Sequence of points in a nested sequence of sets converges to the point in the limit of the sequence of sets.

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.

• – Martin R May 9 at 12:47

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.
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 $$b \in B=\cap_{n\ge 1} B_k$$.