Does factor-wise continuity imply continuity?
$begingroup$
Let $f$ denote a map from a product space $X times Y$ to $Z$. If for every $xin X$, the map $f(x,-)$ is continuous, and the same holds for every $y in Y$, then is $f$ continuous in general? If not, is there any condition to be imposed to make $f$ continuous?
general-topology continuity product-space
edited Dec 3 '18 at 16:56
Martin Sleziak
44.7k9117272
asked Dec 12 '12 at 6:31
Ash GXAsh GX
536318
$endgroup$
add a comment |
$begingroup$
Let $f$ denote a map from a product space $X times Y$ to $Z$. If for every $xin X$, the map $f(x,-)$ is continuous, and the same holds for every $y in Y$, then is $f$ continuous in general? If not, is there any condition to be imposed to make $f$ continuous?
general-topology continuity product-space
edited Dec 3 '18 at 16:56
Martin Sleziak
44.7k9117272
asked Dec 12 '12 at 6:31
Ash GXAsh GX
536318
$endgroup$
3
$begingroup$
A categorical reason why you would not believe this statement holds is that you are asking of a product to satisfy a coproduct property. This is why at first glance it must feel false. The product behaves well when you project it on its components, i.e. $X times Y to X$ ; the coproduct behaves well when you inject its components in, i.e. $X to X times Y$. Notice that the maps $f(x,-)$ are nothing more than composing $pi_Y : Y to X times Y$ defined by $pi_Y(y) = (x,y)$ with $f$, i.e. $f(x,-) = f circ pi_Y$.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:48
$begingroup$
Interesting question. Does the converse hold by chance? (I'm guessing not.)
$endgroup$
– Noldorin
Apr 13 '14 at 2:39
add a comment |
$begingroup$
Let $f$ denote a map from a product space $X times Y$ to $Z$. If for every $xin X$, the map $f(x,-)$ is continuous, and the same holds for every $y in Y$, then is $f$ continuous in general? If not, is there any condition to be imposed to make $f$ continuous?
general-topology continuity product-space
edited Dec 3 '18 at 16:56
Martin Sleziak
44.7k9117272
asked Dec 12 '12 at 6:31
Ash GXAsh GX
536318
$endgroup$
Let $f$ denote a map from a product space $X times Y$ to $Z$. If for every $xin X$, the map $f(x,-)$ is continuous, and the same holds for every $y in Y$, then is $f$ continuous in general? If not, is there any condition to be imposed to make $f$ continuous?
general-topology continuity product-space
general-topology continuity product-space
edited Dec 3 '18 at 16:56
Martin Sleziak
44.7k9117272
asked Dec 12 '12 at 6:31
Ash GXAsh GX
536318
edited Dec 3 '18 at 16:56
Martin Sleziak
44.7k9117272
asked Dec 12 '12 at 6:31
Ash GXAsh GX
536318
edited Dec 3 '18 at 16:56
Martin Sleziak
44.7k9117272
edited Dec 3 '18 at 16:56
Martin Sleziak
44.7k9117272
edited Dec 3 '18 at 16:56
Martin Sleziak
44.7k9117272
44.7k9117272
asked Dec 12 '12 at 6:31
Ash GXAsh GX
536318
asked Dec 12 '12 at 6:31
Ash GXAsh GX
536318
asked Dec 12 '12 at 6:31
Ash GXAsh GX
536318
536318
3
$begingroup$
A categorical reason why you would not believe this statement holds is that you are asking of a product to satisfy a coproduct property. This is why at first glance it must feel false. The product behaves well when you project it on its components, i.e. $X times Y to X$ ; the coproduct behaves well when you inject its components in, i.e. $X to X times Y$. Notice that the maps $f(x,-)$ are nothing more than composing $pi_Y : Y to X times Y$ defined by $pi_Y(y) = (x,y)$ with $f$, i.e. $f(x,-) = f circ pi_Y$.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:48
$begingroup$
Interesting question. Does the converse hold by chance? (I'm guessing not.)
