Give an example of Tietze extension theorem?

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Give an example of Tietze extension theorem?



I know the theorem definition, but i could not able to find the example



Pliz help me






general-topology continuity







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    up vote
    2
    down vote

    favorite












    Give an example of Tietze extension theorem?



    I know the theorem definition, but i could not able to find the example



    Pliz help me






    general-topology continuity







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Give an example of Tietze extension theorem?



      I know the theorem definition, but i could not able to find the example



      Pliz help me






      general-topology continuity







      share|cite|improve this question















      Give an example of Tietze extension theorem?



      I know the theorem definition, but i could not able to find the example



      Pliz help me






      general-topology continuity




      general-topology continuity






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited 3 hours ago









      José Carlos Santos

      132k17107194




      132k17107194










      asked 3 hours ago









      jasmine

      1,139213




      1,139213




















          3 Answers
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          accepted










          Let $ell^infty$ be the space of all bounded sequences of complex numbers, endowed with the metric$$d_inftybigl((x_n)_ninmathbb N,(y_n)_ninmathbb Nbigr)=supbigllvert x_n-y_nvert,.$$Let $C$ be the subspace of $ell^infty$ which consists of the convergent sequences. The map$$beginarrayrcccLcolon&C&longrightarrow&mathbb C\&(x_n)_ninmathbb N&mapsto&displaystylelim_ntoinftyx_nendarray$$is continuous and $C$ is a closed subspace of $ell^infty$. Therefore, according to the Tietze extension theorem, we can extend $L$ to a continuous function from $ell^infty$ into $mathbb C$.






          share|cite|improve this answer




















          • carlos sir..thanks .Is there any easy example ?
            – jasmine
            3 hours ago






          • 3




            Sure. Take $fcolon0longrightarrowmathbb R$ defined by $f(0)=0$. Then you can extend it to a continuous function $FcolonmathbbRlongrightarrowmathbb R$.
            – José Carlos Santos
            3 hours ago










          • thanks @Jose carlos sir
            – jasmine
            3 hours ago






          • 1




            I'm glad I could help.
            – José Carlos Santos
            3 hours ago

















          up vote
          1
          down vote













          One dimensional example:



          Consider $[1,2]$ as a subspace of $BbbR$ with the induced standard topology and consider $$f:[1,2] ni x mapsto frac1x in BbbR$$ Then $f$ is continuous and $[1,2]$ is closed and $BbbR$ is normal so by Tietze extension theorem, $f$ can be extended to $g$ on $BbbR$ so that $g$ is continuous and $f(x)=g(x)$ for all $x in [1,2]$. One such extension is $$g(x)=begincases f(x)=frac1x & textif ;x in [1,2]\\1 & textif ;x in (-infty,1)\\frac12 & textif;xin (2, infty)endcases$$






          share|cite|improve this answer




















          • thanks u @Chinnapparaj
            – jasmine
            19 mins ago

















          up vote
          1
          down vote













          An application: Let $S$ be a closed bounded subset of $Bbb R^n$ ( with $nin Bbb N$) and let $f:Sto Bbb R$ be continuous. We can extend $f$ to a continuous $f:Bbb R^nto Bbb R$ such that $xin Bbb R^n: f(x)ne 0$ is bounded.



          Let $B$ be a bounded open ball of positive radius, such that $Ssubset B.$ Observe that if we let $f(x)=0$ for all $xin Bbb R^nsetminus B$ then $f$ is continuous on the closed set $B^*=Scup (Bbb R^nsetminus B).$ By the Tietze Extension Theorem, the domain of ( continuous) $f$ can be extended from $B^*$ to all of $Bbb R^n.$



          Remark: The intermediate step, that $f:B^*to Bbb R$ is continuous, can be done by showing that if $Y$ is any closed subset of $Bbb R$ then $(f^-1Y) cap B^*$ is closed in $Bbb R^n$ so, a fortiori, it is closed in $B^*$.






          share|cite|improve this answer




















          • thanks u @Daniel
            – jasmine
            19 mins ago










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          3 Answers
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          3 Answers
          3






          active

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          up vote
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          accepted










          Let $ell^infty$ be the space of all bounded sequences of complex numbers, endowed with the metric$$d_inftybigl((x_n)_ninmathbb N,(y_n)_ninmathbb Nbigr)=supbigllvert x_n-y_nvert,.$$Let $C$ be the subspace of $ell^infty$ which consists of the convergent sequences. The map$$beginarrayrcccLcolon&C&longrightarrow&mathbb C\&(x_n)_ninmathbb N&mapsto&displaystylelim_ntoinftyx_nendarray$$is continuous and $C$ is a closed subspace of $ell^infty$. Therefore, according to the Tietze extension theorem, we can extend $L$ to a continuous function from $ell^infty$ into $mathbb C$.






