(ZF) Prove 'the set of all subsequential limits of a sequence in a metric space is closed.

Question

Let $X$ be a metric space. Let $\{p_n\}$ be a sequence in $X$. Let $E$ be a set of all subsequential limits of $\{p_n\}$.

How do i prove that $E$ is closed in ZF?

Is there a well-ordering of convergent subsequences? I can't think of one..

Answer 1

Let us denote $\{ p_n : n \in \mathbb{N} \}$ by $B$.

Clearly $x \in \overline{ B }$ iff $x = p_n$ for some $n$, or $( \forall m ) ( \exists n ) ( 0 < d (x,p_n ) \leq \frac{1}{m} )$. In the latter case we can construct, via induction and without any choice, a subsequence converging to $x$. (In fact, the latter condition is easily seen to be equivalent to $x$ being a subsequential limit of $(p_n)_{n \in \mathbb{N}}$.)

Suppose $x \in \overline{ B }$ is not a subsequential limit. This means that $x = p_n$ for some $n$. But also that there is an $m \in \mathbb{N}$ such that $d ( x , p_n ) > \frac{1}{m}$ for all $n$. Therefore $x$ is an isolated point of $B$, and thence it is also an isolated point of $\overline{ B }$.

It thus suffices to show that $\overline{ B } \setminus \{ x : x\text{ is an isolated point of }B \}$ is closed, and this follows from the following:

Claim: Suppose $F \subseteq X$ is closed and $A \subseteq F$ is a set of isolated points of $F$. Then $F \setminus A$ is closed.

Proof: It suffices to show that $X \setminus ( F \setminus A)$ is open. If $x \in X \setminus ( F \setminus A )$ there are two cases: Either $x \notin F$, in which case $x \in X \setminus F$, and this is a neighbourhood of $x$ disjoint from $F \setminus A$. Otherwise $x \in A$, but as $x$ is isolated there is a neighbourhood $U$ of $x$ such that $F \cap U = \{ x \}$, and therefore $( F \setminus A ) \cap U = \emptyset$.) $\dashv$

Edit: There is a minor issue in what I have done above. I seem to have assumed that the sequence $( p_n )_{n \in \mathbb{N}}$ is one-to-one (silly me). This means that some of what I said above is not quite true in general. The relevant facts we still have are:

• If $x \in \overline{B}$ then either $x \in B$ or there is a subsequence of $( p_n )_{n \in \mathbb{N}}$ converging to $a$.
• If there is a subsequence of $( p_n )_{n \in \mathbb{N}}$ converging to $x$, then either $( \forall m ) ( \exists n ) ( 0 < d (x,p_n ) \leq \frac{1}{m} )$, or $( \forall N ) ( \exists n \geq N ) ( p_n = x )$.

The set we wish to show is closed is therefore $\overline{B} \setminus A$ where $$A = \{ x \in B : x\text{ is an isolated point of }A\text{ and }( \exists N ) ( \forall n \geq N ) ( p_n \neq x ) \}.$$ As $A$ is a set of isolated points, by the above $\overline{B} \setminus A$ is closed.

Answer 2

Check if this is ok: $ \forall $ $ y $ in the closure of $ E $, let $ N_{y} $ be any neighbourhood of $ y $. Then, by definition of an element of the closure, for any neighbourhood of $ y $, $ N_{y} \bigcap E \neq \emptyset $.

By the same token, for any z in E, $ N_{z} $ (any neighbourhood of z) is such that $ N_{z} \bigcap \left(p_{n}\right) \neq \emptyset $.

Therefore, there is a $ z \in N_{y} \bigcap E $ so that $ N_{z} \bigcap \left(p_{n}\right) \neq \emptyset $.

Let $ N = N_{z} \bigcap N_{y} $. Then N is a neighbourhood of z and $ N \subset N_{y} $ (intersections of neighbourhoods are neighbourhoods and z is in both)

Therefore $ N_{y} \bigcap \left(p_{n}\right) \neq \emptyset $. And $ y $ is in $ E $, so that $ E $ is closed, since $ N_{y} $ is arbitrary

Since in this construction there are only intersections of sets, I think it is independent from AC.