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Is some version of the Axiom of Choice required to show that a complete subspace $A$ of a metric space $X$ is closed? By closed I mean that the set's complement is open, or equivalently that it contains all of its accumulation points.

I've been messing around it like this:

Let $l$ be a limit point of $A$. For each $\delta_n$ with $\delta_n \to 0$ as $n \to \infty$ there exists some $a_n \in A$ with $0 \lt d(a_n, l) \lt \delta_n$. This sequence $\{a_n\}$ a Cauchy sequence in $A$, and so converges to something in $A$. The limit has to be $l$.

If creating $\{a_n\}$ is using some variety of choice, then is there a way around it?

Thank you.

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Are you thinking about axiom of choice? –  tessellation Sep 4 '13 at 18:27
    
If you define "closed" by "any limit of a sequence of points from $A$ is again in $A$", then no choice at all seems to be needed. If you define "closed" by "the complement is open", I don't see how you can avoid countable choice. –  Etienne Sep 4 '13 at 18:33
    
@tessellation - yes. I just tweaked the question. –  bryanj Sep 4 '13 at 18:56

2 Answers 2

up vote 2 down vote accepted

This is a slight expansion of part of Disaster $4.53$ in Herrlich’s Axiom of Choice, Lecture Notes in Mathematics $1876$.

Take a model of $\mathsf{ZF}$ in which there is a bounded, infinite, Dedekind finite set $X\subseteq\Bbb R$. One can still prove that $X$ has a limit point $p\in\Bbb R$, and by replacing $X$ by $X\setminus\{p\}$ if necessary we may assume that $p\notin X$. Then $X$ is not closed in $\Bbb R$. However, $X$ is sequentially closed and therefore complete: since $X$ is Dedekind finite, every sequence in $X$ is eventually constant. Thus, some choice is needed even in $\Bbb R$.

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IIRC, we must assume Dependent Choice in order to establish the equivalence "continuous $\iff$ sequentially continuous". –  Matemáticos Chibchas Sep 4 '13 at 21:06
    
Thanks a bunch! –  bryanj Sep 5 '13 at 1:50

Caveat: This is an attempt (pretty much just for my own benefit) to flesh out some of the details in @Brian M. Scott's answer.

I feel pretty solid about what comes next, but please feel free to point out errors (especially in Step 2)...

Goal: Show that if you have a model of ZF without Choice, then Cauchy complete does not imply closed.

Step 1)
Obtain a bounded, infinte, Dedekind-finite set $X \subseteq \mathbb{R}$.
See http://mathoverflow.net/questions/68037/dedekinds-theorem.

Step 2)
X contains a limit point $p$ in $\mathbb{R}$.
You can do this without Choice. The set $X$ is assumed to be bounded, so is contained in some closed interval. $X$ is infinite, so you use repeated bisection to create a sequence of nested intervals, each of which contains an infinite # of points of $X$, whose endpoints converge to a single point. The limit of the endpoints is a limit point of $X$.
Another way is to first show that the closed interval $[0, 1]$ in $\mathbb{R}$ is compact, by considering (for some open cover of the interval) $\sup \{x: 0 \le x < 1 \text{ and } [0, x] \text{ is contained in a finite subcover} \}$. Then you show using compactness that if $X$ doesn't have a limit point in the interval then it must be finite. For each point in the interval, find the open interval around that point of radius $1/n$, with integer $n$ minimal, such that the interval intersects $X$ in at most one point, etc...

Step 3)
By replacing $X$ with $X \setminus \{p\}$, we can assume that $p \notin X$.
Why? If $X$ is Dedekind-finite, then so is $X\setminus \{p\}$. This is because an injection $f: X \setminus \{p\} \to A \subset X \setminus \{p\} $ yields an injection $g: X \to A \cup \{p\}$, with $g(x) = f(x)$ when $x \in X\setminus \{p\}$, and $g(p) = p$.

Step 4)
So $X$ has a limit point which is not contained in $X$, and so is not closed.

Step 5)
If $\{x_n\}$ is a sequence in $X$, then the set $\{x_n : n \in \mathbb{N}\}$ is finite.
If that were not true, then we can construct a subsequence $\{y_n\}$ where all values are distinct. Do this inductively, and without Choice, by defining $y_1$ to be $x_1$, and $y_n$ to be the sequence element $x_k$ of smallest index $k$ is at least as big as any index used so far, and which is not contained in the set $\{y_j : j < n\}$. This sequence $\{y_n\}$ provides an injection of $\mathbb{N}$ into $X$. That would mean that $X$ is not Dedekind-finite (shift the elements in the sequence to get the proper injection $X \to X$).

Step 6)
If $\{x_n\}$ is a Cauchy sequence in $X$, then it is eventually constant.
Why? The sequence only hits a finite number of values. The distances between these points is bounded away from zero.

Step 7)
So $X$ is a Cauchy complete space.

Steps 4) and 7) give what we wanted.

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