Let $(X,d)$ be a metric space, and $K\subset X$ compact. Define for $x\in X$, $\rho(x, K)=\inf_{y\in K}d(x,y).$ Let $(x_n)_n\subset X$ be a sequence in $X$ such that $\rho(x_n,K)\to 0$. Is it true that $(x_n)_n$ has a convergent subsequence with limit $x_0$ in $K$?
Distance between a sequences and compact sets
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real-analysis
general-topology
metric-spaces
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1@Matt N, Norbert I forgot to mention that $\rho(x_n,K)\to 0$. I've just fixed it – 2019-05-27
1 Answers
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The map $y\mapsto d(x,y)$ is continuous for each $x\in X$ fixed, and since $K$ is compact, for each $x$ we can find $y(x)$ such that $d(x,y(x))=d(x,K)$.
Let $y_n\in K$ such that $d(x_n,y_n)=d(x_n,K)$. Then by Bolzano-Weierstrass theorem, we can find a subsequence $\{y_{n_k}\}$ which converges to some $y$ ($\in K$, since $K$ is closed). Since $d(x_{n_k},y)\leq d(x_{n_k},y_{n_k})+d(y_{n_k},y),$ and the two terms in the right hand side converge to $0$ as $k\to +\infty$, we have found a subsequence of $\{x_n\}$ which converges to some $y\in K$.
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3In a metric space, compactness is equivalent to sequential compactness (so every sequence in a compact set has a convergent subsequence). – 2012-07-10