The Closedness in the hypothesis of Tietze extension theorem is necessary

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Consider $B= (x,y) in BbbR^2:x^2+y^2 leq 1$ and $D= (x,y) in BbbR^2:x^2+y^2<1$.



Prove or disprove:



1) Given a continuous function $f:B rightarrow BbbR$, there exist a continuous function $g:BbbR^2 rightarrow BbbR$ such that $g=f$ on $B$.



2) Given a continuous function $s:D rightarrow BbbR$, there exist a continuous function $t:BbbR^2 rightarrow BbbR$ such that $t=s$ on $D$.




I know 1) follows from Tietze extension theorem, since $B$ is closed and $BbbR$ is a normal space.



What about 2? Any hint must be appreciated!




general-topology




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    Part 2 is incorrect. To disprove this, construct a function that becomes unbounded near the boundary of $D$.
    – Hans Engler
    Jul 20 at 15:39














up vote
0
down vote

favorite













Consider $B= (x,y) in BbbR^2:x^2+y^2 leq 1$ and $D= (x,y) in BbbR^2:x^2+y^2<1$.



Prove or disprove:



1) Given a continuous function $f:B rightarrow BbbR$, there exist a continuous function $g:BbbR^2 rightarrow BbbR$ such that $g=f$ on $B$.



2) Given a continuous function $s:D rightarrow BbbR$, there exist a continuous function $t:BbbR^2 rightarrow BbbR$ such that $t=s$ on $D$.




I know 1) follows from Tietze extension theorem, since $B$ is closed and $BbbR$ is a normal space.



What about 2? Any hint must be appreciated!




general-topology




share|cite|improve this question

















  • 3




    Part 2 is incorrect. To disprove this, construct a function that becomes unbounded near the boundary of $D$.
    – Hans Engler
    Jul 20 at 15:39












up vote
0
down vote

favorite









up vote
0
down vote

favorite












Consider $B= (x,y) in BbbR^2:x^2+y^2 leq 1$ and $D= (x,y) in BbbR^2:x^2+y^2<1$.



Prove or disprove:



1) Given a continuous function $f:B rightarrow BbbR$, there exist a continuous function $g:BbbR^2 rightarrow BbbR$ such that $g=f$ on $B$.



2) Given a continuous function $s:D rightarrow BbbR$, there exist a continuous function $t:BbbR^2 rightarrow BbbR$ such that $t=s$ on $D$.




I know 1) follows from Tietze extension theorem, since $B$ is closed and $BbbR$ is a normal space.



What about 2? Any hint must be appreciated!




general-topology




share|cite|improve this question














Consider $B= (x,y) in BbbR^2:x^2+y^2 leq 1$ and $D= (x,y) in BbbR^2:x^2+y^2<1$.



Prove or disprove:



1) Given a continuous function $f:B rightarrow BbbR$, there exist a continuous function $g:BbbR^2 rightarrow BbbR$ such that $g=f$ on $B$.



2) Given a continuous function $s:D rightarrow BbbR$, there exist a continuous function $t:BbbR^2 rightarrow BbbR$ such that $t=s$ on $D$.




I know 1) follows from Tietze extension theorem, since $B$ is closed and $BbbR$ is a normal space.



What about 2? Any hint must be appreciated!





general-topology





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 21 at 9:57
























asked Jul 20 at 15:35









Learning Mathematics

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  • 3




    Part 2 is incorrect. To disprove this, construct a function that becomes unbounded near the boundary of $D$.
    – Hans Engler
    Jul 20 at 15:39












  • 3




    Part 2 is incorrect. To disprove this, construct a function that becomes unbounded near the boundary of $D$.
    – Hans Engler
    Jul 20 at 15:39







3




3




Part 2 is incorrect. To disprove this, construct a function that becomes unbounded near the boundary of $D$.
– Hans Engler
Jul 20 at 15:39




Part 2 is incorrect. To disprove this, construct a function that becomes unbounded near the boundary of $D$.
– Hans Engler
Jul 20 at 15:39










1 Answer
1






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As a simple example where 2. fails: let $s: D to mathbbR$ be given by $s(x) = frac1x^2 + y^2 - 1$. You cannot even continuously extend it to a single point of $B setminus D =S^1$.






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  • And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
    – Rafael Gonzalez Lopez
    Jul 21 at 10:05











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










As a simple example where 2. fails: let $s: D to mathbbR$ be given by $s(x) = frac1x^2 + y^2 - 1$. You cannot even continuously extend it to a single point of $B setminus D =S^1$.






share|cite|improve this answer























  • And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
    – Rafael Gonzalez Lopez
    Jul 21 at 10:05















up vote
3
down vote



accepted










As a simple example where 2. fails: let $s: D to mathbbR$ be given by $s(x) = frac1x^2 + y^2 - 1$. You cannot even continuously extend it to a single point of $B setminus D =S^1$.






share|cite|improve this answer























  • And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
    – Rafael Gonzalez Lopez
    Jul 21 at 10:05













up vote
3
down vote



accepted







up vote
3
down vote



accepted






As a simple example where 2. fails: let $s: D to mathbbR$ be given by $s(x) = frac1x^2 + y^2 - 1$. You cannot even continuously extend it to a single point of $B setminus D =S^1$.






share|cite|improve this answer















As a simple example where 2. fails: let $s: D to mathbbR$ be given by $s(x) = frac1x^2 + y^2 - 1$. You cannot even continuously extend it to a single point of $B setminus D =S^1$.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 20 at 17:31


























answered Jul 20 at 16:53









Henno Brandsma

91.6k342100




91.6k342100











  • And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
    – Rafael Gonzalez Lopez
    Jul 21 at 10:05

















  • And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
    – Rafael Gonzalez Lopez
    Jul 21 at 10:05
















And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
– Rafael Gonzalez Lopez
Jul 21 at 10:05





And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
– Rafael Gonzalez Lopez
Jul 21 at 10:05













 

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