Is a function $X times Y to Z$ continuous iff precomposition with continuous functions $Xto Xtimes Y$ and $Yto Xtimes Y$ are continuous Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why are such functions not always necessarily uniformly continuous?image of continuous function is compact and connected.If two points in a space are connected by a path, is there an injective map of an interval?If M is connected and f is continuous, prove that the graph of f is connectedRelation between continuous maps and convergence of sequencesIs a pathwise-continuous function continuous?Continuous on paths implies continuous if space is locally path-connected?Uniformly continuous functions and the preservation of Cauchy sequencesSequence converges iff function is continuous on extension of natural numbers (topology)About continuous functions on $p$-adic fields
Is the Standard Deduction better than Itemized when both are the same amount?
Fundamental Solution of the Pell Equation
Delete nth line from bottom
Compare a given version number in the form major.minor.build.patch and see if one is less than the other
First console to have temporary backward compatibility
Is there a kind of relay only consumes power when switching?
If my PI received research grants from a company to be able to pay my postdoc salary, did I have a potential conflict interest too?
How could we fake a moon landing now?
Is there any way for the UK Prime Minister to make a motion directly dependent on Government confidence?
Can a party unilaterally change candidates in preparation for a General election?
Significance of Cersei's obsession with elephants?
What are the out-of-universe reasons for the references to Toby Maguire-era Spider-Man in ITSV
Is grep documentation wrong?
Has negative voting ever been officially implemented in elections, or seriously proposed, or even studied?
What does "lightly crushed" mean for cardamon pods?
How does the math work when buying airline miles?
Can you use the Shield Master feat to shove someone before you make an attack by using a Readied action?
How come Sam didn't become Lord of Horn Hill?
Do square wave exist?
Is it a good idea to use CNN to classify 1D signal?
Integration Help
How to compare two different files line by line in unix?
Closed form of recurrent arithmetic series summation
Is it fair for a professor to grade us on the possession of past papers?
Is a function $X times Y to Z$ continuous iff precomposition with continuous functions $Xto Xtimes Y$ and $Yto Xtimes Y$ are continuous
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why are such functions not always necessarily uniformly continuous?image of continuous function is compact and connected.If two points in a space are connected by a path, is there an injective map of an interval?If M is connected and f is continuous, prove that the graph of f is connectedRelation between continuous maps and convergence of sequencesIs a pathwise-continuous function continuous?Continuous on paths implies continuous if space is locally path-connected?Uniformly continuous functions and the preservation of Cauchy sequencesSequence converges iff function is continuous on extension of natural numbers (topology)About continuous functions on $p$-adic fields
$begingroup$
Is a function from a product space $X times Y to Z$ continuous if and only if precomposition with all continuous functions $Xto Xtimes Y$ and $Yto Xtimes Y$ are continuous? Clearly one direction holds for all spaces and I think I've shown that the other direction is true for functions $f:mathbbR^n to mathbbR^m$ but I don't know if this is true generally or just true for nice enough spaces. Any counterexamples or special cases where this is true would be greatly appreciated.
Update: I have shown this is true if your base spaces are path connected or locally path connected metric spaces using the Tietze extension theorem but I still don't know if this is true generally.
Update: In retrospect, this argument only really works for spaces where there are short enough paths between close points, such as a length space.
Suppose $X$ and $Y$ are path connected metric spaces. In metric spaces continuity is the same as sequential continuity. Take a convergent sequence $x_n$ and its limit $x_infty$ in $X$ (or $Y$) and define a continuous map from $A=x_ncup x_infty$ to $mathbbR$ which sends $x_n$ to $1/n$ and $x_infty$ to 0. $A$ is a closed subset of $X$ so by the Tietze extension theorem it extends to a map $Xto mathbbR$. Then given a convergent sequence $z_n$ which converges to $z_infty$ in $Xtimes Y$ we can construct a map from $ frac1n cup 0$ to our sequence which sends $frac1n$ to $z_n$ and extend it to a map from $[0, 1]$ to $Xtimes Y$ by connecting the paths between $z_n$ and $z_n+1$. We then extend that function to the entirety of $mathbbR$, which when composed with our function $X to mathbbR$ gives us a continuous function which takes a convergent sequence in $X$ to a convergent sequence in $Xtimes Y$. If the image of the convergent sequence in $X$ under this map is convergent in $Z$, then the image of the convergent sequence in $Xtimes Y$ will also be convergent in $Z$.
