If $A$ and $B$ are compact sets in $M$, show that $A\cap B$ is compact

Question

If $A$ and $B$ are compact sets in $M$, show that $A\cap B$ is compact.

Could someone please check my proof for this problem? My working definition of "compact" is

$P$ is compact iff every sequence in $P$ has a subsequence that converges to a point in $P$.

My attempt:

Let $(x_n)$ be a sequence in $A\cap B$.

$x_n\in A\cap B$

$\implies x_n\in A$ and $x_n\in B$

So, $(x_n)$ is a sequence in $A$. Since $A$ is compact, $(x_n)$ has a subsequence $(x_{n_i})$ that converges to a point $y\in A$.

$(x_n)$ is a sequence in $B$. Since $B$ is compact, $(x_n)$ has a subsequence $(x_{m_i})$ that converges to a point $z\in B$.

This is where I'm stuck. How do I proceed? I considered taking the common terms of the 2 subsequences but there need not be any.

Answer 1

I assume the space involved are Hausdorff since you use subsequences. The idea is to consider $x_{n_i}$as a subsequence of $B$ since $x_{n_i}$ is also in $B$ which has a subsequence $x_{n_{i_j}}$ which converges and the limit is also $y$.

Answer 2

You want to show that every open cover of $A\cap B$ has a finite sub-cover.

Take an open cover of $A\cap B$ and augment it with the complement of $A\cap B$

Note that this complement is open and it covers everything which in not in $A\cap B$

This augmented covering will cover both $A$ and $B$

Thus it has a finite sub-cover for $A$ and a finite sub-cover for $B$

The union of these two covering is a finite open covering for $A\cap B $

Thus $A\cap B $ is compact.

Answer 3

You take the sequence $(x_n) \subset A \cap B$, then $(x_n) \subset A$, so there exists $(x_{n_i})$ subsequence that converges to a $y \in A$

Then $(x_{n_i}) \subset A \cap B$, so it has a subsequence $(x_{n_{i_j}}) \subset B$ that conveges a point $z \in B$

Then $y = z$

Sorry for my bad english