We are given a metric space $(M,d(x,y))$ and a sequence $\{x_j\}_{j=1}^\infty$ of elements of a compact set $L\subseteq M$. Also K is the set of $x\in M$ for which there is a subsequence, $\{x_{j_n}\}_{n=1}^\infty$ that converges to $x$. If $K$ has exactly one element, then we have to show that $\{x_j\}_{j=1}^\infty$ converges to that element.
Is this problem the same as proving that if all possible subsequences of a given sequence of elements of a compact set converge to an element, then the sequence also converges to that same element?
And if not, how do we prove this result? By contradiction?
Best Answer
If $(x_j)$ doesn't converge to that single subsequential limit $x\in K$, then there is a subsequence $(y_j)$ of $(x_j)$ such that for some $\epsilon>0$, for all $j$, we have $d(y_j,x)\geq \epsilon$. Because $(y_j)$ is a sequence in the compact set $L$, it has a convergent subsequence, call it $(z_j)$, such that $(z_j)$ converges to some $y\in L$ where $y\neq x$. Since $(z_j)$ is a subsequence of $(x_j)$, we get $y\in K$, a contradiction.
Your saying that all possible subsequences of a given sequence of elements of a compact set converge to an element doesn't make sense here because that just means the sequence itself converges since it is a subsequence of itself.