Criterion for a compactness (Bolzano-Weierstrass condition for compactness I believe):
A Set S $\in \Bbb R^n$ is compact if every sequence in $S$ has a convergent subsequence to a point in $S$.
Using the statement above, prove that the intersection of two compact sets is compact.
My attempt at the proof:
Let compact sets $A$ and $B$ be given. By the criterion above, we know:
- Every sequence in $A$ has a convergent subsequence to a point in $A$.
- Every sequence in $B$ has a convergent subsequence to a point in $B$.
Consider the subsequence $a_{n_k}$ in $ A \cap B$
By the criterion above, the subsequence of $a_{n_k}$, lets call it $a_{n_{k_j}}$, converges to a point $p_A$ in $A$ and a point $p_B$ in $B$.
Since the limit of a convergent sequence is unique, $p = p_A = p_B$.
Therefore $p \in A\cap B$ and $a_{n_{k_j}}$ converges to a point in $A\cap B$.
Let me know if/where I made a fatal mistake, or how one might approach the problem differently.
Thanks!
The problem with your approach is that you have two different subsequences of $(a_n)$, one converging to a point in $A$, and another subsequence converging to some point in $A\cap B$. Then the "limits are unique" argument doesn't work, because you're looking at different (sub)sequences.
To fix this, start with a subsequence $(a_{n_k})$ converging to some point in $A$, and then take a subsequence of this subsequence, say $(a_{n_{k_j}})$. Then your argument works.
Aside: it's simpler to use Heine-Borel to show this is true.