Topology and axiom of choice

Question

It it an easy exercise to show that if $X$ is first-countable then for every point $x$ and every subset $A$ we have $x \in \text{cl}A$ iff there exists a sequence $(x_n)_n$ that converges to $x$.

Well, this uses the axiom of choice to create the sequence (I think). What would happen if we don't have that? (I know that in topology it is much better to have AC but I want to figure out what happens).

score 12 · Accepted Answer · answered Feb 08 '11 at 12:53

It is consistent with the ZF axioms that there is a dense set of reals $D\subset\mathbb{R}$ having no countable subset. Such a set is infinite, but Dedekind finite. It follows that any point in $\mathbb{R}-D$ is in the closure of $D$, but not a limit of any sequence from $D$, since any such sequence would give rise to a countable subset of $D$.

Meanwhile, your argument does not require full AC, but only countable AC, since you are making countably many choices of points closer and closer to $x$.

score 12 · Answer 2 · answered Feb 08 '11 at 15:47

12

Some papers: Disasters in metric topology without choice

Continuing horrors of topology without choice

and references therein.

answered Feb 08 '11 at 15:47

Henno Brandsma

242,131

2

I must add Apollo Hogan's thesis, "General topology under the axiom of determinacy: the beauty of topology without choice", PhD dissertation, UC Berkeley, Fall 2004, Advisor John Steel. Of course, his choice of title was a response to the two titles you mentioned. – Andrés E. Caicedo Feb 08 '11 at 16:30
The link to sciencedirect.com is broken, but the article can be found at doi:10.1016/0166-8641(95)90010-1. – The Amplitwist May 08 '23 at 17:52

score 7 · Answer 3 · answered Feb 08 '11 at 16:34

Note that depending on the way you've defined the topology and how you want to use it, you may need choice to get the countable base at each point in your space. Thus even if you have dependent (countable) choice, there may be subtleties.

For example, suppose we work in ZF+DC+AD. Then $\omega_1$ with the usual topology is first-countable and we can even exhibit a countable local base at each point $\aleph\in\omega_1$, namely the collection of half-open intervals $\{(\beta,\alpha] : \beta<\alpha\}$ --- we can even order this in order-type $\omega$. However, we cannot uniformly order all the bases in order-type $\omega$. That is, there is no function $f:\omega_1\times\omega\to P(\omega_1)$ such that $\{f(\alpha,n) : n\in\omega\}$ is a local base at $\alpha$. (Recall that AD implies that there is no sequence $\{C_\lambda\subseteq\lambda\}_{\lambda\in\omega_1}$ such that $C_\lambda$ is a cofinal subset of $\lambda$ with order-type $\omega$.

Thanks Andres! "General topology under the axiom of determinacy" can be found at www(dot)math(dot)berkeley(dot)edu/(tilde)apollo/gen_top_ad.ps.gz for those curious. — Apollo, Feb 08 '11 at 16:39

Topology and axiom of choice

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