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\begin{document}

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\section*{Urysohn&#39;s lemma}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{topology}{}\paragraph*{{Topology}}\label{topology}

[[!include topology - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{statement}{Statement}\dotfill \pageref*{statement} \linebreak
\noindent\hyperlink{Proof}{Proof}\dotfill \pageref*{Proof} \linebreak
\noindent\hyperlink{Implications}{Implications}\dotfill \pageref*{Implications} \linebreak
\noindent\hyperlink{related_statements}{Related statements}\dotfill \pageref*{related_statements} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

\emph{Urysohn's lemma} (prop. \ref{UrysohnLemma} below) states that on a [[normal topological space]] disjoint [[closed subsets]] may be separated by [[continuous functions]] in the sense that a continuous function exists which takes value 0 on one of the two subsets and value 1 on the other (called an ``Urysohn function'', def. \ref{UrysohnFunction}) below. In fact the existence of such functions is equivalent to a space being normal (remark \ref{Equivalence} below).

Urysohn's lemma is a key ingredient for instance in the proof of the [[Tietze extension theorem]] and in the proof of the existence of [[partitions of unity]] on [[paracompact topological spaces]]. See the list of implications \hyperlink{Implications}{below}.

\hypertarget{statement}{}\subsection*{{Statement}}\label{statement}

\begin{defn}
\label{UrysohnFunction}\hypertarget{UrysohnFunction}{}
\textbf{(Urysohn function)}

Let $X$ be a [[topological space]], and let $A,B \subset X$ be disjoint [[closed subsets]]. Then an \emph{Urysohn function} for this situation is

\begin{itemize}%
\item a [[continuous function]] $f \colon X \to [0,1]$

\end{itemize}
to the [[closed interval]] equipped with its [[Euclidean space|Euclidean]] [[metric topology]], such that

\begin{itemize}%
\item it takes the value 0 on $A$ and the value 1 on $B$:

\begin{displaymath}
f(A) = \{0\}
  \phantom{AAA}
  \text{and}
  \phantom{AAA}
  f(B) = \{1\}
  \,.
\end{displaymath}


\end{itemize}
\end{defn}
\begin{prop}
\label{UrysohnLemma}\hypertarget{UrysohnLemma}{}
\textbf{(Urysohn's lemma)}

Assuming [[excluded middle]] then:

Let $X$ be a [[normal topological space|normal]] (or $T_4$) [[topological space]], and let $A,B \subset X$ be two disjoint [[closed subsets]] of $X$. Then there exists an Urysohn function (def. \ref{UrysohnFunction}).

\end{prop}
\begin{remark}
\label{}\hypertarget{}{}
Beware that the function in prop. \ref{UrysohnLemma} may take the values 0 or 1 even outside of the two subsets. The condition that the function takes value 0 or 1, respectively, \emph{precisely} on the two subsets corresponds to \emph{[[perfectly normal spaces]]}.

\end{remark}
\begin{remark}
\label{Equivalence}\hypertarget{Equivalence}{}
It is immediate that, conversely, the existence of an Urysohn function (def. \ref{UrysohnFunction}) implies that the topological space is [[normal topological space|normal]]. For let $A, B \subset X$ be disjoint closed subsets, and consider a continuous function $f \colon X \to [0,1]$ with $f(A) = \{0\}$ and $f(B) = \{1\}$ then

\begin{displaymath}
U_A \coloneqq f^{-1}([0,1/3)
  \phantom{AAA}
  U_B \coloneqq f^{-1}((2/3,1])
\end{displaymath}
are disjoint open neighbourhoods of these subsets.

Hence Urysohn's lemma shows that a topological space being normal is \emph{equivalent} to it admitting Urysohn functions.

\end{remark}
\hypertarget{Proof}{}\subsection*{{Proof}}\label{Proof}

\begin{proof}
of Urysohn's lemma, prop. \ref{UrysohnLemma}

Set

\begin{displaymath}
C_0 \coloneqq A
  \phantom{AAA}
  U_1 \coloneqq X \backslash B
  \,.
\end{displaymath}
Since by assumption

\begin{displaymath}
A \cap B = \emptyset
  \,.
\end{displaymath}
we have

\begin{displaymath}
C_0 \subset U_1
  \,.
\end{displaymath}
Notice that (by \href{separation+axioms#T4InTermsOfTopologicalClosures}{this lemma}) if a space is normal then every open neighbourhood $U \supset C$ of a closed subset $C$ contains a smaller neighbourhood $V$ together with its closure $Cl(V)$

\begin{displaymath}
C \subset V \subset Cl(V) \subset U
  \,.
\end{displaymath}
Apply this fact successively to the above situation to obtain the following infinite sequence of nested open subsets $U_r$ and closed subsets $C_r$

