All Questions
Tagged with axiom-of-choice metric-spaces
35
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Axiom of Choice in characterizing openness in subspace
Below is the typical characterization of open sets in a subspace $Y$ of a metric space $X$.
$E$ is $Y$-open iff there exists an $X$-open $S$ such that $E = S \cap Y$.
The forwards direction usually ...
- 119
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1
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Is AC invoked in the claim "the product/sum of a sequence of separable metrizable spaces is separable"?
The Claim:The product/sum of a sequence of separable metrizable spaces is separable.
Here the sum of a family $((X_i, d_i))_{i\in I}$ of metric spaces is defined (up to isometry) as follows: By ...
- 1,467
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Is it possible to construct a $\sigma-$locally finite basis of $\mathbb{R}^\omega$ in the uniform topology
I've been reading Nagata-Smirnov metrization theorem these days. It states that a space is metrizable if and only if it is regular(and Hausdorff) and has a $\sigma-$locally finite basis. I've searched ...
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Choice in the non-smooth Whitney Embedding Theorem
Introduction
In Munkres' Topology, he presents a precursor of what seems to be called the non-smooth Whitney Embedding Theorem in Section 50:
Theorem 50.5 (The imbedding theorem). Every compact ...
- 1,249
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Is there an explicit example of a complete norm on $C[0,1]$ that is not equivalent to $\|\cdot\|_\infty$?
Does anyone have an constructive, explicit example for a norm $||.||$ on $C[0,1]$, such that $(C[0,1], ||\cdot||)$ is a Banach space, but such that $||\cdot||$ is not equivalent to $||\cdot||_{\infty}$...
- 2,813
2
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1
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280
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Vitali covering lemma and choice
Wikipedia proof of Vitali covering lemma uses Zorn's lemma. I was able to replace part of the argument which uses Zorn's lemma with another argument(I′ll provide the idea without the details in the ...
- 2,104
11
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Can we define a norm on $\Bbb{R^\omega}$ in a basis free way?
Let $\Bbb{R^\omega}=\{(x_n)_{n\in \mathbb{N}}: x_n \in \Bbb{R}\}$.
Then, $(\Bbb{R^\omega}, +, \cdot) $ is a linear space.
I know , if $(x_n) $ are $p$- summable, then we can define norm , $\ell_p$-...
- 13k
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A trick to avoid choice in proper metrization of countable disjoint union of proper metric spaces
A proper metric space is a metric space where closed balls are compact. Equivalently, it's a metric space such that bounded sets are compact.
In this answer user Saucy O'Path describes, given a ...
user239203
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Question about sequence convergent to a limit point and Axiom of (Countable) Choice
I read this question about how the Axiom of Countable Choice is both necessary and sufficient to show the following:
If a point $a$ in a metric space $X$ is a limit point of $A\subseteq X$, then ...
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Nowhere dense sets and metric space [closed]
Excuse me can you see this question
Prove that in a metric space the frontier of an open set is the set of accumulation points of a discrete set ...
It wrote that " this requires the axioms of choice ...
- 93
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242
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Characterization of open sets [duplicate]
I have a question concerning the characterization of open sets in a metric space.
Per definition a set O in a metric space (X,d) is open, if it is a neighborhood of each of its points.
In the book "...
- 163
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Closed iff sequentially closed in metric spaces and the Axiom of Countable Choice [duplicate]
Since I couldn't find no similar question already asked here, I launch this question (I'm sorry if what I'm going to ask has been already posted. I would like to know if so in order to delete this ...
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A method to choose a point from a nonempty closed set to avoid the use of $\mathsf{AC}$ [duplicate]
Fix a point $z \in \mathbb{C}$. Let $S \subset \mathbb{C}$ be a nonempty closed set. Let $$T = \{ w \mid d(z, w) = d(z, S)\}.$$ Then $S \cap T$ is a nonempty closed set. Let $f\colon [0, 1] \to T$ be ...
- 1,292
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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 ...
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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 ...
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