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Let $x_n$ converge in a metric space to $x$, and $F_n=\{x, x_n,x_{n+1}, x_{n+1}, ...\}$.

Why is $F_n$ compact? I was going to invoke the following theorem "A subset $A$ of a metric space is compact iff. every infinite subset of $A$ has an accumulation point in $A$". But it doesn't even seem to me now that the sequence limit $x$ must be contained in every infinite subset of $A$: for example $\{x_n,x_{n+1}, x_{n+1}, ...\}$ is an infinite subset of $F_n$, but without knowing if this subset is closed or not, how do we know whether $x$ is in this subset?

EDIT: Oops, I just realised that I had misread the theorem: $x$ needs to lie in $F_n$, not in the infinite subset of $F_n$, for $F_n$ to be compact.

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Use the definition of compact sets. Let $\mathfrak{U} = \{ U_\alpha : \alpha \in A\}$ an open covering of $F_n$. There is an $\alpha_0$ with $x \in U_{\alpha_0}$. Then ... –  Daniel Fischer Aug 18 '13 at 18:29

2 Answers 2

up vote 5 down vote accepted

Given any $\epsilon >0$; all but finitely many $x_i$ are in $B(x,\epsilon)$ by the definition of convergence. Pick an open set $U$ in the cover containing $x$, pick a ball inside it, and you're done.

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Let $\mathscr{U}$ be any open cover of $F_n$. There is some $U_0\in\mathscr{U}$ that contains $x$. The sequence converges to $x$, so there is an $m\in\Bbb N$ such that $x_k\in U_0$ for all $k\ge m$. Use this to show that $U_0\setminus F_n$ must be finite and therefore requires only finitely many more members of $\mathscr{U}$ to cover it.

Alternatively, for $n\ge 1$ let $S_n=\{0\}\cup\left\{\frac1k:k\ge n\right\}$; verify that $S_n$ is a compact subset of $\Bbb R$. Let

$$f_n:S_n\to F_n:x\mapsto\begin{cases} 0,\text{if }x=0\\ x_k,\text{if }x=\frac1k\;, \end{cases}$$

and prove that $f_n$ is continuous; it follows at once that $F_n$ is compact.

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