Urysohn's lemma

<h2 id="discussion">Discussion</h2>

<p>Two subsets 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 of a <a href="/facts/Topological_space/v2ut7CbK">topological space</a> 
  
    
      
        X
      
    
    {\displaystyle X}
  
 are said to be <a href="/facts/Separated_by_neighbourhoods/NxVgxIF9">separated by neighbourhoods</a> if there are <a href="/facts/Neighbourhood_(topology)/XHKgrpkp">neighbourhoods</a> 
  
    
      
        U
      
    
    {\displaystyle U}
  
 of 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        V
      
    
    {\displaystyle V}
  
 of 
  
    
      
        B
      
    
    {\displaystyle B}
  
 that are disjoint. In particular 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 are necessarily disjoint.
</p><p>Two plain subsets 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 are said to be <a href="/facts/Separated_by_a_function/NxVgxIF9">separated by a continuous function</a> if there exists a <a href="/facts/Continuous_function_(topology)/sKbl02pB">continuous function</a> 
  
    
      
        f
        :
        X
        →
        [
        0
        ,
        1
        ]
      
    
    {\displaystyle f:X\to [0,1]}
  
 from 
  
    
      
        X
      
    
    {\displaystyle X}
  
 into the <a href="/facts/Unit_interval/W4dP8IJo">unit interval</a> 
  
    
      
        [
        0
        ,
        1
        ]
      
    
    {\displaystyle [0,1]}
  
 such that 
  
    
      
        f
        (
        a
        )
        =
        0
      
    
    {\displaystyle f(a)=0}
  
 for all 
  
    
      
        a
        ∈
        A
      
    
    {\displaystyle a\in A}
  
 and 
  
    
      
        f
        (
        b
        )
        =
        1
      
    
    {\displaystyle f(b)=1}
  
 for all 
  
    
      
        b
        ∈
        B
        .
      
    
    {\displaystyle b\in B.}
  
 Any such function is called a Urysohn function for 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
        .
      
    
    {\displaystyle B.}
  
 In particular 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 are necessarily disjoint.
</p><p>It follows that if two subsets 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 are separated by a function then so are their closures. Also it follows that if two subsets 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 are separated by a function then 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 are separated by neighbourhoods.
</p><p>A <a href="/facts/Normal_space/t4DLqLCd">normal space</a> is a topological space in which any two disjoint closed sets can be separated by neighbourhoods. Urysohn's lemma states that a topological space is normal if and only if any two disjoint closed sets can be separated by a continuous function.
</p><p>The sets 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 need not be <a href="/facts/Precisely_separated_by_a_function/NxVgxIF9">precisely separated by 
  
    
      
        f
      
    
    {\displaystyle f}
  
</a>, i.e., it is not necessary and guaranteed that 
  
    
      
        f
        (
        x
        )
        ≠
        0
      
    
    {\displaystyle f(x)\neq 0}
  
 and 
  
    
      
        ≠
        1
      
    
    {\displaystyle \neq 1}
  
 for 
  
    
      
        x
      
    
    {\displaystyle x}
  
 outside 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
        .
      
    
    {\displaystyle B.}
  
 A topological space 
  
    
      
        X
      
    
    {\displaystyle X}
  
 in which every two disjoint closed subsets 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 are precisely separated by a continuous function is <a href="/facts/Normal_space/t4DLqLCd">perfectly normal</a>.
</p><p>Urysohn's lemma has led to the formulation of other topological properties such as the 'Tychonoff property' and 'completely Hausdorff spaces'. For example, a corollary of the lemma is that normal <a href="/facts/T1_space/K4exfuf7">T1 spaces</a> are <a href="/facts/Tychonoff_space/jyCgpKB5">Tychonoff</a>.
</p>
<h2 id="formal-statement">Formal statement</h2>
<p>A topological space 
  
    
      
        X
      
    
    {\displaystyle X}
  
 is normal if and only if, for any two non-empty closed disjoint subsets 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 of 
  
    
      
        X
        ,
      
    
    {\displaystyle X,}
  
 there exists a continuous map 
  
    
      
        f
        :
        X
        →
        [
        0
        ,
        1
        ]
      
    
    {\displaystyle f:X\to [0,1]}
  
 such that 
  
    
      
        f
        (
        A
        )
        =
        {
        0
        }
      
    
    {\displaystyle f(A)=\{0\}}
  
 and 
  
    
      
        f
        (
        B
        )
        =
        {
        1
        }
        .
      