          share|cite|improve this answer




















          • carlos sir..thanks .Is there any easy example ?
            – jasmine
            3 hours ago






          • 3




            Sure. Take $fcolon0longrightarrowmathbb R$ defined by $f(0)=0$. Then you can extend it to a continuous function $FcolonmathbbRlongrightarrowmathbb R$.
            – José Carlos Santos
            3 hours ago










          • thanks @Jose carlos sir
            – jasmine
            3 hours ago






          • 1




            I'm glad I could help.
            – José Carlos Santos
            3 hours ago














          up vote
          3
          down vote



          accepted










          Let $ell^infty$ be the space of all bounded sequences of complex numbers, endowed with the metric$$d_inftybigl((x_n)_ninmathbb N,(y_n)_ninmathbb Nbigr)=supbigllvert x_n-y_nvert,.$$Let $C$ be the subspace of $ell^infty$ which consists of the convergent sequences. The map$$beginarrayrcccLcolon&C&longrightarrow&mathbb C\&(x_n)_ninmathbb N&mapsto&displaystylelim_ntoinftyx_nendarray$$is continuous and $C$ is a closed subspace of $ell^infty$. Therefore, according to the Tietze extension theorem, we can extend $L$ to a continuous function from $ell^infty$ into $mathbb C$.






          share|cite|improve this answer




















          • carlos sir..thanks .Is there any easy example ?
            – jasmine
            3 hours ago






          • 3




            Sure. Take $fcolon0longrightarrowmathbb R$ defined by $f(0)=0$. Then you can extend it to a continuous function $FcolonmathbbRlongrightarrowmathbb R$.
            – José Carlos Santos
            3 hours ago










          • thanks @Jose carlos sir
            – jasmine
            3 hours ago






          • 1




            I'm glad I could help.
            – José Carlos Santos
            3 hours ago












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          Let $ell^infty$ be the space of all bounded sequences of complex numbers, endowed with the metric$$d_inftybigl((x_n)_ninmathbb N,(y_n)_ninmathbb Nbigr)=supbigllvert x_n-y_nvert,.$$Let $C$ be the subspace of $ell^infty$ which consists of the convergent sequences. The map$$beginarrayrcccLcolon&C&longrightarrow&mathbb C\&(x_n)_ninmathbb N&mapsto&displaystylelim_ntoinftyx_nendarray$$is continuous and $C$ is a closed subspace of $ell^infty$. Therefore, according to the Tietze extension theorem, we can extend $L$ to a continuous function from $ell^infty$ into $mathbb C$.






          share|cite|improve this answer












          Let $ell^infty$ be the space of all bounded sequences of complex numbers, endowed with the metric$$d_inftybigl((x_n)_ninmathbb N,(y_n)_ninmathbb Nbigr)=supbigllvert x_n-y_nvert,.$$Let $C$ be the subspace of $ell^infty$ which consists of the convergent sequences. The map$$beginarrayrcccLcolon&C&longrightarrow&mathbb C\&(x_n)_ninmathbb N&mapsto&displaystylelim_ntoinftyx_nendarray$$is continuous and $C$ is a closed subspace of $ell^infty$. Therefore, according to the Tietze extension theorem, we can extend $L$ to a continuous function from $ell^infty$ into $mathbb C$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          José Carlos Santos

          132k17107194




          132k17107194











          • carlos sir..thanks .Is there any easy example ?
            – jasmine
            3 hours ago






          • 3




            Sure. Take $fcolon0longrightarrowmathbb R$ defined by $f(0)=0$. Then you can extend it to a continuous function $FcolonmathbbRlongrightarrowmathbb R$.
            – José Carlos Santos
            3 hours ago










          • thanks @Jose carlos sir
            – jasmine
            3 hours ago






          • 1




            I'm glad I could help.
            – José Carlos Santos
            3 hours ago
















          • carlos sir..thanks .Is there any easy example ?
            – jasmine
            3 hours ago






          • 3




            Sure. Take $fcolon0longrightarrowmathbb R$ defined by $f(0)=0$. Then you can extend it to a continuous function $FcolonmathbbRlongrightarrowmathbb R$.
            – José Carlos Santos
            3 hours ago










