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

Let $\{p_n\}$ be a sequence in a metric space $X$.

Let $B=\{p_n|n\in\mathbb{N}\}$ and $p$ be a limit point of $B$.

It first seemed obvious that there exists a subsequence convergent to $p$, but i realized that

I can construct a 'infinite subset' of $\{p_n\}$ and form a new sequence convergent to $p$, but can't construct a 'subsequence'.

Help me how to construct a subsequence.

Additional Question; What is the precise definition of convergent? I think the definition should mention ordering of $\mathbb{N}$, but every definition i saw doesn't mention this. For example, let $p_n = 1/n$. Say $G$ is the usual ordering of $\mathbb{N}$ (i.e. Well-ordered by $\in$) If i follow this ordering, $\{p_n\}$ is convergent to 0. But if i follow $G^{-1}$, it's convergent to 1.

Or is it defined by the point where is the most, unofficially speaking, 'dense'?

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2 Answers 2

up vote 0 down vote accepted

Let $B = \{p_n : n \in\Bbb N\}$. Let $p$ be a limit point of $p$. Define by recursion (possible in ZF):

Suppose you have $p_{n_1}, p_{n_2}, ..., p_{n_k}$ such that $n_1 < n_2 < ... < n_{k}$ and $d(p, n_j) < \frac{1}{j}$ for all $1 \leq j \leq k$, then find the least $l > n_k$ such that $d(p, p_l) < \frac{1}{k + 1}$. Let $l = n_{k + 1}$.

This does not require any choice. You just start computing $d(p, p_l)$ for $l > n_k$. By assuming $p$ is a limit point of $B$, you sure to find an $l$.

Then $(p_{n_k})_{k \in\Bbb N}$ is a subsequence converging to $p$.


The definition of convergence does use the ordering of $\mathbb{N}$. The definition of $(p_n)_{n \in \mathbb{N}}$ converging to $p$ is :

For all $\epsilon > 0$, there exists a $N \in \mathbb{N}$ such that for all $n > N$, $d(p_n, p) < \epsilon$.

The part "$n > N$" refers to the usual well-ordering of ordering of $\mathbb{N}$.

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The definition of convergence is the same as with the axiom of choice:

Let $\{x_n\}$ be a sequence; we say that $x_n$ converges to $x$ if for every $\varepsilon>0$ there exists some $n_0$ such that for all $n>n_0$ we have $d(x,x_n)<\varepsilon$.

Now it is clear that if $\{p_n\}$ is a sequence and $p$ is a limit point of the sequence there is a subsequence converging to it:

We define by induction over $\mathbb N$:

  • Let $q_0=p_0$.
  • Suppose that $q_n=p_k$ was defined, let $q_{n+1}=p_m$ be such that $m=\min\{i\in\mathbb N\mid i>k, d(p,p_i)<\frac12d(p,q_n)\}$. We did not use Dependent Choice because we use the well-ordering of $\mathbb N$ to choose $q_{n+1}$.

    We therefore just need to verify that such $i$ exists, but this is obvious since $p$ is a limit point of $\{p_n\}$, for every $\varepsilon>0$ there is some $k$ such that $p_k\in B(p,\varepsilon)$. In particular for $\varepsilon=\frac12d(p,q_n)$.

I claim that $\{q_n\}$ converges to $p$, since for every $\varepsilon>0$ there is some $n$ such that $d(p,q_n)<\varepsilon$ (note that we take $q_n$'s to have distance which closes in on $p$ exponentially or faster), and moreover all the choices of $q_k$ for $k>n$ are such that $d(p,q_k)<d(p,q_n)<\varepsilon$, as wanted.

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