# Existence of sequence converging to infimum in a choiceless universe

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 $a_n \to I$?

Proof:

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?

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

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 $0$.