Urysohn is refining-preserved

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 $\tau$, and if $\tau '$ is a finer topology than $\tau$, then $X$ is a Urysohn space with topology $\tau '$.

Related facts

• Hausdorffness is refining-preserved
• Regularity is not refining-preserved
• Complete regularity is not refining-preserved

Proof

Given: A topological space $(X,\tau )$. $\tau '$ is a finer topology than $\tau$. $X$ is a Urysohn space with topology $\tau$.

To prove: $(X,\tau ')$ 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 $\tau$ such that $\!f'(x)=0$ and $\!f'(y)=1$.

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