$endgroup$
– Noldorin
Apr 13 '14 at 2:39
add a comment |
3
$begingroup$
A categorical reason why you would not believe this statement holds is that you are asking of a product to satisfy a coproduct property. This is why at first glance it must feel false. The product behaves well when you project it on its components, i.e. $X times Y to X$ ; the coproduct behaves well when you inject its components in, i.e. $X to X times Y$. Notice that the maps $f(x,-)$ are nothing more than composing $pi_Y : Y to X times Y$ defined by $pi_Y(y) = (x,y)$ with $f$, i.e. $f(x,-) = f circ pi_Y$.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:48
$begingroup$
Interesting question. Does the converse hold by chance? (I'm guessing not.)
$endgroup$
– Noldorin
Apr 13 '14 at 2:39
3
3
$begingroup$
A categorical reason why you would not believe this statement holds is that you are asking of a product to satisfy a coproduct property. This is why at first glance it must feel false. The product behaves well when you project it on its components, i.e. $X times Y to X$ ; the coproduct behaves well when you inject its components in, i.e. $X to X times Y$. Notice that the maps $f(x,-)$ are nothing more than composing $pi_Y : Y to X times Y$ defined by $pi_Y(y) = (x,y)$ with $f$, i.e. $f(x,-) = f circ pi_Y$.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:48
$begingroup$
A categorical reason why you would not believe this statement holds is that you are asking of a product to satisfy a coproduct property. This is why at first glance it must feel false. The product behaves well when you project it on its components, i.e. $X times Y to X$ ; the coproduct behaves well when you inject its components in, i.e. $X to X times Y$. Notice that the maps $f(x,-)$ are nothing more than composing $pi_Y : Y to X times Y$ defined by $pi_Y(y) = (x,y)$ with $f$, i.e. $f(x,-) = f circ pi_Y$.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:48
$begingroup$
Interesting question. Does the converse hold by chance? (I'm guessing not.)
$endgroup$
– Noldorin
Apr 13 '14 at 2:39
$begingroup$
Interesting question. Does the converse hold by chance? (I'm guessing not.)
$endgroup$
– Noldorin
Apr 13 '14 at 2:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, separate continuity does not in general imply joint continuity. The function
$$f:Bbb R^2toBbb R:langle x,yranglemapstobegin{cases}frac{xy}{x^2+y^2}&,text{if }langle x,yranglenelangle0,0rangle\\0,&text{if }langle x,yrangle=langle 0,0rangleend{cases}$$
is continuous everywhere except at the origin. At the origin it is continuous in each variable separately, but it is not continuous.
Added: Useful keywords on which to search are separate continuity and joint continuity. Z. Piotrowski, a former colleague of mine, did quite a lot of work in this area; you’ll find many of his papers here, and in them both results and further references. Be warned, though, that the links don’t always match the text: the link to the comprehensive (if now dated) survey Separate and joint continuity is actually at number $17$, not at number $19$. As I recall, there are more results giving conditions under which the set of points of continuity contains a dense $G_delta$-set in $Xtimes Y$ than there are giving conditions that guarantee that a separately continuous function is jointly continuous.
Added 8 March 2015: Number $51$ is an updated survey.
answered Dec 12 '12 at 6:40
Brian M. ScottBrian M. Scott
456k38507908
$endgroup$
$begingroup$
Curiosity : Why did you use $langle x,y rangle$ instead of $(x,y)$ for the coordinates? This doesn't feel natural, I've never seen that before.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:41
$begingroup$
Well, in set theory, bivariate functions are actually functions from ordered pairs to the result set, i.e. $f(x,y) := f(langle x,yrangle)$.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:44
1
$begingroup$
@Mario Carneiro : You are suggesting that $langle x,y rangle$ is a set-theoretic notation for the ordered pair? I know ordered pairs are defined something like $langle x,y rangle = {x, {x,y } }$ so I understand the need for a notation but just let me know if I am wrong.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:50
3
$begingroup$
@Patrick: Yes, the notation comes from set theory. I habitually use it, partly because a lot of my background is in set theory, and partly simply because I prefer it: ordinary parentheses have enough work to do!
$endgroup$
– Brian M. Scott
Dec 12 '12 at 6:51
1
$begingroup$
I'm pretty sure $langle x,yrangle$ is the standard notation for ordered pairs in general. The definition of the ordered pair in set theory is $langle x,yrangle := {{x},{x,y}}$, which is what I think you are thinking of.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:52
|
show 5 more comments
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$begingroup$
No, separate continuity does not in general imply joint continuity. The function
$$f:Bbb R^2toBbb R:langle x,yranglemapstobegin{cases}frac{xy}{x^2+y^2}&,text{if }langle x,yranglenelangle0,0rangle\\0,&text{if }langle x,yrangle=langle 0,0rangleend{cases}$$
is continuous everywhere except at the origin. At the origin it is continuous in each variable separately, but it is not continuous.
