Urysohn is refining-preserved: Difference between revisions

Latest revision as of 02:17, 25 January 2012

This article gives the statement, and possibly proof, of a topological space property (i.e., Urysohn space) satisfying a topological space metaproperty (i.e., refining-preserved property of topological spaces)
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Statement

If $X$ is a Urysohn space with a topology $τ$ , and if $τ^{'}$ is a finer topology than $τ$ , then $X$ is a Urysohn space with topology $τ^{'}$ .

Related facts

Proof

Given: A topological space $(X, τ)$ . $τ^{'}$ is a finer topology than $τ$ . $X$ is a Urysohn space with topology $τ$ .

To prove: $(X, τ^{'})$ is a Urysohn space: for distinct points $x, y \in X$ , there exists a function $f^{'} : X \to [0, 1]$ that is continuous with respect to $τ$ such that $f^{'} (x) = 0$ and $f^{'} (y) = 1$ .

Proof: We have a continuous function $f : (X, τ) \to [0, 1]$ such that $f (x) = 0$ and $f (y) = 1$ , continuous with topology $τ$ . Since $τ^{'}$ is finer than $τ$ , the identity map $(X, τ^{'}) \to (X, τ)$ is continuous. Composing with $f$ , we obtain a map $f^{'} : X \to [0, 1]$ such that $f (x) = 0$ and $f (y) = 1$ .

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 ==Statement==
-If <math>X</math> is a [[Urysohn space]] with a topology <math>\tau<math>, and if <math>\tau'</math> is a [[finer topology]] than <math>\tau</math>, then <math>X</math> is a Urysohn space with topology <math>\tau'</math>.
+If <math>X</math> is a [[Urysohn space]] with a topology <math>\tau</math>, and if <math>\tau'</math> is a [[finer topology]] than <math>\tau</math>, then <math>X</math> is a Urysohn space with topology <math>\tau'</math>.
 ==Related facts==
@@ Line 17: / Line 17: @@
 '''Given''': A topological space <math>(X,\tau)</math>. <math>\tau'</math> is a finer topology than <math>\tau</math>. <math>X</math> is a Urysohn space with topology <math>\tau</math>.
-'''To prove''': <math>(X,\tau')</math> is a Urysohn space: for distinct points <math>x,y \in X</math>, there exists a function <math>f':X \to [0,1]</math> such that <math>f'(x) = 0</math> and <math>f'(y) = 1</math>.
+'''To prove''': <math>(X,\tau')</math> is a Urysohn space: for distinct points <math>x,y \in X</math>, there exists a function <math>f':X \to [0,1]</math> that is continuous with respect to <math>\tau</math> such that <math>\! f'(x) = 0</math> and <math>\! f'(y) = 1</math>.
-'''Proof''': We have a continuous function <math>f:(X,\tau) \to [0,1]</math> such that <math>f(x) = 0</math> and <math>f(y) = 1</math>, continuous with topology <math>\tau</math>. Since <math>\tau'</math> is finer than <math>\tau</math>, the identity map <math>(X,\tau') \to (X,\tau)</math> is continuous. Composing with <math>f</math>, we obtain a map <math>f':X \to [0,1]</math> such that <math>f(x) = 0</math> and <math>f(y) = 1</math>.
+'''Proof''': We have a continuous function <math>\! f:(X,\tau) \to [0,1]</math> such that <math>\! f(x) = 0</math> and <math>\! f(y) = 1</math>, continuous with topology <math>\tau</math>. Since <math>\tau'</math> is finer than <math>\tau</math>, the identity map <math>(X,\tau') \to (X,\tau)</math> is continuous. Composing with <math>f</math>, we obtain a map <math>\! f':X \to [0,1]</math> such that <math>\! f(x) = 0</math> and <math>\! f(y) = 1</math>.