Prove that $d(x,F)>0$, where F is a family of closed subsets with an empty intersection

Question

I have come across a problem in a compact metric space chapter of a Real Analysis text:

It states that $P_i$ is a family of closed, nonempty subsets of X with an empty intersection. Where X is a compact metric space.

It asks for proof that $\forall x\in X$, there is a positive distance between $x$ and $P_i$ for some $i\in I$.

I understand the suggested proof, which uses the fact that convergent sequences in a compact metric space have convergent sub-sequences, to show that $x\in P_i\ \forall i$, if $d(x,P_i)=0$ (which is a contradiction).

However, I fail to see why we need to require X to be compact.

It seems to me, that we could simply state if $d(x,P_i)=0\ \forall i$, Then there is an x in one of the members of $P_i$ and a y in a different member of $P_i$ such that $d(x,y)=0$ but because these sets have an empty intersection, $x\neq y$ contradicting the distance axiom of ANY metric space, not just a compact metric space.

Any suggestions?

Answer 1

You are right, you don't need $X$ to be compact. But I don't understand your proof. Instead, I provide a proof for the statement.
Suppose that there is a point $x\in X$ such that $d(x,P_i)=0$ for all $i.$ This means that, for each $j,$ either $x$ is a point of $P_j$ or $x$ is a limit point of $P_j$ and since $P_j$ is a closed set, then necessarily $x\in P_j,$ which implies that $x\in\bigcap\limits_iP_i$ and hence $\bigcap\limits_iP_i\neq\varnothing.$ The result follows taking the contrapositive.

Answer 2

Remember that the distance between a point $x\in X$ and a set $A\subset X$ is defined by

$$d(x,A) = \inf\{d(x,y) : y \in A\}.$$

But as you probably know, the infimum of a set is the limit of a sequence of points of the set. In that case, this means there is a sequence of points $(y_n)$ of $A$ such that $d(x,y_n)\to d(x,A)$.

If the distance $d(x,A)$ is zero, this means that $d(x,y_n)\to 0$ and it is easy to see that this means $y_n\to x$.

Now, about your problem, the space need not be compact. The only important thing is that the $P_i$ are closed and their intersection is empty. In that case you want to show that for some $i\in I$, $d(x,P_i)\neq 0$. In that case, suppose, for the sake of contradiction, that for all $i\in I$, $d(x,P_i)=0$.

In that case, by the remarks I've made, there is a sequence of elements of $P_i$, say $(y_{n,i})$ such that $(y_{n,i})\to x$. But since $P_i$ is closed, the limit of a sequence of points of $P_i$ is in $P_i$, so that $x\in P_i$ for all $i$, which is a contradiction. In that case, there is $i\in I$ such that $d(x,P_i)\neq 0$.