54.3k
2
votes
1
answer
135
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Only base in definition of compactness - do we need AC?
This is the usual definition of a compact space: A topological space is compact iff every open cover has a finite subcover.
It is possible to consider only covers by sets from a fixed base $\mathcal ...
- 54.3k
1
vote
1
answer
316
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Prove that Sequentially Compact Metric Spaces are Lindelöf without the Axiom of Choice.
A proof can be found here, but it seems that it uses AC. I would like to know if there is a proof for this fact without AC.
I came up with this question after seeing a proof of sequential compactness ...
- 2,503
3
votes
1
answer
99
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Possible use of choice in proving "Compactness implies limit point compactness"
A standard proof can be found here. Basically, the idea is to prove the contrapositive:
Let $A\subseteq X$. If $X$ is compact and $A$ doesn't have any limit
point, then A is finite.
Since A has ...
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