The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.
0
votes
3answers
75 views
Existence of function (AC)
I want to make such a construction:
Let $X$ be an infinite set. Put:
$Y_0 = X \\ y_0 \in Y_0$
$Y_1 = Y_0 - \{y_0\} \\ y_1 \in Y_1$
...
$Y_n = Y_{n-1} - \{y_{n-1}\} \\ y_n \in Y_n$
...
for every ...
-4
votes
2answers
389 views
Dedekind's theorem on an integrally closed algebra over a field without Axiom of Choice
Motivation
This question came from my efforts to solve this problem presented by Andre Weil in 1951.
Can we prove the following theorem without Axiom of Choice?
Theorem
Let $A$ be an integrally ...
4
votes
1answer
155 views
Totally bouded,sequentially compact,complete,bounded,closed,equicontinuous$\Rightarrow$ compact?
Related; When $K$ is compact, if $S\subset C_b(K)$ is closed,bounded and equicontinuous, then $S$ is compact? (ZF)
I just edited my whole question since i think it was a bit messy.
Here is my ...
1
vote
1answer
48 views
Some properties of Polish space
Let $X$ be a separable complete metric space.
I wonder if following properties hold in ZF.
Limit Compact ⇒ Compact
Does there exist a function$f$ such that $f(E)$ is closed and $f(E)\subset E$, for ...
1
vote
1answer
227 views
Existence of valuation rings in a finite extension of the field of fractions of a weakly Artinian domain without Axiom of Choice
Can we prove the following theorem without Axiom of Choice?
This is a generalization of this problem.
Theorem
Let $A$ be a weakly Artinian domain.
Let $K$ be the field of fractions of $A$.
Let $L$ ...
24
votes
0answers
421 views
Explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$
Do you know an explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$? Using the axiom of choice, every vector space admits a norm but have you an explicit formula on ...
6
votes
0answers
105 views
What are 'weak' forms of Urysohn's lemma, which do not require choice?
Reference: http://web.mat.bham.ac.uk/C.Good/research/pdfs/horror.pdf
It is well known that the original proof of Urysohn's lemma uses a choice principle. (DC)
What are weak forms of the Urysohn's ...
6
votes
0answers
64 views
Why can countable recursive constructions not be done using countable choice?
Why can countable recursive constructions not be done using countable choice? For example, replace $Z$ by $\omega$ in the following theorem
Why can't one prove it using countable choice? (The proof ...
5
votes
0answers
89 views
Is dependent choice necessary to prove every perfect compact Hausdorff space is uncountable?
The answer to Cardinality of a locally compact space without isolated point shows that AC is required to show that if $X$ is a compact Hausdorff space with no isolated points then $|X| \ge ...
5
votes
0answers
36 views
Understanding a proof of Diaconescu's theorem
I am trying to walk through the proof of Diaconescu's theorem that the axiom of choice implies the law of excluded middle at http://plato.stanford.edu/entries/intuitionism/#ChoAxi. To paraphrase:
...
4
votes
0answers
58 views
Axiom of Choice-esque argument to show that a proof of a statement exists without actually giving a proof
What if the set of all well-formed statements in ZFC formed a kind of pseudo-category where a morphism f between objects A, B represented a formal proof that A implied B? What if that category could ...
4
votes
0answers
76 views
When $K$ is compact, if $S\subset C_b(K)$ is closed,bounded and equicontinuous, then $S$ is compact? (ZF)
Let $K$ be a compact metric space and $S\subset C(K,\mathbb{C})$.
Let $S$ be closed,bounded and equicontinuous. The usual proof for this is, using Arzela-Ascoli Theorem and Axiom of countable choice, ...
3
votes
0answers
39 views
Follow up on “Proof of $X \times X \hookrightarrow X$ implies $[X]^2 \hookrightarrow X$”
This is a follow up on this earlier question of mine.
We have the following statements:
(HSO) For every infinite set $X$ there exists an injection $f: X \times X \hookrightarrow X$
(HSU) For every ...
3
votes
0answers
68 views
Does the non-emptiness of the spectrum of an element of a Banach algebra depend on the Axiom of Choice?
One of the most basic results in functional analysis states that the spectrum of any element of a Banach algebra is non-empty. The proof, as most people might have seen, makes use of Liouville's ...
2
votes
0answers
84 views
Does the existence proof of $\mu$-completion require choice?
Let $(X,\mathfrak{M},\mu)$ be a measure space and $\mathfrak{M}^*=\{E\subset X|\exists A,B\in\mathfrak{M} \text{ such that} A\subset E \subset B \text{ and} \mu(B\setminus A)=0\}$.
How do i show that ...
2
votes
0answers
70 views
Theorem's relying on algebraic closures
When working with fields, it's a usual method to work on an algebraic closure of a field to obtain results about that field. In general (i. e. unless you're explicitly considering "well-behaved" ...
2
votes
0answers
94 views
Connected subset of a separable metric space is separable?
Continuity and the Axiom of Choice
I have proved a small generalization of Brian's argument, that is, "If $f:X\rightarrow Y$ is sequentially continuous on $X$ and $X$ is separable, then $f$ is ...
2
votes
0answers
134 views
Noetherian condition on the ring of formal power series without Axiom of Choice
I use the definition of a Noetherian ring given by Qiaochu in this:
A commutative ring is Noetherian if, for any nonempty collection of ideals $\mathcal{I}$, there is some $I \in \mathcal{I}$ which is ...
0
votes
0answers
26 views
(ZF) Sequential Continuity of Derivatives and Darboux
Let $C$ be a connected set of reals and $f:C\rightarrow \mathbb{R}$ be differentiable on $C$.
It can be proven that $f$ is sequentially continuous on $C$ iff $f$ is continuous on $C$.
First, i want ...
0
votes
0answers
222 views
Every k-cell is compact (in ZF)
This is the part of proof on my book.
Let $I$ be a k-cell.
Then for every $x\in I$ and $j\in k$, $a_j ≦ x(j) ≦ b_j$ for some $a_j, b_j \in \mathbb{R}$.
Let {$G_{\alpha}$} be an open cover of I and ...