Added: Useful keywords on which to search are separate continuity and joint continuity. Z. Piotrowski, a former colleague of mine, did quite a lot of work in this area; you’ll find many of his papers here, and in them both results and further references. Be warned, though, that the links don’t always match the text: the link to the comprehensive (if now dated) survey Separate and joint continuity is actually at number $17$, not at number $19$. As I recall, there are more results giving conditions under which the set of points of continuity contains a dense $G_delta$-set in $Xtimes Y$ than there are giving conditions that guarantee that a separately continuous function is jointly continuous.
Added 8 March 2015: Number $51$ is an updated survey.
answered Dec 12 '12 at 6:40
Brian M. ScottBrian M. Scott
456k38507908
$endgroup$
$begingroup$
Curiosity : Why did you use $langle x,y rangle$ instead of $(x,y)$ for the coordinates? This doesn't feel natural, I've never seen that before.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:41
$begingroup$
Well, in set theory, bivariate functions are actually functions from ordered pairs to the result set, i.e. $f(x,y) := f(langle x,yrangle)$.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:44
1
$begingroup$
@Mario Carneiro : You are suggesting that $langle x,y rangle$ is a set-theoretic notation for the ordered pair? I know ordered pairs are defined something like $langle x,y rangle = {x, {x,y } }$ so I understand the need for a notation but just let me know if I am wrong.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:50
3
$begingroup$
@Patrick: Yes, the notation comes from set theory. I habitually use it, partly because a lot of my background is in set theory, and partly simply because I prefer it: ordinary parentheses have enough work to do!
$endgroup$
– Brian M. Scott
Dec 12 '12 at 6:51
1
$begingroup$
I'm pretty sure $langle x,yrangle$ is the standard notation for ordered pairs in general. The definition of the ordered pair in set theory is $langle x,yrangle := {{x},{x,y}}$, which is what I think you are thinking of.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:52
|
show 5 more comments
$begingroup$
No, separate continuity does not in general imply joint continuity. The function
$$f:Bbb R^2toBbb R:langle x,yranglemapstobegin{cases}frac{xy}{x^2+y^2}&,text{if }langle x,yranglenelangle0,0rangle\\0,&text{if }langle x,yrangle=langle 0,0rangleend{cases}$$
is continuous everywhere except at the origin. At the origin it is continuous in each variable separately, but it is not continuous.
Added: Useful keywords on which to search are separate continuity and joint continuity. Z. Piotrowski, a former colleague of mine, did quite a lot of work in this area; you’ll find many of his papers here, and in them both results and further references. Be warned, though, that the links don’t always match the text: the link to the comprehensive (if now dated) survey Separate and joint continuity is actually at number $17$, not at number $19$. As I recall, there are more results giving conditions under which the set of points of continuity contains a dense $G_delta$-set in $Xtimes Y$ than there are giving conditions that guarantee that a separately continuous function is jointly continuous.
Added 8 March 2015: Number $51$ is an updated survey.
answered Dec 12 '12 at 6:40
Brian M. ScottBrian M. Scott
456k38507908
$endgroup$
$begingroup$
Curiosity : Why did you use $langle x,y rangle$ instead of $(x,y)$ for the coordinates? This doesn't feel natural, I've never seen that before.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:41
$begingroup$
Well, in set theory, bivariate functions are actually functions from ordered pairs to the result set, i.e. $f(x,y) := f(langle x,yrangle)$.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:44
1
$begingroup$
@Mario Carneiro : You are suggesting that $langle x,y rangle$ is a set-theoretic notation for the ordered pair? I know ordered pairs are defined something like $langle x,y rangle = {x, {x,y } }$ so I understand the need for a notation but just let me know if I am wrong.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:50
3
$begingroup$
@Patrick: Yes, the notation comes from set theory. I habitually use it, partly because a lot of my background is in set theory, and partly simply because I prefer it: ordinary parentheses have enough work to do!
