Finite Intersection Property - Applications

Applications

Theorem

Let X be a compact Hausdorff space that satisfies the property that no one-point set is open. If X has more than one point, then X is uncountable.

Before proving this, we give some examples:

1. We cannot eliminate the Hausdorff condition; a countable set with the indiscrete topology is compact, has more than one point, and satisfies the property that no one point sets are open, but is not uncountable.

2. We cannot eliminate the compactness condition as the set of all rational numbers shows.

3. We cannot eliminate the condition that one point sets cannot be open as a finite space given the discrete topology shows.

Proof of theorem:

Let X be a compact Hausdorff space. We will show that if U is a nonempty, open subset of X and if x is a point of X, then there is a neighbourhood V contained in U whose closure doesn’t contain x (x may or may not be in U). First of all, choose y in U different from x (if x is in U, then there must exist such a y for otherwise U would be an open one point set; if x isn’t in U, this is possible since U is nonempty). Then by the Hausdorff condition, choose disjoint neighbourhoods W and K of x and y respectively. Then (K ∩ U) will be a neighbourhood of y contained in U whose closure doesn’t contain x as desired.

Now suppose ƒ is a bijective function from Z (the positive integers) to X. Denote the points of the image of Z under ƒ as {x₁, x₂, ……}. Let X be the first open set and choose a neighbourhood U₁ contained in X whose closure doesn’t contain x₁. Secondly, choose a neighbourhood U₂ contained in U₁ whose closure doesn’t contain x₂. Continue this process whereby choosing a neighbourhood U_n+1 contained in U_n whose closure doesn’t contain x_n+1. Note that the collection {U_i} for i in the positive integers satisfies the finite intersection property and hence the intersection of their closures is nonempty (by the compactness of X). Therefore there is a point x in this intersection. No x_i can belong to this intersection because x_i doesn’t belong to the closure of U_i. This means that x is not equal to x_i for all i and ƒ is not surjective; a contradiction. Therefore, X is uncountable.

Corollary

Every closed interval (a < b) is uncountable. Therefore, the set of real numbers is uncountable.

Corollary

Every locally compact Hausdorff space that is also perfect is uncountable.

Proof

Suppose X is a locally compact Hausdorff space that is perfect and compact. Then it immediately follows that X is uncountable (from the theorem). If X is a locally compact Hausdorff, perfect space that is not compact, then the one-point compactification of X is a compact Hausdorff space that is also perfect. It follows that the one point compactification of X is uncountable. Therefore X is uncountable (deleting a point from an uncountable set still retains the uncountability of that set).