Urysohn's lemma: Difference between revisions

Latest revision as of 00:10, 26 December 2010

This article gives the statement, and possibly proof, of a basic fact in topology.

Statement

Let $X$ be a Normal space (?) (i.e., a topological space that is T1 and where disjoint closed subsets can be separated by disjoint open subsets). Suppose $A,B$ are disjoint closed subsets of $X$ . Then, there exists a continuous function $f:X\to [0,1]$ such that $f(a)=0$ for all $a\in A$ , and $f(b)=1$ for all $b\in B$ .

Note that the T1 assumption is not necessary, so Urysohn's lemma also holds for normal-minus-Hausdorff spaces, which is what many point set topologists are referring to when they use the term normal space.

Related facts

@@ Line 3: / Line 3: @@
 ==Statement==
-Let <math>X</math> be a [[normal space]] (i.e., a topological space that is [[T1 space|T1]] and where disjoint closed subsets can be separated by disjoint open subsets). Suppose <math>A,B</math> are disjoint [[closed subset]]s of <math>X</math>. Then, there exists a continuous function <math>f:X \to [0,1]</math> such that <math>f(a) = 0</math> for all <math>a \in A</math>, and <math>f(b) = 1</math> for all <math>b \in B</math>.
+Let <math>X</math> be a [[fact about::normal space]] (i.e., a topological space that is [[T1 space|T1]] and where disjoint closed subsets can be separated by disjoint open subsets). Suppose <math>A,B</math> are disjoint [[closed subset]]s of <math>X</math>. Then, there exists a continuous function <math>f:X \to [0,1]</math> such that <math>f(a) = 0</math> for all <math>a \in A</math>, and <math>f(b) = 1</math> for all <math>b \in B</math>.
+Note that the [[T1 space|T1]] assumption is not necessary, so Urysohn's lemma also holds for [[normal-minus-Hausdorff space]]s, which is what many point set topologists are referring to when they use the term ''normal space''.
 ==Related facts==