$endgroup$
– Brian M. Scott
Dec 12 '12 at 6:51
1
$begingroup$
I'm pretty sure $langle x,yrangle$ is the standard notation for ordered pairs in general. The definition of the ordered pair in set theory is $langle x,yrangle := {{x},{x,y}}$, which is what I think you are thinking of.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:52
|
show 5 more comments
$begingroup$
No, separate continuity does not in general imply joint continuity. The function
$$f:Bbb R^2toBbb R:langle x,yranglemapstobegin{cases}frac{xy}{x^2+y^2}&,text{if }langle x,yranglenelangle0,0rangle\\0,&text{if }langle x,yrangle=langle 0,0rangleend{cases}$$
is continuous everywhere except at the origin. At the origin it is continuous in each variable separately, but it is not continuous.
Added: Useful keywords on which to search are separate continuity and joint continuity. Z. Piotrowski, a former colleague of mine, did quite a lot of work in this area; you’ll find many of his papers here, and in them both results and further references. Be warned, though, that the links don’t always match the text: the link to the comprehensive (if now dated) survey Separate and joint continuity is actually at number $17$, not at number $19$. As I recall, there are more results giving conditions under which the set of points of continuity contains a dense $G_delta$-set in $Xtimes Y$ than there are giving conditions that guarantee that a separately continuous function is jointly continuous.
Added 8 March 2015: Number $51$ is an updated survey.
answered Dec 12 '12 at 6:40
Brian M. ScottBrian M. Scott
456k38507908
$endgroup$
No, separate continuity does not in general imply joint continuity. The function
$$f:Bbb R^2toBbb R:langle x,yranglemapstobegin{cases}frac{xy}{x^2+y^2}&,text{if }langle x,yranglenelangle0,0rangle\\0,&text{if }langle x,yrangle=langle 0,0rangleend{cases}$$
is continuous everywhere except at the origin. At the origin it is continuous in each variable separately, but it is not continuous.
Added: Useful keywords on which to search are separate continuity and joint continuity. Z. Piotrowski, a former colleague of mine, did quite a lot of work in this area; you’ll find many of his papers here, and in them both results and further references. Be warned, though, that the links don’t always match the text: the link to the comprehensive (if now dated) survey Separate and joint continuity is actually at number $17$, not at number $19$. As I recall, there are more results giving conditions under which the set of points of continuity contains a dense $G_delta$-set in $Xtimes Y$ than there are giving conditions that guarantee that a separately continuous function is jointly continuous.
Added 8 March 2015: Number $51$ is an updated survey.
answered Dec 12 '12 at 6:40
Brian M. ScottBrian M. Scott
456k38507908
edited Mar 9 '15 at 0:48
answered Dec 12 '12 at 6:40
Brian M. ScottBrian M. Scott
456k38507908
answered Dec 12 '12 at 6:40
Brian M. ScottBrian M. Scott
456k38507908
answered Dec 12 '12 at 6:40
Brian M. ScottBrian M. Scott
456k38507908
456k38507908
$begingroup$
Curiosity : Why did you use $langle x,y rangle$ instead of $(x,y)$ for the coordinates? This doesn't feel natural, I've never seen that before.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:41
$begingroup$
Well, in set theory, bivariate functions are actually functions from ordered pairs to the result set, i.e. $f(x,y) := f(langle x,yrangle)$.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:44
1
$begingroup$
@Mario Carneiro : You are suggesting that $langle x,y rangle$ is a set-theoretic notation for the ordered pair? I know ordered pairs are defined something like $langle x,y rangle = {x, {x,y } }$ so I understand the need for a notation but just let me know if I am wrong.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:50
3
$begingroup$
@Patrick: Yes, the notation comes from set theory. I habitually use it, partly because a lot of my background is in set theory, and partly simply because I prefer it: ordinary parentheses have enough work to do!
