Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
up vote 10 down vote accepted

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 $x$.

share|cite|improve this answer
Clear. Thank you. – Jonas Teuwen Feb 8 '11 at 14:59

Some papers: Disasters in metric topology without choice

Continuing horrors of topology without choice

and references therein.

share|cite|improve this answer
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 8 '11 at 16:30

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 $\omega$.

share|cite|improve this answer
Thanks Andres! "General topology under the axiom of determinacy" can be found at www(dot)math(dot)berkeley(dot)edu/(tilde)apollo/ for those curious. – Apollo Feb 8 '11 at 16:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.