Mixing
Real-valued function $f : mathbbR to overlinemathbbR$ is continuous iff its graph is closed
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
I'm considering functions $f : mathbbR to mathbbR$ with the codomain extended to the extended reals $mathbbR cup pminfty.$ Now if $f$ is continuous then the function $g : mathbbR times overlinemathbbR to mathbbR : (x, y) mapsto y - f(x)$ is also continuous. Then the graph of $f$ is the set $g^-1(0)$ and is therefore closed.
It seems plausible that the converse is true - that is, that if the graph of $f : mathbbR to overlinemathbbR$ is closed then $f$ is continuous - but I wouldn't be too surprised if there's an annoying counterexample.
Is the converse true?
general-topology continuity
asked Jul 30 at 12:32
Andrew
871211
add a comment |Â
up vote
4
down vote
favorite
I'm considering functions $f : mathbbR to mathbbR$ with the codomain extended to the extended reals $mathbbR cup pminfty.$ Now if $f$ is continuous then the function $g : mathbbR times overlinemathbbR to mathbbR : (x, y) mapsto y - f(x)$ is also continuous. Then the graph of $f$ is the set $g^-1(0)$ and is therefore closed.
It seems plausible that the converse is true - that is, that if the graph of $f : mathbbR to overlinemathbbR$ is closed then $f$ is continuous - but I wouldn't be too surprised if there's an annoying counterexample.
Is the converse true?
general-topology continuity
asked Jul 30 at 12:32
Andrew
871211
The closest related statement (sans compactness) that is actually true (and I can think of) is that a function is Borel if and only if its graph is analytic.
– tomasz
Jul 30 at 15:17
1
Well, unless you assume algebraic regularity or something. I've heard of automatic continuity theorems that could give you something like that (but I don't remember the details). See also various closed graph theorems.
– tomasz
Jul 30 at 15:19
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm considering functions $f : mathbbR to mathbbR$ with the codomain extended to the extended reals $mathbbR cup pminfty.$ Now if $f$ is continuous then the function $g : mathbbR times overlinemathbbR to mathbbR : (x, y) mapsto y - f(x)$ is also continuous. Then the graph of $f$ is the set $g^-1(0)$ and is therefore closed.
It seems plausible that the converse is true - that is, that if the graph of $f : mathbbR to overlinemathbbR$ is closed then $f$ is continuous - but I wouldn't be too surprised if there's an annoying counterexample.
Is the converse true?
general-topology continuity
asked Jul 30 at 12:32
Andrew
871211
I'm considering functions $f : mathbbR to mathbbR$ with the codomain extended to the extended reals $mathbbR cup pminfty.$ Now if $f$ is continuous then the function $g : mathbbR times overlinemathbbR to mathbbR : (x, y) mapsto y - f(x)$ is also continuous. Then the graph of $f$ is the set $g^-1(0)$ and is therefore closed.
It seems plausible that the converse is true - that is, that if the graph of $f : mathbbR to overlinemathbbR$ is closed then $f$ is continuous - but I wouldn't be too surprised if there's an annoying counterexample.
Is the converse true?
general-topology continuity
asked Jul 30 at 12:32
Andrew
871211
asked Jul 30 at 12:32
Andrew
871211
asked Jul 30 at 12:32
Andrew
871211
asked Jul 30 at 12:32
Andrew
871211
871211
The closest related statement (sans compactness) that is actually true (and I can think of) is that a function is Borel if and only if its graph is analytic.
– tomasz
Jul 30 at 15:17
1
Well, unless you assume algebraic regularity or something. I've heard of automatic continuity theorems that could give you something like that (but I don't remember the details). See also various closed graph theorems.
– tomasz
Jul 30 at 15:19
add a comment |Â
The closest related statement (sans compactness) that is actually true (and I can think of) is that a function is Borel if and only if its graph is analytic.
– tomasz
Jul 30 at 15:17
1
Well, unless you assume algebraic regularity or something. I've heard of automatic continuity theorems that could give you something like that (but I don't remember the details). See also various closed graph theorems.
