I am following a real analysis course and I would like to discuss some definitions and results about metric spaces.
First of all, the definition of compact set. I the course we defined it as follows:
A subset $A$ of a metric space $(X,d)$ is called compact if for every open cover of $A$ there exist a finite sub-cover.
Why such a definition ? It doesn't seem natural to me.
The next question is about an exercise. Let $(X,d)$ be a metric space and consider a decreasing sequence of compact subsets $X \supset A_0 \supset A_1 \supset …$, then show that ${\bigcap}_{i=0}^{\infty} A_i$ is again compact.
Is it sufficient to show that ${\bigcap}_{i=0}^{\infty} A_i$ is bounded and closed ? If yes, then ${\bigcap}_{i=0}^{\infty} A_i$ is closed because any intersection of closed sets is again closed and ${\bigcap}_{i=0}^{\infty} A_i$ is bounded because it is contained in $A_0$. But if this is true I didn't use the decreasing hypothesis. Where am I wrong ?
Thank you for your help and comments.
Best Answer
No, this is not sufficient. There exist sets which are bounded and closed, yet they are not compact. For example, the set $(0,1)$ is abounded closed subset of the space $(0,1)$, yet the set is not compact.
There are two ways I see that you can solve the question:
Option 1: There is a theorem that states that a closed subset of a compact set is compact. This means all you need to prove is that the intersection in question is closed, and since it is a subset of $A_1$, a compact set, you are done.
Option 2: If you cannot yet use the above theorem (this is probably the case, since otherwise, the problem is way too easy), you can just go from definitions. That is, start with an open cover for ${\bigcap}_{i=0}^{\infty} A_i$, and try to find some finite subcover. In order to do that, think about this fact:
Any open cover of ${\bigcap}_{i=0}^{\infty} A_i$, along with the set $X\setminus {\bigcap}_{i=0}^{\infty} A_i$, is an open cover of $A_1$...