Final answer:
To prove that a compact Hausdorff space is metrizable if and only if it has a countable basis, we need to show both directions of the implication. In the forward direction, if a compact Hausdorff space is metrizable, then it has a countable basis. In the reverse direction, if a compact Hausdorff space has a countable basis, then it is metrizable.
Step-by-step explanation:
In order to prove that a compact Hausdorff space is metrizable if and only if it has a countable basis, we need to show both directions of the implication. Let's start with the forward direction:
Forward Direction:
If a compact Hausdorff space, x, is metrizable, then it has a countable basis.
Since x is metrizable, there exists a metric space Y and a homeomorphism f: x -> Y. The collection of open sets in Y, which form a basis for the topology of Y, is countable. Let's denote this collection as B_Y.
Now, we can define a countable basis for x as B_x = U is an open set in B_Y. Since f is a homeomorphism, f^(-1)(U) is open in x for any U in B_Y. Therefore, B_x is countable and is a basis for the topology of x.
Reverse Direction:
If x is a compact Hausdorff space with a countable basis, then x is metrizable.
Let B be a countable basis for x. For each pair of distinct points p,q in x, there exists open sets A_p and A_q in B such that p is in A_p and q is in A_q.
Since x is Hausdorff, A_p and A_q are disjoint open sets. We can define a metric space Y with the set of all ordered pairs (A_p,A_q) as its underlying set, and a metric d: Y x Y -> R defined as d((A_p,A_q),(A_r,A_s)) = 1/n, where n is the smallest positive integer such that A_p and A_r are contained in the union of the first n open sets in B and A_q and A_s are contained in the union of the first n open sets in B.
It can be shown that this metric space Y is complete and induces the same topology as x, making x metrizable.