– tomasz
Jul 30 at 15:19
The closest related statement (sans compactness) that is actually true (and I can think of) is that a function is Borel if and only if its graph is analytic.
– tomasz
Jul 30 at 15:17
The closest related statement (sans compactness) that is actually true (and I can think of) is that a function is Borel if and only if its graph is analytic.
– tomasz
Jul 30 at 15:17
1
1
Well, unless you assume algebraic regularity or something. I've heard of automatic continuity theorems that could give you something like that (but I don't remember the details). See also various closed graph theorems.
– tomasz
Jul 30 at 15:19
Well, unless you assume algebraic regularity or something. I've heard of automatic continuity theorems that could give you something like that (but I don't remember the details). See also various closed graph theorems.
– tomasz
Jul 30 at 15:19
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Suppose $f$ is not continuous at some $x_0$. Then there is a sequence $(x_n)$ with $x_n to x_0$, but $|f(x_n) - f(x_0)|geq epsilon >0$ for all $n$.
Switching to a subsequence, we can assume $f(x_n)to y in overlinemathbbR$, since $overlinemathbbR$ is sequentially compact. Since th graph of $f$ is closed, this means $(x_0,y)in mathrmgraph f$ and hence $y=f(x_0)$. Hence, (along the subsequence) $f(x_n)to y =f(x_0)$, in contradiction to $|f(x_n) - f(x_0)|geq epsilon >0$ for all $n$.
answered Jul 30 at 15:51
PhoemueX
26.4k22353
Thank you! I should have remembered that the extended reals were compact. 😀
– Andrew
Jul 30 at 18:30
add a comment |Â
up vote
2
down vote
This is true. Given $[a,b] subset mathbbR$, $mathrmGr(f) cap ([a,b] times overlinemathbbR)$ is compact (since it is a closed set inside a compact one). It follows that $pi_1 |_mathrmGr(f) cap ([a,b] times overlinemathbbR)$ is a homeomorphism with its image, and hence $f|_[a,b]=pi_2 circ (pi_1 |_mathrmGr(f) cap ([a,b] times overlinemathbbR))^-1$ is continuous. Therefore $f|_(a,b)$ is continuous. Since this holds for arbitrary $(a,b) subset mathbbR$, $f$ is continuous in the whole real line.
Note that this generalizes to the following statement:
Let $X$ be locally compact and Hausdorff, and $Y$ be compact. If $f:X to Y$ is a function such that $mathrmGr(f)$ is closed, then $f$ is continuous.
answered Jul 30 at 16:50
Aloizio Macedo
22.5k23283
Thank you! I accepted the other answer as I found it a bit easier to follow, but I do like how the result generalises. 🙂
– Andrew
Jul 30 at 18:31
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Suppose $f$ is not continuous at some $x_0$. Then there is a sequence $(x_n)$ with $x_n to x_0$, but $|f(x_n) - f(x_0)|geq epsilon >0$ for all $n$.
Switching to a subsequence, we can assume $f(x_n)to y in overlinemathbbR$, since $overlinemathbbR$ is sequentially compact. Since th graph of $f$ is closed, this means $(x_0,y)in mathrmgraph f$ and hence $y=f(x_0)$. Hence, (along the subsequence) $f(x_n)to y =f(x_0)$, in contradiction to $|f(x_n) - f(x_0)|geq epsilon >0$ for all $n$.
answered Jul 30 at 15:51
PhoemueX
26.4k22353
Thank you! I should have remembered that the extended reals were compact. 😀
– Andrew
Jul 30 at 18:30
add a comment |Â
up vote
1
down vote
accepted
Suppose $f$ is not continuous at some $x_0$. Then there is a sequence $(x_n)$ with $x_n to x_0$, but $|f(x_n) - f(x_0)|geq epsilon >0$ for all $n$.