          • thanks @Jose carlos sir
            – jasmine
            3 hours ago






          • 1




            I'm glad I could help.
            – José Carlos Santos
            3 hours ago















          carlos sir..thanks .Is there any easy example ?
          – jasmine
          3 hours ago




          carlos sir..thanks .Is there any easy example ?
          – jasmine
          3 hours ago




          3




          3




          Sure. Take $fcolon0longrightarrowmathbb R$ defined by $f(0)=0$. Then you can extend it to a continuous function $FcolonmathbbRlongrightarrowmathbb R$.
          – José Carlos Santos
          3 hours ago




          Sure. Take $fcolon0longrightarrowmathbb R$ defined by $f(0)=0$. Then you can extend it to a continuous function $FcolonmathbbRlongrightarrowmathbb R$.
          – José Carlos Santos
          3 hours ago












          thanks @Jose carlos sir
          – jasmine
          3 hours ago




          thanks @Jose carlos sir
          – jasmine
          3 hours ago




          1




          1




          I'm glad I could help.
          – José Carlos Santos
          3 hours ago




          I'm glad I could help.
          – José Carlos Santos
          3 hours ago










          up vote
          1
          down vote













          One dimensional example:



          Consider $[1,2]$ as a subspace of $BbbR$ with the induced standard topology and consider $$f:[1,2] ni x mapsto frac1x in BbbR$$ Then $f$ is continuous and $[1,2]$ is closed and $BbbR$ is normal so by Tietze extension theorem, $f$ can be extended to $g$ on $BbbR$ so that $g$ is continuous and $f(x)=g(x)$ for all $x in [1,2]$. One such extension is $$g(x)=begincases f(x)=frac1x & textif ;x in [1,2]\\1 & textif ;x in (-infty,1)\\frac12 & textif;xin (2, infty)endcases$$






          share|cite|improve this answer




















          • thanks u @Chinnapparaj
            – jasmine
            19 mins ago














          up vote
          1
          down vote













          One dimensional example:



          Consider $[1,2]$ as a subspace of $BbbR$ with the induced standard topology and consider $$f:[1,2] ni x mapsto frac1x in BbbR$$ Then $f$ is continuous and $[1,2]$ is closed and $BbbR$ is normal so by Tietze extension theorem, $f$ can be extended to $g$ on $BbbR$ so that $g$ is continuous and $f(x)=g(x)$ for all $x in [1,2]$. One such extension is $$g(x)=begincases f(x)=frac1x & textif ;x in [1,2]\\1 & textif ;x in (-infty,1)\\frac12 & textif;xin (2, infty)endcases$$






          share|cite|improve this answer




















          • thanks u @Chinnapparaj
            – jasmine
            19 mins ago












          up vote
          1
          down vote










          up vote
          1
          down vote









          One dimensional example:



          Consider $[1,2]$ as a subspace of $BbbR$ with the induced standard topology and consider $$f:[1,2] ni x mapsto frac1x in BbbR$$ Then $f$ is continuous and $[1,2]$ is closed and $BbbR$ is normal so by Tietze extension theorem, $f$ can be extended to $g$ on $BbbR$ so that $g$ is continuous and $f(x)=g(x)$ for all $x in [1,2]$. One such extension is $$g(x)=begincases f(x)=frac1x & textif ;x in [1,2]\\1 & textif ;x in (-infty,1)\\frac12 & textif;xin (2, infty)endcases$$






          share|cite|improve this answer












          One dimensional example:



          Consider $[1,2]$ as a subspace of $BbbR$ with the induced standard topology and consider $$f:[1,2] ni x mapsto frac1x in BbbR$$ Then $f$ is continuous and $[1,2]$ is closed and $BbbR$ is normal so by Tietze extension theorem, $f$ can be extended to $g$ on $BbbR$ so that $g$ is continuous and $f(x)=g(x)$ for all $x in [1,2]$. One such extension is $$g(x)=begincases f(x)=frac1x & textif ;x in [1,2]\\1 & textif ;x in (-infty,1)\\frac12 & textif;xin (2, infty)endcases$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Chinnapparaj R

          3,991725




          3,991725











          • thanks u @Chinnapparaj
            – jasmine
            19 mins ago
















          • thanks u @Chinnapparaj
            – jasmine
            19 mins ago















          thanks u @Chinnapparaj
          – jasmine
          19 mins ago




          thanks u @Chinnapparaj
          – jasmine
          19 mins ago










          up vote
          1
          down vote













          An application: Let $S$ be a closed bounded subset of $Bbb R^n$ ( with $nin Bbb N$) and let $f:Sto Bbb R$ be continuous. We can extend $f$ to a continuous $f:Bbb R^nto Bbb R$ such that $xin Bbb R^n: f(x)ne 0$ is bounded.