\begin{displaymath}
\itexarray{
    C_0 &&  &&  &&  &\subset&  &&  &&  && U_1
    \\
    C_0 &&  &\subset&  && U_{1/2} &\subset& C_{1/2} &&  &\subset&  && U_1
    \\
    C_0 &\subset& U_{1/4} &\subset& C_{1/4} &\subset& U_{1/2} &\subset& C_{1/2} &\subset& U_{3/4} &\subset& C_{3/4} &\subset& U_1
  }
\end{displaymath}
and so on, labeled by the [[dyadic rational numbers]] $\mathbb{Q}_{dy} \subset \mathbb{Q}$ within $(0,1]$

\begin{displaymath}
\{ U_{r} \subset X \}_{r \in (0,1] \cap \mathbb{Q}_{dy}}
\end{displaymath}
with the property

\begin{displaymath}
\underset{r_1,r_2 \in (0,1] \cap \mathbb{Q}_{dy}}{\forall}
  \left(
    \left(
       r_1 \lt r_2
    \right)
     \Rightarrow
    \left(
      U_{r_1} \subset Cl(U_{r_1}) \subset U_{r_2}
    \right)
  \right)
  \,.
\end{displaymath}
Define then the function

\begin{displaymath}
f \;\colon\; X \longrightarrow [0,1]
\end{displaymath}
to assign to a point $x \in X$ the [[infimum]] of the labels of those open subsets in this sequence that contain $x$:

\begin{displaymath}
f(x) 
     \coloneqq
  \underset{U_r \supset \{x\}}{\lim} r
\end{displaymath}
Here the [[limit of a net|limit]] is over the [[directed set]] of those $U_r$ that contain $x$, ordered by reverse inclusion.

This function clearly has the property that $f(A) = \{0\}$ and $f(B) = \{1\}$. It only remains to see that it is continuous.

To this end, first observe that

\begin{displaymath}
\itexarray{
    (\star)
    &&
    \left( 
      x \in Cl(U_r) 
    \right)
      &\Rightarrow&
    \left(
       f(x) \leq r
    \right)
    \\
    (\star\star)
    &&
    \left(
      x \in U_r
    \right)
      &\Leftarrow&    
    \left(
      f(x) \lt r
    \right)
  }
  \,.
\end{displaymath}
Here it is immediate from the definition that $(x \in U_r) \Rightarrow (f(x) \leq r)$ and that $(f(x) \lt r) \Rightarrow (x \in U_r \subset Cl(U_r))$. For the remaining implication, it is sufficient to observe that

\begin{displaymath}
(x \in \partial U_r) \Rightarrow (f(x) = r)
  \,,
\end{displaymath}
where $\partial U_r \coloneqq Cl(U_r) \backslash U_r$ is the [[boundary]] of $U_r$.

This holds because the [[dyadic numbers]] are [[dense subset|dense]] in $\mathbb{R}$. (And this would fail if we stopped the above decomposition into $U_{a/2^n}$-s at some finite $n$.) Namely, in one direction, if $x \in \partial U_r$ then for every small positive real number $\epsilon$ there exists a dyadic rational number $r'$ with $r \lt r' \lt r + \epsilon$, and by construction $U_{r'} \supset Cl(U_r)$ hence $x \in U_{r'}$. This implies that $\underset{U_r \supset \{x\}}{\lim} = r$.

 Now we claim that for all $\alpha \in [0,1]$ then

\begin{enumerate}%
\item $f^{-1}(\,(\alpha, 1]\,) = \underset{r \gt \alpha}{\cup} \left( X \backslash Cl(U_r) \right)$


\item $f^{-1}(\,[0,\alpha)\,) = \underset{r \lt \alpha}{\cup} U_r$



\end{enumerate}
Thereby $f^{-1}(\,(\alpha, 1]\,)$ and $f^{-1}(\,[0,\alpha)\,)$ are exhibited as unions of open subsets, and hence they are open.