general-topology
asked Mar 27 at 3:31
LiquidLiquid
364
$endgroup$
|
show 4 more comments
$begingroup$
Is a function from a product space $X times Y to Z$ continuous if and only if precomposition with all continuous functions $Xto Xtimes Y$ and $Yto Xtimes Y$ are continuous? Clearly one direction holds for all spaces and I think I've shown that the other direction is true for functions $f:mathbbR^n to mathbbR^m$ but I don't know if this is true generally or just true for nice enough spaces. Any counterexamples or special cases where this is true would be greatly appreciated.
Update: I have shown this is true if your base spaces are path connected or locally path connected metric spaces using the Tietze extension theorem but I still don't know if this is true generally.
Update: In retrospect, this argument only really works for spaces where there are short enough paths between close points, such as a length space.
Suppose $X$ and $Y$ are path connected metric spaces. In metric spaces continuity is the same as sequential continuity. Take a convergent sequence $x_n$ and its limit $x_infty$ in $X$ (or $Y$) and define a continuous map from $A=x_ncup x_infty$ to $mathbbR$ which sends $x_n$ to $1/n$ and $x_infty$ to 0. $A$ is a closed subset of $X$ so by the Tietze extension theorem it extends to a map $Xto mathbbR$. Then given a convergent sequence $z_n$ which converges to $z_infty$ in $Xtimes Y$ we can construct a map from $ frac1n cup 0$ to our sequence which sends $frac1n$ to $z_n$ and extend it to a map from $[0, 1]$ to $Xtimes Y$ by connecting the paths between $z_n$ and $z_n+1$. We then extend that function to the entirety of $mathbbR$, which when composed with our function $X to mathbbR$ gives us a continuous function which takes a convergent sequence in $X$ to a convergent sequence in $Xtimes Y$. If the image of the convergent sequence in $X$ under this map is convergent in $Z$, then the image of the convergent sequence in $Xtimes Y$ will also be convergent in $Z$.
general-topology
asked Mar 27 at 3:31
LiquidLiquid
364
$endgroup$
$begingroup$
it is not clear what exactly you are pre-composing? Is it the inclusion map? OR do you want the statement to hold for all continuous functions $Xto Xtimes Y$, as mentioned in your post? Later seems very strong requirement!
$endgroup$
– Mike V.D.C.
Mar 27 at 4:02
$begingroup$
I am asking about if precomposition with every continuous function from your base spaces are continuous, not just an inclusion map.
$endgroup$
– Liquid
Mar 27 at 4:16
$begingroup$
You should add your proof for the path connected case.
$endgroup$
– Paul Frost
Mar 27 at 16:52
$begingroup$
I added it but while I was writing it up I realized it only really works if the paths in $Xtimes Y$ can be sufficiently short.
$endgroup$
– Liquid
Mar 27 at 18:17
$begingroup$
@Liquid Your function constructed by "connecting paths between $z_n$ and $z_n+1$" need not be continuous at $0$. I assume that you chunk $[0,1]$ into $[1/(n+1), 1/n]$ intervals and take path at each one? Assume that the glueing is continuous. Let $lambda_n:[1/(n+1), 1/n]to X$ be such paths. Note that any sequence $x_n$ such that $x_nin[1/(n+1), 1/n]$ converges to $0$. And thus the sequence $lambda_n(x_n)$ has to converge to $z_infty$ by continuity. With that property you can easily constuct a counterexample: paths with mid-points converging somewhere else.
