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If $A\subset \mathbb R^n$ is compact, $x\in A$, and let $\{x_i\}$ be a sequence in $A$ such that every convergent subsequence of $\{x_i\}$ converge to $x$. Prove that $\{x_i\}$ also converge.

How to even approach this problem, by contradiction?

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up vote 3 down vote accepted

Assume that $(x_n)$ does not converge to $x$, which means that there exists $\epsilon>0$ such that for all $N$ there exists $n\geq N$ with $|x_n-x|\geq \epsilon$.

By a proper induction, this is in turn equivalent to the existence of a subsequence $(x_{n_k})$ such that $|x_{n_k}-x|\geq \epsilon$ for all $k$.

Now by compactness, there exists a subsequence $(x_{n_{k_m}})$ of the latter which converges. By assumption, it converges to $x$.

Since $|x_{n_{k_m}}-x|\geq \epsilon$ for all $m$, we can pass to the limit as $m$ tends to $+\infty$ and find $$ 0=|x-x|\geq \epsilon $$ a contradiction.

So $(x_n)$ converges to $x$.

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1) Show that every subsequence of $(x_i)$ has a further subsequence that converges to $x$.

2) Show that the condition in 1) implies that $(x_i)$ converges to $x$ (the contrapositive is easy to prove).

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Given $\epsilon>0$, we have to show that almost all $x_i$ are in $B(x,\epsilon)$.

We are given that for $y\ne x$ there is no subsequence converging to $y$, hence for some $r(y)>0$ there are only finitely many $x_i\in U_y:=B(y,r(y))$. Then $$ A\subseteq B(x,\epsilon)\cup\bigcup_{y\in A\setminus\{x\}}U_y$$ is an open cover of $A$. By compactness, there is a finite subcover, hence $$ A\subseteq B(x,\epsilon)\cup \bigcup_{i=1}^nU_{y_i}.$$ Each $U_{y_i}$ contains only sinitely many members of the sequence, hence $B(x,\epsilon)$ must contain almost all members of the sequence, as was to be shown.

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