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Let $(X,d)$ be a metric space and let {$p_n$} be a sequence of points in $X$ with $\lim_{n\to ∞}p_n = p_0$. Prove that the set $K =$ {$p_0, p_1, p_2,...$} is a compact subset of $X$.

I have absolutely no idea how this is supposed to work, so an answer would be greatly appreciated!

Edit: Definition of compact is given as: A set S is compact if and only if every open cover of S has a finite subcover.

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    $\begingroup$ What is your definition of compactness? $\endgroup$ – user99914 Oct 20 '14 at 21:40
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    $\begingroup$ Do you know that subset of a metric space is compact iff any infinite sequence has a convergent subsequence? $\endgroup$ – Timbuc Oct 20 '14 at 21:41
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We shall do it according to the most standard definition of compactness:

A subset $K$ of a metric space $X$ is compact if every open cover of $K$ has a finite subcover.

Let $\{U_i\}_{i\in I}$ be an open cover of $K$. Then $p_0\in U_{i_0}$, for some $i_0\in I$. But as $U_{i_0}$ is open, then a whole ball $B(p_0,\varepsilon)\subset U_{i_0}$. Next, as $p_n\to p_0$, there exists an $n_0$, such that $n\ge n_0$ implies that $\varrho(p_n,p_0)<\varepsilon$ or equivalently $$ n\ge n_0 \quad\Longrightarrow\quad p_n\in B(p_0,\varepsilon)\subset U_{i_0}. $$ Now all but (at most) the first $n_0-1$ terms belong to $U_{i_0}$. Next choose $U_1,\ldots, U_{n_0-1}$, so that $p_j\in U_j$, for $j=1,\ldots,n_0-1$. Finally now $$ U_1,\ldots,U_{n_0-1}\,\&\, U_{i_0}, $$ is a finite subcover of $K$.

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Hint: Fix any element of an open cover of $K$ that contains $p_0$. Finitely many remain out there.

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