$endgroup$
– freakish
Mar 27 at 22:01
|
show 4 more comments
$begingroup$
Is a function from a product space $X times Y to Z$ continuous if and only if precomposition with all continuous functions $Xto Xtimes Y$ and $Yto Xtimes Y$ are continuous? Clearly one direction holds for all spaces and I think I've shown that the other direction is true for functions $f:mathbbR^n to mathbbR^m$ but I don't know if this is true generally or just true for nice enough spaces. Any counterexamples or special cases where this is true would be greatly appreciated.
Update: I have shown this is true if your base spaces are path connected or locally path connected metric spaces using the Tietze extension theorem but I still don't know if this is true generally.
Update: In retrospect, this argument only really works for spaces where there are short enough paths between close points, such as a length space.
Suppose $X$ and $Y$ are path connected metric spaces. In metric spaces continuity is the same as sequential continuity. Take a convergent sequence $x_n$ and its limit $x_infty$ in $X$ (or $Y$) and define a continuous map from $A=x_ncup x_infty$ to $mathbbR$ which sends $x_n$ to $1/n$ and $x_infty$ to 0. $A$ is a closed subset of $X$ so by the Tietze extension theorem it extends to a map $Xto mathbbR$. Then given a convergent sequence $z_n$ which converges to $z_infty$ in $Xtimes Y$ we can construct a map from $ frac1n cup 0$ to our sequence which sends $frac1n$ to $z_n$ and extend it to a map from $[0, 1]$ to $Xtimes Y$ by connecting the paths between $z_n$ and $z_n+1$. We then extend that function to the entirety of $mathbbR$, which when composed with our function $X to mathbbR$ gives us a continuous function which takes a convergent sequence in $X$ to a convergent sequence in $Xtimes Y$. If the image of the convergent sequence in $X$ under this map is convergent in $Z$, then the image of the convergent sequence in $Xtimes Y$ will also be convergent in $Z$.
general-topology
asked Mar 27 at 3:31
LiquidLiquid
364
$endgroup$
Is a function from a product space $X times Y to Z$ continuous if and only if precomposition with all continuous functions $Xto Xtimes Y$ and $Yto Xtimes Y$ are continuous? Clearly one direction holds for all spaces and I think I've shown that the other direction is true for functions $f:mathbbR^n to mathbbR^m$ but I don't know if this is true generally or just true for nice enough spaces. Any counterexamples or special cases where this is true would be greatly appreciated.
Update: I have shown this is true if your base spaces are path connected or locally path connected metric spaces using the Tietze extension theorem but I still don't know if this is true generally.
Update: In retrospect, this argument only really works for spaces where there are short enough paths between close points, such as a length space.
Suppose $X$ and $Y$ are path connected metric spaces. In metric spaces continuity is the same as sequential continuity. Take a convergent sequence $x_n$ and its limit $x_infty$ in $X$ (or $Y$) and define a continuous map from $A=x_ncup x_infty$ to $mathbbR$ which sends $x_n$ to $1/n$ and $x_infty$ to 0. $A$ is a closed subset of $X$ so by the Tietze extension theorem it extends to a map $Xto mathbbR$. Then given a convergent sequence $z_n$ which converges to $z_infty$ in $Xtimes Y$ we can construct a map from $ frac1n cup 0$ to our sequence which sends $frac1n$ to $z_n$ and extend it to a map from $[0, 1]$ to $Xtimes Y$ by connecting the paths between $z_n$ and $z_n+1$. We then extend that function to the entirety of $mathbbR$, which when composed with our function $X to mathbbR$ gives us a continuous function which takes a convergent sequence in $X$ to a convergent sequence in $Xtimes Y$. If the image of the convergent sequence in $X$ under this map is convergent in $Z$, then the image of the convergent sequence in $Xtimes Y$ will also be convergent in $Z$.