    
    {\displaystyle f(B)=\{1\}.}

</p>
<h2 id="proof-sketch">Proof sketch</h2>

<p>The proof proceeds by repeatedly applying the following alternate characterization of normality. If 
  
    
      
        X
      
    
    {\displaystyle X}
  
 is a normal space, 
  
    
      
        Z
      
    
    {\displaystyle Z}
  
 is an <a href="/facts/Open_set/QfTCIqMu">open</a> subset of 
  
    
      
        X
      
    
    {\displaystyle X}
  
, and 
  
    
      
        Y
        ⊆
        Z
      
    
    {\displaystyle Y\subseteq Z}
  
 is closed, then there exists an open 
  
    
      
        U
      
    
    {\displaystyle U}
  
 and a closed 
  
    
      
        V
      
    
    {\displaystyle V}
  
 such that 
  
    
      
        Y
        ⊆
        U
        ⊆
        V
        ⊆
        Z
      
    
    {\displaystyle Y\subseteq U\subseteq V\subseteq Z}
  
. 
</p><p>Let 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        B
      
    
    {\displaystyle B}
  
 be disjoint closed subsets of 
  
    
      
        X
      
    
    {\displaystyle X}
  
. The main idea of the proof is to repeatedly apply this characterization of normality to 
  
    
      
        A
      
    
    {\displaystyle A}
  
 and 
  
    
      
        
          B
          
            ∁
          
        
      
    
    {\displaystyle B^{\complement }}
  
, continuing with the new sets built on every step. 
</p><p>The sets we build are indexed by <a href="/facts/Dyadic_rational/LPwRYKaD">dyadic fractions</a>. For every dyadic fraction 
  
    
      
        r
        ∈
        (
        0
        ,
        1
        )
      
    
    {\displaystyle r\in (0,1)}
  
, we construct an open subset 
  
    
      
        U
        (
        r
        )
      
    
    {\displaystyle U(r)}
  
 and a closed subset 
  
    
      
        V
        (
        r
        )
      
    
    {\displaystyle V(r)}
  
 of 
  
    
      
        X
      
    
    {\displaystyle X}
  
 such that:
</p>
<ul><li>
  
    
      
        A
        ⊆
        U
        (
        r
        )
      
    
    {\displaystyle A\subseteq U(r)}
  
 and 
  
    
      
        V
        (
        r
        )
        ⊆
        
          B
          
            ∁
          
        
      
    
    {\displaystyle V(r)\subseteq B^{\complement }}
  
 for all 
  
    
      
        r
      
    
    {\displaystyle r}
  
,</li>
<li>
  
    
      
        U
        (
        r
        )
        ⊆
        V
        (
        r
        )
      
    
    {\displaystyle U(r)\subseteq V(r)}
  
 for all 
  
    
      
        r
      
    
    {\displaystyle r}
  
,</li>
<li>For 
  
    
      
        r
        <
        s
      
    
    {\displaystyle r<s}
  
, 
  
    
      
        V
        (
        r
        )
        ⊆
        U
        (
        s
        )
      
    
    {\displaystyle V(r)\subseteq U(s)}
  
.</li></ul>
<p>Intuitively, the sets 
  
    
      
        U
        (
        r
        )
      
    
    {\displaystyle U(r)}
  
 and 
  
    
      
        V
        (
        r
        )
      
    
    {\displaystyle V(r)}
  
 expand outwards in layers from 
  
    
      
        A
      
    
    {\displaystyle A}
  
:
</p>

A
              
              
              
              
              
              
              
              
                ⊆
              
              
              
              
              
              
              
              
                
                  B
                  
                    ∁
                  
                
              
            
            
