Irreducible and Hausdorff implies one-point space: Difference between revisions

Latest revision as of 19:59, 26 January 2012

Suppose $X$ is a non-empty topological space that is both an irreducible space and a Hausdorff space. Then, $X$ must be a one-point space.

We prove a somewhat modified form: we prove that an Hausdorff space containing two distinct points cannot be irreducible.

Given: A non-empty topological space $X$ that is Hausdorff and has at least two points. Say $x, y$ are two distinct points of $X$ .

To prove: $X$ is not irreducible, i.e., it contains two non-empty open subsets with empty intersection.

Proof: Use the definition of Hausdorff to find non-empty open subsets $U ∋ x, V ∋ y$ that have empty intersection. This completes the proof.

@@ Line 10: / Line 10: @@
 ==Proof==
-We prove a somewhat modified form: we prove that an irreducible space containing two distinct points cannot be Hausdorff.
+We prove a somewhat modified form: we prove that an Hausdorff space containing two distinct points cannot be irreducible.
-{{fillin}}
+'''Given''': A non-empty topological space <math>X</math> that is Hausdorff and has at least two points. Say <math>x,y</math> are two distinct points of <math>X</math>.
+'''To prove''': <math>X</math> is not irreducible, i.e., it contains two non-empty open subsets with empty intersection.
+'''Proof''': Use the definition of Hausdorff to find non-empty open subsets <math>U \ni x, V \ni y</math> that have empty intersection. This completes the proof.