Let $A \subseteq \Bbb R$. Let $\inf A$ exist and let $I = \inf A$. Then, must there be a sequence $a_n$ in $A$ such that $a_n \to I$?


Let $A_n = \left\{ a \in A ~\middle|~ a < I + \dfrac1n \right\}$ for $n \in \Bbb N \setminus \{0\}$.

Then, $A_n$ is non-empty. For if it were empty, then $I+\dfrac1n$ would be a lower bound of $A$, contradicting the fact that $I$ is the greatest lower bound of $A$.

Therefore, by the axiom of countable choice, there is a sequence $a_n$ in $A$ such that $a_n \in A_n$ for all $n$.

For all $\varepsilon>0$, consider $N=\left\lceil \dfrac1\varepsilon \right\rceil$. Then, for all $n>N$, we have $a_n \in A_n$, so $a_n < I+\dfrac1n$, whence $a_n - I < \dfrac1n \le \dfrac1N = \varepsilon$.

Therefore, $a_n \to I$.

If the axiom of countable choice fails, must this still be true? Is there any counter-example, presumably with sets like amorphous sets or infinite Dedekind-finite sets?

  • $\begingroup$ You ask "must there be a sequence...", isn´t that already usage of countable choice in the formulation of a question? $\endgroup$ – user480281 Sep 30 '17 at 10:12
  • $\begingroup$ @Antoine No, of course not. $\endgroup$ – Andrés E. Caicedo Sep 30 '17 at 13:35

Yes, some choice is needed.

Suppose that $A\subseteq(0,1)$ is a dense Dedekind-finite set (this was shown consistent with the failure of the axiom of choice by Cohen).

Now $\inf A$ is $0$, but since any decreasing sequence from $A$ is eventually constant, so there is no sequence in $A$ converging to $0$.

As for the equivalence with countable choice, the answer is negative for two reasons:

  1. You only talk about sets of reals here. So of course this is not going to be equivalent to something which affects the entire universe of sets.

  2. The statement "Every Dedekind-finite set is finite" is weaker than the axiom of countable choice. Even more so if we consider only sets of real numbers. So there is no reason to expect the two things are going to be equivalent.


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