              
                A
              
              
              
              
                ⊆
              
              
              
              
                 
                U
                (
                1
                
                  /
                
                2
                )
              
              
                ⊆
              
              
                V
                (
                1
                
                  /
                
                2
                )
              
              
              
              
                ⊆
              
              
              
              
                
                  B
                  
                    ∁
                  
                
              
            
            
              
                A
              
              
                ⊆
              
              
                U
                (
                1
                
                  /
                
                4
                )
              
              
                ⊆
              
              
                V
                (
                1
                
                  /
                
                4
                )
              
              
                ⊆
              
              
                U
                (
                1
                
                  /
                
                2
                )
              
              
                ⊆
              
              
                V
                (
                1
                
                  /
                
                2
                )
              
              
                ⊆
              
              
                U
                (
                3
                
                  /
                
                4
                )
              
              
                ⊆
              
              
                V
                (
                3
                
                  /
                
                4
                )
              
              
                ⊆
              
              
                
                  B
                  
                    ∁
                  
                
              
            
          
        
      
    
    {\displaystyle {\begin{array}{ccccccccccccccc}A&&&&&&&\subseteq &&&&&&&B^{\complement }\\A&&&\subseteq &&&\ U(1/2)&\subseteq &V(1/2)&&&\subseteq &&&B^{\complement }\\A&\subseteq &U(1/4)&\subseteq &V(1/4)&\subseteq &U(1/2)&\subseteq &V(1/2)&\subseteq &U(3/4)&\subseteq &V(3/4)&\subseteq &B^{\complement }\end{array}}}

<p>This construction proceeds by <a href="/facts/Mathematical_induction/KxAPqNrH">mathematical induction</a>. For the base step, we define two extra sets 
  
    
      
        U
        (
        1
        )
        =
        
          B
          
            ∁
          
        
      
    
    {\displaystyle U(1)=B^{\complement }}
  
 and 
  
    
      
        V
        (
        0
        )
        =
        A
      
    
    {\displaystyle V(0)=A}
  
. 
</p><p>Now assume that 
  
    
      
        n
        ≥
        0
      
    
    {\displaystyle n\geq 0}
  
 and that the sets 
  
    
      
        U
        
          (
          
            k
            
              /
            
            
              2
              
                n
              
            
          
          )
        
      
    
    {\displaystyle U\left(k/2^{n}\right)}
  
 and 
  
    
      
        V
        
          (
          
            k
            
              /
            
            
              2
              
                n
              
            
          
          )
        
      
    
    {\displaystyle V\left(k/2^{n}\right)}
  
 have already been constructed for 
  
    
      
        k
        ∈
        {
        1
        ,
        …
        ,
        
          2
          
            n
          
        
        −
        1
        }
      
    
    {\displaystyle k\in \{1,\ldots ,2^{n}-1\}}
  
. Note that this is <a href="/facts/Vacuous_truth/QBsue3pP">vacuously</a> satisfied for 
  
    
      
        n
        =
        0
      
    
    {\displaystyle n=0}
  
. Since 
  
    
      
        X
      
    
    {\displaystyle X}
  
 is normal, for any 
  
    
      
        a
        ∈
        
          {
          
            0
            ,
            1
            ,
            …
            ,
            
              2
              
                n
              
            
            −
            1
          
          }
        
      
    
    {\displaystyle a\in \left\{0,1,\ldots ,2^{n}-1\right\}}
  
, we can find an open set and a closed set such that
</p>

V
        
          (
          
            
              a
              
                2
                
                  n
                
              
            
          
          )
        
        ⊆
        U
        
          (
          
            
              
                2
                a
                +
                1
              
              
                2
                
                  n
                  +
                  1
                
              
            
          
          )
        
        ⊆
        V
        
          (
          
            
              
                2
                a
                +
                1
              
              
                2
                
                  n
                  +
                  1
                
              
            
          
          )
        
        ⊆
        U
        
          (
          
            
              
                a
                +
                1
              
              
                2
                
                  n
                
              
            
          
          )
        
      
    