          Let $B$ be a bounded open ball of positive radius, such that $Ssubset B.$ Observe that if we let $f(x)=0$ for all $xin Bbb R^nsetminus B$ then $f$ is continuous on the closed set $B^*=Scup (Bbb R^nsetminus B).$ By the Tietze Extension Theorem, the domain of ( continuous) $f$ can be extended from $B^*$ to all of $Bbb R^n.$



          Remark: The intermediate step, that $f:B^*to Bbb R$ is continuous, can be done by showing that if $Y$ is any closed subset of $Bbb R$ then $(f^-1Y) cap B^*$ is closed in $Bbb R^n$ so, a fortiori, it is closed in $B^*$.






          share|cite|improve this answer




















          • thanks u @Daniel
            – jasmine
            19 mins ago














          up vote
          1
          down vote













          An application: Let $S$ be a closed bounded subset of $Bbb R^n$ ( with $nin Bbb N$) and let $f:Sto Bbb R$ be continuous. We can extend $f$ to a continuous $f:Bbb R^nto Bbb R$ such that $xin Bbb R^n: f(x)ne 0$ is bounded.



          Let $B$ be a bounded open ball of positive radius, such that $Ssubset B.$ Observe that if we let $f(x)=0$ for all $xin Bbb R^nsetminus B$ then $f$ is continuous on the closed set $B^*=Scup (Bbb R^nsetminus B).$ By the Tietze Extension Theorem, the domain of ( continuous) $f$ can be extended from $B^*$ to all of $Bbb R^n.$



          Remark: The intermediate step, that $f:B^*to Bbb R$ is continuous, can be done by showing that if $Y$ is any closed subset of $Bbb R$ then $(f^-1Y) cap B^*$ is closed in $Bbb R^n$ so, a fortiori, it is closed in $B^*$.






          share|cite|improve this answer




















          • thanks u @Daniel
            – jasmine
            19 mins ago












          up vote
          1
          down vote










          up vote
          1
          down vote









          An application: Let $S$ be a closed bounded subset of $Bbb R^n$ ( with $nin Bbb N$) and let $f:Sto Bbb R$ be continuous. We can extend $f$ to a continuous $f:Bbb R^nto Bbb R$ such that $xin Bbb R^n: f(x)ne 0$ is bounded.



          Let $B$ be a bounded open ball of positive radius, such that $Ssubset B.$ Observe that if we let $f(x)=0$ for all $xin Bbb R^nsetminus B$ then $f$ is continuous on the closed set $B^*=Scup (Bbb R^nsetminus B).$ By the Tietze Extension Theorem, the domain of ( continuous) $f$ can be extended from $B^*$ to all of $Bbb R^n.$



          Remark: The intermediate step, that $f:B^*to Bbb R$ is continuous, can be done by showing that if $Y$ is any closed subset of $Bbb R$ then $(f^-1Y) cap B^*$ is closed in $Bbb R^n$ so, a fortiori, it is closed in $B^*$.






          share|cite|improve this answer












          An application: Let $S$ be a closed bounded subset of $Bbb R^n$ ( with $nin Bbb N$) and let $f:Sto Bbb R$ be continuous. We can extend $f$ to a continuous $f:Bbb R^nto Bbb R$ such that $xin Bbb R^n: f(x)ne 0$ is bounded.



          Let $B$ be a bounded open ball of positive radius, such that $Ssubset B.$ Observe that if we let $f(x)=0$ for all $xin Bbb R^nsetminus B$ then $f$ is continuous on the closed set $B^*=Scup (Bbb R^nsetminus B).$ By the Tietze Extension Theorem, the domain of ( continuous) $f$ can be extended from $B^*$ to all of $Bbb R^n.$



          Remark: The intermediate step, that $f:B^*to Bbb R$ is continuous, can be done by showing that if $Y$ is any closed subset of $Bbb R$ then $(f^-1Y) cap B^*$ is closed in $Bbb R^n$ so, a fortiori, it is closed in $B^*$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          DanielWainfleet

          32.8k31645




          32.8k31645











          • thanks u @Daniel
            – jasmine
            19 mins ago
















          • thanks u @Daniel
            – jasmine
            19 mins ago















          thanks u @Daniel
          – jasmine
          19 mins ago




          thanks u @Daniel
          – jasmine
          19 mins ago

















           

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