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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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
3
Part 2 is incorrect. To disprove this, construct a function that becomes unbounded near the boundary of $D$.
â Hans Engler
Jul 20 at 15:39
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up vote
0
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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
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
edited Jul 21 at 9:57
asked Jul 20 at 15:35
Learning Mathematics
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522313
3
Part 2 is incorrect. To disprove this, construct a function that becomes unbounded near the boundary of $D$.
â Hans Engler
Jul 20 at 15:39
add a comment |Â
3
Part 2 is incorrect. To disprove this, construct a function that becomes unbounded near the boundary of $D$.
â 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 $D$.
â Hans Engler
Jul 20 at 15:39
Part 2 is incorrect. To disprove this, construct a function that becomes unbounded near the boundary of $D$.
â Hans Engler
Jul 20 at 15:39
add a comment |Â
1 Answer
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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 $B setminus D =S^1$.
And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
â Rafael Gonzalez Lopez
Jul 21 at 10:05
add a comment |Â
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 $B setminus D =S^1$.
And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
â Rafael Gonzalez Lopez
Jul 21 at 10:05
add a comment |Â
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 $B setminus D =S^1$.
And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
â Rafael Gonzalez Lopez
Jul 21 at 10:05
add a comment |Â
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 $B setminus D =S^1$.
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 $B setminus D =S^1$.
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 $f(x,y) = (x^2 +y^2-2)^-1$
â Rafael Gonzalez Lopez
Jul 21 at 10:05
add a comment |Â
And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
â Rafael Gonzalez Lopez
Jul 21 at 10:05
And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
â Rafael Gonzalez Lopez
Jul 21 at 10:05
And another trivial counterexample for 1 is $f(x,y) = (x^2 +y^2-2)^-1$
â Rafael Gonzalez Lopez
Jul 21 at 10:05
add a comment |Â
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3
Part 2 is incorrect. To disprove this, construct a function that becomes unbounded near the boundary of $D$.
â Hans Engler
Jul 20 at 15:39