    {\displaystyle V\left({\frac {a}{2^{n}}}\right)\subseteq U\left({\frac {2a+1}{2^{n+1}}}\right)\subseteq V\left({\frac {2a+1}{2^{n+1}}}\right)\subseteq U\left({\frac {a+1}{2^{n}}}\right)}

<p>The above three conditions are then verified.
</p><p>Once we have these sets, we define 
  
    
      
        f
        (
        x
        )
        =
        1
      
    
    {\displaystyle f(x)=1}
  
 if 
  
    
      
        x
        ∉
        U
        (
        r
        )
      
    
    {\displaystyle x\not \in U(r)}
  
 for any 
  
    
      
        r
      
    
    {\displaystyle r}
  
; otherwise 
  
    
      
        f
        (
        x
        )
        =
        inf
        {
        r
        :
        x
        ∈
        U
        (
        r
        )
        }
      
    
    {\displaystyle f(x)=\inf\{r:x\in U(r)\}}
  
 for every 
  
    
      
        x
        ∈
        X
      
    
    {\displaystyle x\in X}
  
, where 
  
    
      
        inf
      
    
    {\displaystyle \inf }
  
 denotes the <a href="/facts/Infimum/oXCKVopz">infimum</a>. Using the fact that the dyadic rationals are <a href="/facts/Dense_set/nPT9Yxao">dense</a>, it is then not too hard to show that 
  
    
      
        f
      
    
    {\displaystyle f}
  
 is continuous and has the property 
  
    
      
        f
        (
        A
        )
        ⊆
        {
        0
        }
      
    
    {\displaystyle f(A)\subseteq \{0\}}
  
 and 
  
    
      
        f
        (
        B
        )
        ⊆
        {
        1
        }
        .
      
    
    {\displaystyle f(B)\subseteq \{1\}.}
  
 This step requires the 
  
    
      
        V
        (
        r
        )
      
    
    {\displaystyle V(r)}
  
 sets in order to work.
</p><p>The <a href="/facts/Mizar_system/lX1GUsUG">Mizar project</a> has completely formalised and automatically checked a proof of Urysohn's lemma in the <a href="http://mizar.org/JFM/Vol13/urysohn3.html">URYSOHN3 file</a>.
</p>
<h2 id="see-also">See also</h2>
<ul><li><a href="/facts/Mollifier/aQx68Ksc">Mollifier</a></li></ul>
<h2 id="notes">Notes</h2>

<ul><li>Willard, Stephen (2004) [1970]. <a href="https://books.google.com/books?id=-o8xJQ7Ag2cC"><i>General Topology</i></a>. <a href="/facts/Mineola,_N.Y./GWYvEokQ">Mineola, N.Y.</a>: <a href="/facts/Dover_Publications/RwjvlSFv">Dover Publications</a>. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-486-43479-7. <a href="/facts/OCLC_(identifier)/Yqad8waq">OCLC</a> <a href="https://search.worldcat.org/oclc/115240">115240</a>.</li>
<li>Willard, Stephen (1970). <i>General Topology</i>. Dover Publications. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 0-486-43479-6.</li></ul>
<h2 id="external-links">External links</h2>
<ul><li><a href="https://www.encyclopediaofmath.org/index.php?title=Urysohn_lemma">"Urysohn lemma"</a>, <i><a href="/facts/Encyclopedia_of_Mathematics/WC6mGtPm">Encyclopedia of Mathematics</a></i>, <a href="/facts/European_Mathematical_Society/B3h7b672">EMS Press</a>, 2001 [1994]</li>
<li><a href="/facts/Mizar_system/lX1GUsUG">Mizar system</a> proof: <a href="https://www.mizar.org/version/current/html/urysohn3.html#T20">https://www.mizar.org/version/current/html/urysohn3.html#T20</a></li></ul>

<h2 id="references">References</h2>

<ol>
<li id="fn:1"><p>Willard 1970 Section 15. - Willard, Stephen (1970). General Topology. Dover Publications. ISBN 0-486-43479-6. <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li>
</ol>

Urysohn's lemma open-in-new

Urysohn's lemma