Editing Finite intersection property (section)

=== Uncountability of perfect spaces ===
Another common application is to prove that the [[Real number|real numbers]] are [[Uncountable set|uncountable]].  {{math theorem | math_statement = Let <math>X</math> be a non-empty [[Compact space|compact]] [[Hausdorff space]] that  satisfies the property that no one-point set is [[Open set|open]]. Then <math>X</math> is [[uncountable]].}}All the conditions in the statement of the theorem are necessary:
# We cannot eliminate the Hausdorff condition; a countable set (with at least two points) 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.
# We cannot eliminate the compactness condition, as the set of [[rational number]]s shows.
# We cannot eliminate the condition that one point sets cannot be open, as any finite space with the [[discrete topology]] shows.

{{math proof | proof =
We will show that if <math>U \subseteq X</math> is non-empty and open, and if <math>x</math> is a point of <math>X,</math> then there is a [[Neighbourhood (mathematics)|neighbourhood]] <math>V \subset U</math> whose [[Closure (topology)|closure]] does not contain <math>x</math> (<math>x</math>' may or may not be in <math>U</math>). Choose <math>y \in U</math> different from <math>x</math> (if <math>x \in U</math> then there must exist such a <math>y</math> for otherwise <math>U</math> would be an open one point set; if <math>x \notin U,</math> this is possible since <math>U</math> is non-empty). Then by the Hausdorff condition, choose disjoint neighbourhoods <math>W</math> and <math>K</math> of <math>x</math> and <math>y</math> respectively. Then <math>K \cap U</math> will be a neighbourhood of <math>y</math> contained in <math>U</math> whose closure doesn't contain <math>x</math> as desired.

Now suppose <math>f : \N \to X</math> is a [[bijection]], and let <math>\left\{ x_i : i \in \N \right\}</math> denote the [[Image (mathematics)#Image of a function|image]] of <math>f.</math> Let <math>X</math> be the first open set and choose a neighbourhood <math>U_1 \subset X</math> whose closure does not contain <math>x_1.</math> Secondly, choose a neighbourhood <math>U_2 \subset U_1</math> whose closure does not contain <math>x_2.</math> Continue this process whereby choosing a neighbourhood <math>U_{n+1} \subset U_n</math> whose closure does not contain <math>x_{n+1}.</math> Then the collection <math>\left\{ U_i : i \in \N \right\}</math> satisfies the finite intersection property and hence the intersection of their closures is non-empty by the compactness of <math>X.</math> Therefore, there is a point <math>x</math> in this intersection. No <math>x_i</math> can belong to this intersection because <math>x_i</math> does not belong to the closure of <math>U_i.</math> This means that <math>x</math> is not equal to <math>x_i</math> for all <math>i</math> and <math>f</math> is not [[Surjection|surjective]]; a contradiction. Therefore, <math>X</math> is uncountable.
}}{{math theorem | name=Corollary | math_statement=Every [[closed interval]] <math>[a, b]</math> with <math>a < b</math> is uncountable. Therefore, <math>\R</math> is uncountable.}}

{{math theorem | name=Corollary | math_statement=Every [[Perfect set|perfect]], [[Locally compact space|locally compact]] [[Hausdorff space]] is uncountable.}}

{{math proof | proof =
Let <math>X</math> be a perfect, compact, Hausdorff space, then the theorem immediately implies that <math>X</math> is uncountable. If <math>X</math> is a perfect, locally compact Hausdorff space that is not compact, then the [[one-point compactification]] of <math>X</math> is a perfect, compact Hausdorff space. Therefore, the one point compactification of <math>X</math> is uncountable. Since removing a point from an uncountable set still leaves an uncountable set, <math>X</math> is uncountable as well.
}}