I have following task.
If $A$ and $B$ are subsets of a normed linear space and both $A$ and $B$ are compact, are $A \cup B$ or $A \cap B$ compact?
The union is compact because of:
It holds that $A \subset A \cup B$ and $B \subset A \cup B$. Hence every open covering of $A \cup B$ is especially an open covering of $A$ and $B$. Since both $A$ and $B$ are compact, there are finite coverings $\{C_A^i\}$, $i=1,...,n$ and $\{C_B^j\}$, $j=1,...,m$ of $A$ and $B$ respectively. Now take the union of these two finite coverings and get again a finite covering which is a covering of $A\cup B$, hence $A \cup B$ is compact.
For the intersection I need some help. For $\mathbb{R}^n$ I know that compactness is equivalent to closedness and boundedness. But since I only have a normed linear space here, nothing is said about the dimension, I can't use this characterization. I also thought about the characterization of compactness via sequences, where it is said that a set $C$ is compact if every sequence $\{x_n\}\subset C$ has a convergent subsequence in $C$.
If I now take an arbitrary sequence $\{x_n\} \subset A \cap B$, I know that this sequence is in both $A$ and $B$. Now since $A$ and $B$ are both compact, this sequence has a convergent subsequence $x_{n_i}^A$ in $A$ and a convergent subsequence $x_{n_i}^B$ in $B$. Now I would like to say something like: Choose the elements which are in both of the subsequences and get a convergent subsequence in $A \cap B$, but I don't know whether there are common elements and elements of the subsequence in $A$ can "lie outside of $A\cap B$", or am I wrong here?
Some hints or solutions would be very nice.
Well, a normed space is a Hausdorff topological space, hence every compact set is closed. So $A\cap B$ is also closed (as an intersection of closed sets), and it is true in any topological space that a closed subset of a compact set is compact. So that way we indeed get that $A\cap B$ is compact.
If you want to do it with sequences it goes like that: let $x_n$ be a sequence in $A\cap B$. Since $A$ is compact it has a subsequence $x_{n_k}$ which converges to an element $a\in A$. But remember that all the elements of $x_{n_k}$ are in $B$, and since $B$ is compact we know that $x_{n_k}$ has a subsequence $x_{n_{k_l}}$ which converges to an element $b\in B$. But it is also a subsequence of a sequence which converges to $a$, hence it must also converge to $a$. So $a=b\in A\cap B$.