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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.

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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.

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You can also prove it directly just by noting that, since the sets are nested, you have that $(b_n)_{n \ge k}$ are convergent sequences in $B_k$ and, since all the $B_k$ are closed, you conclude that $b \in B_k, \forall k$, which is the same as saying that $b \in B=\cap_{n\ge 1} B_k$.

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  • $\begingroup$ Thanks. Very easy and I couldnt get it...I am cursing myself $\endgroup$ – Uday Kumar May 10 at 6:22

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