Switching to a subsequence, we can assume $f(x_n)to y in overlinemathbbR$, since $overlinemathbbR$ is sequentially compact. Since th graph of $f$ is closed, this means $(x_0,y)in mathrmgraph f$ and hence $y=f(x_0)$. Hence, (along the subsequence) $f(x_n)to y =f(x_0)$, in contradiction to $|f(x_n) - f(x_0)|geq epsilon >0$ for all $n$.
answered Jul 30 at 15:51
PhoemueX
26.4k22353
Thank you! I should have remembered that the extended reals were compact. 😀
– Andrew
Jul 30 at 18:30
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Suppose $f$ is not continuous at some $x_0$. Then there is a sequence $(x_n)$ with $x_n to x_0$, but $|f(x_n) - f(x_0)|geq epsilon >0$ for all $n$.
Switching to a subsequence, we can assume $f(x_n)to y in overlinemathbbR$, since $overlinemathbbR$ is sequentially compact. Since th graph of $f$ is closed, this means $(x_0,y)in mathrmgraph f$ and hence $y=f(x_0)$. Hence, (along the subsequence) $f(x_n)to y =f(x_0)$, in contradiction to $|f(x_n) - f(x_0)|geq epsilon >0$ for all $n$.
answered Jul 30 at 15:51
PhoemueX
26.4k22353
Suppose $f$ is not continuous at some $x_0$. Then there is a sequence $(x_n)$ with $x_n to x_0$, but $|f(x_n) - f(x_0)|geq epsilon >0$ for all $n$.
Switching to a subsequence, we can assume $f(x_n)to y in overlinemathbbR$, since $overlinemathbbR$ is sequentially compact. Since th graph of $f$ is closed, this means $(x_0,y)in mathrmgraph f$ and hence $y=f(x_0)$. Hence, (along the subsequence) $f(x_n)to y =f(x_0)$, in contradiction to $|f(x_n) - f(x_0)|geq epsilon >0$ for all $n$.
answered Jul 30 at 15:51
PhoemueX
26.4k22353
answered Jul 30 at 15:51
PhoemueX
26.4k22353
answered Jul 30 at 15:51
PhoemueX
26.4k22353
answered Jul 30 at 15:51
PhoemueX
26.4k22353
26.4k22353
Thank you! I should have remembered that the extended reals were compact. 😀
– Andrew
Jul 30 at 18:30
add a comment |Â
Thank you! I should have remembered that the extended reals were compact. 😀
– Andrew
Jul 30 at 18:30
Thank you! I should have remembered that the extended reals were compact. 😀
– Andrew
Jul 30 at 18:30
Thank you! I should have remembered that the extended reals were compact. 😀
– Andrew
Jul 30 at 18:30
add a comment |Â
up vote
2
down vote
This is true. Given $[a,b] subset mathbbR$, $mathrmGr(f) cap ([a,b] times overlinemathbbR)$ is compact (since it is a closed set inside a compact one). It follows that $pi_1 |_mathrmGr(f) cap ([a,b] times overlinemathbbR)$ is a homeomorphism with its image, and hence $f|_[a,b]=pi_2 circ (pi_1 |_mathrmGr(f) cap ([a,b] times overlinemathbbR))^-1$ is continuous. Therefore $f|_(a,b)$ is continuous. Since this holds for arbitrary $(a,b) subset mathbbR$, $f$ is continuous in the whole real line.
Note that this generalizes to the following statement:
Let $X$ be locally compact and Hausdorff, and $Y$ be compact. If $f:X to Y$ is a function such that $mathrmGr(f)$ is closed, then $f$ is continuous.
answered Jul 30 at 16:50
Aloizio Macedo
22.5k23283
Thank you! I accepted the other answer as I found it a bit easier to follow, but I do like how the result generalises. 🙂
– Andrew
Jul 30 at 18:31
add a comment |Â
up vote
2
down vote
This is true. Given $[a,b] subset mathbbR$, $mathrmGr(f) cap ([a,b] times overlinemathbbR)$ is compact (since it is a closed set inside a compact one). It follows that $pi_1 |_mathrmGr(f) cap ([a,b] times overlinemathbbR)$ is a homeomorphism with its image, and hence $f|_[a,b]=pi_2 circ (pi_1 |_mathrmGr(f) cap ([a,b] times overlinemathbbR))^-1$ is continuous. Therefore $f|_(a,b)$ is continuous. Since this holds for arbitrary $(a,b) subset mathbbR$, $f$ is continuous in the whole real line.
