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''Urysohn–Brouwer–Tietze lemma''
 
''Urysohn–Brouwer–Tietze lemma''
  
An assertion on the possibility of extending a continuous function from a subspace of a topological space to the whole space. Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u0958601.png" /> be a [[Normal space|normal space]] and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u0958602.png" /> a closed subset of it. Then any continuous function <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u0958603.png" /> can be extended to a continuous function <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u0958604.png" />, i.e. one can find a continuous function <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u0958605.png" /> such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u0958606.png" /> for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u0958607.png" />. Moreover, if <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u0958608.png" /> is bounded, then there exists an extension <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u0958609.png" /> such that
+
An assertion on the possibility of extending a continuous function from a subspace of a topological space to the whole space. Let $  X $
 +
be a [[Normal space|normal space]] and $  F $
 +
a closed subset of it. Then any continuous function $  f : F \rightarrow \mathbf R $
 +
can be extended to a continuous function $  g : X \rightarrow \mathbf R $,  
 +
i.e. one can find a continuous function $  g $
 +
such that $  g ( x) = f ( x) $
 +
for all $  x \in F $.  
 +
Moreover, if $  f $
 +
is bounded, then there exists an extension $  g $
 +
such that
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u09586010.png" /></td> </tr></table>
+
$$
 +
\sup _ {x \in F } \
 +
| f ( x) |  = \sup _ {x \in X }  | g ( x) | .
 +
$$
  
The Urysohn–Brouwer lemma was proved by L.E.J. Brouwer and H. Lebesgue for <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u09586011.png" />, by H. Tietze for an arbitrary metric space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095860/u09586012.png" />, and by P.S. Urysohn in the above formulation (which may be used as a characterization of normal spaces and is thus best possible).
+
The Urysohn–Brouwer lemma was proved by L.E.J. Brouwer and H. Lebesgue for $  X = \mathbf R  ^ {n} $,  
 +
by H. Tietze for an arbitrary metric space $  X $,  
 +
and by P.S. Urysohn in the above formulation (which may be used as a characterization of normal spaces and is thus best possible).
  
 
====References====
 
====References====
 
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  P.S. Urysohn,  "Ueber die Mächtigkeit der zusammenhängenden Mengen"  ''Math. Ann.'' , '''94'''  (1925)  pp. 262–295</TD></TR></table>
 
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  P.S. Urysohn,  "Ueber die Mächtigkeit der zusammenhängenden Mengen"  ''Math. Ann.'' , '''94'''  (1925)  pp. 262–295</TD></TR></table>
 
 
  
 
====Comments====
 
====Comments====

Latest revision as of 08:27, 6 June 2020


Urysohn–Brouwer–Tietze lemma

An assertion on the possibility of extending a continuous function from a subspace of a topological space to the whole space. Let $ X $ be a normal space and $ F $ a closed subset of it. Then any continuous function $ f : F \rightarrow \mathbf R $ can be extended to a continuous function $ g : X \rightarrow \mathbf R $, i.e. one can find a continuous function $ g $ such that $ g ( x) = f ( x) $ for all $ x \in F $. Moreover, if $ f $ is bounded, then there exists an extension $ g $ such that

$$ \sup _ {x \in F } \ | f ( x) | = \sup _ {x \in X } | g ( x) | . $$

The Urysohn–Brouwer lemma was proved by L.E.J. Brouwer and H. Lebesgue for $ X = \mathbf R ^ {n} $, by H. Tietze for an arbitrary metric space $ X $, and by P.S. Urysohn in the above formulation (which may be used as a characterization of normal spaces and is thus best possible).

References

[1] P.S. Urysohn, "Ueber die Mächtigkeit der zusammenhängenden Mengen" Math. Ann. , 94 (1925) pp. 262–295

Comments

This assertion is also known as the Tietze–Urysohn extension theorem, or even as the Tietze extension theorem.

References

[a1] R. Engelking, "General topology" , Heldermann (1989)
How to Cite This Entry:
Urysohn-Brouwer lemma. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Urysohn-Brouwer_lemma&oldid=15692
This article was adapted from an original article by I.G. Koshevnikova (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article