Let $B_n$ be a decreasing sequence of compact subsets of a metric space convergent to compact set $B$. That is $B_{n+1}\subseteq B_n$ for all $n$ and $\bigcap\limits_{n\geq 1}B_n = B$. Let $b_n\in B_n$ be a sequence that is converging to $b$. To prove that $b\in B$.
My approach.
Suppose not, that is $b\notin B$. Since $B$ is compact, this implies, $d(b,B)>0$. Let $\epsilon=d(b,B)$. There exists a $N_1$ such that $d(b_n,b)<\epsilon$ for all $n\geq N_1$.
For all $n$, $b_n\in B_1$ which is a compact set (closed and bounded). Therefore the limit $b$ is also in $B_1$. So, $b\in B_1\setminus B$. That means, there exists $N_2$ such that $b\in B_n$ for $n=1,2,\cdots N_2-1$ and $b\notin B_n$ for all $n\geq N_2$. Choose $N = \max\{N_1,N_2\}$. Then, $d(b_n,b)<\epsilon$ and $b\notin B_n$ for all $n\geq N$.
I know, I am close to prove. But not further able to proceed. Any hint would be grateful.
Best Answer
If $b\notin B$, then, by the definition of $B$, $b\notin B_k$, for some $k\in\mathbb N$. But every $b_n$, with $n\geqslant k$, belongs to $B_k$, and therefore, since $B_k$ is closed, $b\in B_k$ too. So, a contradiction is reached.