Note that this generalizes to the following statement:
Let $X$ be locally compact and Hausdorff, and $Y$ be compact. If $f:X to Y$ is a function such that $mathrmGr(f)$ is closed, then $f$ is continuous.
answered Jul 30 at 16:50
Aloizio Macedo
22.5k23283
Thank you! I accepted the other answer as I found it a bit easier to follow, but I do like how the result generalises. 🙂
– Andrew
Jul 30 at 18:31
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is true. Given $[a,b] subset mathbbR$, $mathrmGr(f) cap ([a,b] times overlinemathbbR)$ is compact (since it is a closed set inside a compact one). It follows that $pi_1 |_mathrmGr(f) cap ([a,b] times overlinemathbbR)$ is a homeomorphism with its image, and hence $f|_[a,b]=pi_2 circ (pi_1 |_mathrmGr(f) cap ([a,b] times overlinemathbbR))^-1$ is continuous. Therefore $f|_(a,b)$ is continuous. Since this holds for arbitrary $(a,b) subset mathbbR$, $f$ is continuous in the whole real line.
Note that this generalizes to the following statement:
Let $X$ be locally compact and Hausdorff, and $Y$ be compact. If $f:X to Y$ is a function such that $mathrmGr(f)$ is closed, then $f$ is continuous.
answered Jul 30 at 16:50
Aloizio Macedo
22.5k23283
This is true. Given $[a,b] subset mathbbR$, $mathrmGr(f) cap ([a,b] times overlinemathbbR)$ is compact (since it is a closed set inside a compact one). It follows that $pi_1 |_mathrmGr(f) cap ([a,b] times overlinemathbbR)$ is a homeomorphism with its image, and hence $f|_[a,b]=pi_2 circ (pi_1 |_mathrmGr(f) cap ([a,b] times overlinemathbbR))^-1$ is continuous. Therefore $f|_(a,b)$ is continuous. Since this holds for arbitrary $(a,b) subset mathbbR$, $f$ is continuous in the whole real line.
Note that this generalizes to the following statement:
Let $X$ be locally compact and Hausdorff, and $Y$ be compact. If $f:X to Y$ is a function such that $mathrmGr(f)$ is closed, then $f$ is continuous.
answered Jul 30 at 16:50
Aloizio Macedo
22.5k23283
edited Jul 30 at 16:56
answered Jul 30 at 16:50
Aloizio Macedo
22.5k23283
answered Jul 30 at 16:50
Aloizio Macedo
22.5k23283
answered Jul 30 at 16:50
Aloizio Macedo
22.5k23283
22.5k23283
Thank you! I accepted the other answer as I found it a bit easier to follow, but I do like how the result generalises. 🙂
– Andrew
Jul 30 at 18:31
add a comment |Â
Thank you! I accepted the other answer as I found it a bit easier to follow, but I do like how the result generalises. 🙂
– Andrew
Jul 30 at 18:31
Thank you! I accepted the other answer as I found it a bit easier to follow, but I do like how the result generalises. 🙂
– Andrew
Jul 30 at 18:31
Thank you! I accepted the other answer as I found it a bit easier to follow, but I do like how the result generalises. 🙂
– Andrew
Jul 30 at 18:31
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866973%2freal-valued-function-f-mathbbr-to-overline-mathbbr-is-continuous-if%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
The closest related statement (sans compactness) that is actually true (and I can think of) is that a function is Borel if and only if its graph is analytic.
– tomasz
Jul 30 at 15:17
1
Well, unless you assume algebraic regularity or something. I've heard of automatic continuity theorems that could give you something like that (but I don't remember the details). See also various closed graph theorems.
– tomasz
Jul 30 at 15:19