Regarding the first point:

\begin{displaymath}
\begin{aligned}
     & x \in f^{-1}( \,(\alpha,1]\, )
     \\
     \Leftrightarrow\,
     & 
     f(x) \gt \alpha
     \\
     \Leftrightarrow\,
     &
     \underset{r \gt \alpha}{\exists} (f(x) \gt r)
     \\
     \overset{(\star)}{\Rightarrow}\,
     &
     \underset{r \gt \alpha}{\exists}
     \left( x \notin Cl(U_r) \right)
     \\
     \Leftrightarrow\,
     &
     x  \in \underset{r \gt \alpha}{\cup} \left(X \backslash Cl(U_r)\right)
  \end{aligned}
\end{displaymath}
and

\begin{displaymath}
\begin{aligned}
     & x  \in \underset{r \gt \alpha}{\cup} \left(X \backslash Cl(U_r)\right)
     \\
     \Leftrightarrow\,
     &
     \underset{r \gt \alpha}{\exists}
     \left( x \notin Cl(U_r) \right)
     \\
     \Rightarrow\,
     &
     \underset{r \gt \alpha}{\exists}
     \left( x \notin U_r \right)     
     \\
     \overset{(\star \star)}{\Rightarrow}\,
     &
     \underset{r \gt \alpha}{\exists}
     \left(
       f(x) \geq r
     \right)
     \\
     \Leftrightarrow\,
     &
     f(x) \gt \alpha
     \\
     \Leftrightarrow\,
     &
     x \in f^{-1}(\, (\alpha,1] \,)
  \end{aligned}
  \,.
\end{displaymath}
Regarding the second point:

\begin{displaymath}
\begin{aligned}
    & 
    x \in f^{-1}(\, [0,\alpha) \,)
    \\
    \Leftrightarrow\,
    &
    f(x) \lt \alpha
    \\
    \Leftrightarrow\,
    &
    \underset{r \lt \alpha}{\exists}( f(x) \lt r )
    \\
    \overset{(\star \star)}{\Rightarrow}\,
    &
    \underset{r \lt \alpha}{\exists }( x \in U_r )
    \\
    \Leftrightarrow\,
     &
    x \in \underset{r \lt \alpha}{\cup} U_r
  \end{aligned}
\end{displaymath}
and

\begin{displaymath}
\begin{aligned}
     &
    x \in \underset{r \lt \alpha}{\cup} U_r
    \\    
    \Leftrightarrow\,
    &
    \underset{r \lt \alpha}{\exists }( x \in U_r )
    \\
    \overset{}{\Rightarrow}\,
    &
    \underset{r \lt \alpha}{\exists }( x \in Cl(U_r) )   
    \\
    \overset{(\star)}{\Rightarrow}\,
     &
    \underset{r \lt \alpha}{\exists }( f(x) \leq r )       
    \\
    \Leftrightarrow\,
     &
    f(x) \lt \alpha
    \\
    \Leftrightarrow\,
    &
    x \in f^{-1}(\, [0,\alpha) \,)
  \end{aligned}
 \,.
\end{displaymath}
(In these derivations we repeatedly use that $(0,1] \cap \mathbb{Q}_{dy}$ is dense in $[0,1]$, and we use the [[contrapositions]] of $(\star)$ and $(\star \star)$, by [[excluded middle]].)

Now since the subsets $\{ [0,\alpha), (\alpha,1]\}_{\alpha \in [0,1]}$ form a [[topological subbase|sub-base]] for the Euclidean metric topology on $[0,1]$, it follows that all pre-images of $f$ are open, hence that $f$ is continuous.

\end{proof}
\hypertarget{Implications}{}\subsection*{{Implications}}\label{Implications}

Urysohn's lemma is key in the proof of many other theorems, for instance

\begin{itemize}%
\item [[Tietze extension theorem]]


\item [[paracompact Hausdorff spaces equivalently admit subordinate partitions of unity]]



\end{itemize}
\hypertarget{related_statements}{}\subsection*{{Related statements}}\label{related_statements}

\begin{itemize}%
\item [[paracompact Hausdorff spaces are normal]]

\end{itemize}
\hypertarget{references}{}\subsection*{{References}}\label{references}

Due to [[Pavel Urysohn]].

\begin{itemize}%
\item [[Terence Tao]], \emph{\href{https://terrytao.wordpress.com/2009/03/02/245b-notes-12-continuous-functions-on-locally-compact-hausdorff-spaces/#more-1844}{245B, Notes 12: Continuous functions on locally compact Hausdorff spaces}}


\item \href{http://planetmath.org/proofofurysohnslemma}{Proof at planetmath}


\item Wikipedia, \emph{\href{https://en.wikipedia.org/wiki/Urysohn%27s_lemma}{Urysohn's lemma}}



\end{itemize}
Lectures notes include

\begin{itemize}%
\item Tarun Chitra, section 2.1 of \emph{The Stone-Cech Compactification} 2009 \href{httP://www.math.cornell.edu/~riley/Teaching/Topology2009/essays/chitra.pdf}{pdf}

\end{itemize}
[[!redirects Urysohn's extension theorem]] [[!redirects Urysohn's theorem]] [[!redirects Urysohn theorem]] [[!redirects Urysohn lemma]]

[[!redirects Urysohn function]] [[!redirects Urysohn functions]]



\end{document}