$endgroup$
– Brian M. Scott
Dec 12 '12 at 6:51
1
$begingroup$
I'm pretty sure $langle x,yrangle$ is the standard notation for ordered pairs in general. The definition of the ordered pair in set theory is $langle x,yrangle := {{x},{x,y}}$, which is what I think you are thinking of.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:52
|
show 5 more comments
$begingroup$
Curiosity : Why did you use $langle x,y rangle$ instead of $(x,y)$ for the coordinates? This doesn't feel natural, I've never seen that before.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:41
$begingroup$
Well, in set theory, bivariate functions are actually functions from ordered pairs to the result set, i.e. $f(x,y) := f(langle x,yrangle)$.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:44
1
$begingroup$
@Mario Carneiro : You are suggesting that $langle x,y rangle$ is a set-theoretic notation for the ordered pair? I know ordered pairs are defined something like $langle x,y rangle = {x, {x,y } }$ so I understand the need for a notation but just let me know if I am wrong.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:50
3
$begingroup$
@Patrick: Yes, the notation comes from set theory. I habitually use it, partly because a lot of my background is in set theory, and partly simply because I prefer it: ordinary parentheses have enough work to do!
$endgroup$
– Brian M. Scott
Dec 12 '12 at 6:51
1
$begingroup$
I'm pretty sure $langle x,yrangle$ is the standard notation for ordered pairs in general. The definition of the ordered pair in set theory is $langle x,yrangle := {{x},{x,y}}$, which is what I think you are thinking of.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:52
$begingroup$
Curiosity : Why did you use $langle x,y rangle$ instead of $(x,y)$ for the coordinates? This doesn't feel natural, I've never seen that before.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:41
$begingroup$
Curiosity : Why did you use $langle x,y rangle$ instead of $(x,y)$ for the coordinates? This doesn't feel natural, I've never seen that before.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:41
$begingroup$
Well, in set theory, bivariate functions are actually functions from ordered pairs to the result set, i.e. $f(x,y) := f(langle x,yrangle)$.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:44
$begingroup$
Well, in set theory, bivariate functions are actually functions from ordered pairs to the result set, i.e. $f(x,y) := f(langle x,yrangle)$.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:44
1
1
$begingroup$
@Mario Carneiro : You are suggesting that $langle x,y rangle$ is a set-theoretic notation for the ordered pair? I know ordered pairs are defined something like $langle x,y rangle = {x, {x,y } }$ so I understand the need for a notation but just let me know if I am wrong.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:50
$begingroup$
@Mario Carneiro : You are suggesting that $langle x,y rangle$ is a set-theoretic notation for the ordered pair? I know ordered pairs are defined something like $langle x,y rangle = {x, {x,y } }$ so I understand the need for a notation but just let me know if I am wrong.
$endgroup$
– Patrick Da Silva
Dec 12 '12 at 6:50
3
3
$begingroup$
@Patrick: Yes, the notation comes from set theory. I habitually use it, partly because a lot of my background is in set theory, and partly simply because I prefer it: ordinary parentheses have enough work to do!
$endgroup$
– Brian M. Scott
Dec 12 '12 at 6:51
$begingroup$
@Patrick: Yes, the notation comes from set theory. I habitually use it, partly because a lot of my background is in set theory, and partly simply because I prefer it: ordinary parentheses have enough work to do!
$endgroup$
– Brian M. Scott
Dec 12 '12 at 6:51
1
1
$begingroup$
I'm pretty sure $langle x,yrangle$ is the standard notation for ordered pairs in general. The definition of the ordered pair in set theory is $langle x,yrangle := {{x},{x,y}}$, which is what I think you are thinking of.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:52
$begingroup$
I'm pretty sure $langle x,yrangle$ is the standard notation for ordered pairs in general. The definition of the ordered pair in set theory is $langle x,yrangle := {{x},{x,y}}$, which is what I think you are thinking of.
$endgroup$
– Mario Carneiro
Dec 12 '12 at 6:52
|
show 5 more comments
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3
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A categorical reason why you would not believe this statement holds is that you are asking of a product to satisfy a coproduct property. This is why at first glance it must feel false. The product behaves well when you project it on its components, i.e. $X times Y to X$ ; the coproduct behaves well when you inject its components in, i.e. $X to X times Y$. Notice that the maps $f(x,-)$ are nothing more than composing $pi_Y : Y to X times Y$ defined by $pi_Y(y) = (x,y)$ with $f$, i.e. $f(x,-) = f circ pi_Y$.
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– Patrick Da Silva
Dec 12 '12 at 6:48
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Interesting question. Does the converse hold by chance? (I'm guessing not.)
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– Noldorin
Apr 13 '14 at 2:39