Criteria for convergence of a sequence in a metric space

Question

Let $X$ be a metric space, $z$ be in $X$ and let $(x_n)$ be a sequence in $X$

Using the fact that, Every open subset of $X$ that contains $z$ includes a tail of $(x_n)$,

I have to prove that

$\{z\}=⋂\{\textrm{cl}\{x_n∣n∈S\}∣S⊆\mathbb{N}$ and $S$ is infinite $\}$.

I started off with the proof by letting, $S$ to be an infinite subset of $\mathbb{N}$.

Then $X\backslash\textrm{cl}\{x_n∣n∈S\}$ is an open subset of $X$.

I know that this should not contain a tail of $(x_n)$, because if it would then $z$ would not belong to $\textrm{cl}\{x_n∣n∈S\}$.

What i don't understand is that, how should i show this ? I know this has got something to do with $S$ being infinite, but i just can't relate the two things.

An infinite subset $S$ of $\mathbb{N}$ must be necessarily unbounded. ie. it must contain a subsequence ${n_k}_{k\in\mathbb{N}}$ — Prahlad Vaidyanathan, Sep 30 '13 at 11:57

score 1 · Accepted Answer · answered Oct 01 '13 at 02:13

You’ve started off just fine. For convenience let $U=X\setminus\operatorname{cl}\{x_n:n\in S\}$. For future reference note that if $n\in S$, then $x_n\notin U$.

Now suppose that $z\in U$; your hypothesis tells you that $U$ contains a tail of the sequence, so there is an $m\in\Bbb N$ such that $x_n\in U$ for all $n\ge m$. Suppose that $n\in S$; then $x_n\notin U$, so $n<m$. But that means that $S$ is finite, which is a contradiction. Thus, $z\notin U$, and therefore $z\in\operatorname{cl}\{x_n:n\in S\}$.

Criteria for convergence of a sequence in a metric space

1 Answers1