Set theory
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5d ago – Martin Sleziak
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In the paper Applications of limited information strategies in Menger’s game by Clontz the following definition was considered:
Def. $\mathcal{A}(\kappa)$ iff for every countable subset $A\subseteq \kappa$ we can choose injective $f_A:A\to \mathbb{N}$ such that $\{x\in A\cap B : f_A(x)\neq f_B(x...
The compactness of the Hilbert cube follows (without AC) from the compactness of $2^\omega$, since $[0,1]$ as well as $[0,1]^\omega$ are continuous images of $2^\omega$.
(Conversely, $2^\omega$ is a closed subset of the Hilbert cube.)
The compactness of $2^\omega$ is just König's lemma for tr...
In Zermelo-Fraenkel set theory $ZF$ consider the following statement defined for every field $K$:
$B_K$ : Every vector space over $K$ has a basis.
It is well-known that $AC \Rightarrow \forall K (B_K)$. A. Blass proved the converse. In his paper from 1984 he said that for a fixed field $K$, th...In predicate logic, existential instantiation (also called existential elimination) is a rule of inference which says that, given a formula of the form
(
∃
x
)
ϕ
(
x
)
{\displaystyle (\exists x)\phi (x)}
, one may infer
ϕ
(
c
)
{\displaystyle \phi (c)}
for a new constant symbol c. The rule has the restrictions that the constant c introduced by the rule must be a new term that has not occurred earlier in the proof, and it also must not...
I learnt that the following theorem (in ZFC): A countable union of countable sets is countable can not be proven in ZF, all proofs must use some choice. This made me wonder whether we can prove: A countable union of finite sets is countable in ZF.
So my question is does ZF prove this?