Let $(X,d)$ be a metric space. Why the following statements are true?

Question

So I have given that $(X,d)$ is a metric space. It is also given that $A \subset X$ and $A$ is a closed set. Moreover $K \subset X$ and $K$ is compact. Why the following statement is true/false?

1. $A \cap K$ is compact. (True)$\\$

2. $X\setminus K$ not compact (False) $\\$

3. $X\setminus A$ is not closed.(Flase) $\\$

Ad 1. Since $K$ is compact it is also closed. We have said that the intersection of two closed sets is closed, but how do we know if the intersection in our case is also bounded, and therefore compact. Or is the intersection of a bounded and a compact set, bounded?

Ad.2 I don't understand why this is false. Since $K$ is compact it is also closed, therefore this set should be not closed and therefore it can't be compact.

Ad.3 I also don't know why this is false. I took the case where $A$ is the set where $1\leq x$. I think this is a closed set, its complement would be open, so also not closed.

I would be very thankful if somebody could shed some light for me in order to help me have a clearer idea of these concepts.

Thank you very much in advance,

Annalisa

Answer 1

1. Do you think that compact $\iff$ closed and bounded? Not true in general metric spaces, although it is true in $\Bbb R^n$ with respect to the usual metric. Anyway, if $K$ is compact and $A$ is closed, then $K\cap A$ is a closed subset of $K$ and every closed subset of a compact set is compact too.
2. It is false because, for instance, if $X=\{-1\}\cup[0,1]$ (with the usual metric) and $K=X$,then $X\setminus K=\{-1\}$, which is compact.
3. The same example applies here.

Answer 2

1. is true because $A \cap K$ is closed, and hence closed in $K$, and a closed subset of a compact subspace is compact. Do not think boundedness has any role here. You need compact implies closed (in metric spaces), intersections of closed sets are closed, and closed subsets of compact spaces are always compact.

2. can be false when $X$ consists of two compact pieces, say $X=[0,1] \cup [2,3]$ as a subspace of $\Bbb R$ and $K=[0,1]$,say, is one of those pieces. The remainder can be compact. It need not be, but not all metric spaces are like $\Bbb R$ or the plane etc.

3. Can be false, when $A$ is both open and closed, so then $X\setminus A$ is closed as $A$ is open. Doesn't happen for $X=\Bbb R$, unless we take $A=\Bbb R$ or $B = \emptyset$ (which are valid examples), but in the example from 2. we can also take $A=[0,1]$ to see that $X\setminus A= [2,3]$ is also closed.