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
2
$\begingroup$
real-analysis
general-topology
metric-spaces
-
0It's not true true without further assumptions on the sequence. For example it fails when the sequence is convergent to a point which in not in $K$. – 2012-07-10
-
0Did you mean $\rho(x_0,K) = 0$? – 2012-07-10
-
0Otherwise here is a counterexample: $X = \mathbb R$, $K = [3,4]$, $x_n = \frac1n$. – 2012-07-10
-
0I don't understand how $\rho$ is related to the question – 2012-07-10
-
0@Norbert If $\rho(x_0, K) > 0$, we can have an open ball around $x_0$ that does not intersect with $K$ hence certainly $x_0 \notin K$. No? – 2012-07-10
-
0But admittedly, I'm slightly confused by the question. – 2012-07-10
-
0@Matt N. You are right, but anyway definition of $\rho$ is useless here. The question can be posed without it. – 2012-07-10
-
1@Matt N, Norbert I forgot to mention that $\rho(x_n,K)\to 0$. I've just fixed it – 2012-07-10