general-topology
general-topology
asked Mar 27 at 3:31
LiquidLiquid
364
asked Mar 27 at 3:31
LiquidLiquid
364
edited Mar 27 at 18:14
Liquid
asked Mar 27 at 3:31
LiquidLiquid
364
asked Mar 27 at 3:31
LiquidLiquid
364
asked Mar 27 at 3:31
LiquidLiquid
364
364
$begingroup$
it is not clear what exactly you are pre-composing? Is it the inclusion map? OR do you want the statement to hold for all continuous functions $Xto Xtimes Y$, as mentioned in your post? Later seems very strong requirement!
$endgroup$
– Mike V.D.C.
Mar 27 at 4:02
$begingroup$
I am asking about if precomposition with every continuous function from your base spaces are continuous, not just an inclusion map.
$endgroup$
– Liquid
Mar 27 at 4:16
$begingroup$
You should add your proof for the path connected case.
$endgroup$
– Paul Frost
Mar 27 at 16:52
$begingroup$
I added it but while I was writing it up I realized it only really works if the paths in $Xtimes Y$ can be sufficiently short.
$endgroup$
– Liquid
Mar 27 at 18:17
$begingroup$
@Liquid Your function constructed by "connecting paths between $z_n$ and $z_n+1$" need not be continuous at $0$. I assume that you chunk $[0,1]$ into $[1/(n+1), 1/n]$ intervals and take path at each one? Assume that the glueing is continuous. Let $lambda_n:[1/(n+1), 1/n]to X$ be such paths. Note that any sequence $x_n$ such that $x_nin[1/(n+1), 1/n]$ converges to $0$. And thus the sequence $lambda_n(x_n)$ has to converge to $z_infty$ by continuity. With that property you can easily constuct a counterexample: paths with mid-points converging somewhere else.
$endgroup$
– freakish
Mar 27 at 22:01
|
show 4 more comments
$begingroup$
it is not clear what exactly you are pre-composing? Is it the inclusion map? OR do you want the statement to hold for all continuous functions $Xto Xtimes Y$, as mentioned in your post? Later seems very strong requirement!
$endgroup$
– Mike V.D.C.
Mar 27 at 4:02
$begingroup$
I am asking about if precomposition with every continuous function from your base spaces are continuous, not just an inclusion map.
$endgroup$
– Liquid
Mar 27 at 4:16
$begingroup$
You should add your proof for the path connected case.
$endgroup$
– Paul Frost
Mar 27 at 16:52
$begingroup$
I added it but while I was writing it up I realized it only really works if the paths in $Xtimes Y$ can be sufficiently short.
$endgroup$
– Liquid
Mar 27 at 18:17
$begingroup$
@Liquid Your function constructed by "connecting paths between $z_n$ and $z_n+1$" need not be continuous at $0$. I assume that you chunk $[0,1]$ into $[1/(n+1), 1/n]$ intervals and take path at each one? Assume that the glueing is continuous. Let $lambda_n:[1/(n+1), 1/n]to X$ be such paths. Note that any sequence $x_n$ such that $x_nin[1/(n+1), 1/n]$ converges to $0$. And thus the sequence $lambda_n(x_n)$ has to converge to $z_infty$ by continuity. With that property you can easily constuct a counterexample: paths with mid-points converging somewhere else.
$endgroup$
– freakish
Mar 27 at 22:01
$begingroup$
it is not clear what exactly you are pre-composing? Is it the inclusion map? OR do you want the statement to hold for all continuous functions $Xto Xtimes Y$, as mentioned in your post? Later seems very strong requirement!
$endgroup$
– Mike V.D.C.
Mar 27 at 4:02
$begingroup$
it is not clear what exactly you are pre-composing? Is it the inclusion map? OR do you want the statement to hold for all continuous functions $Xto Xtimes Y$, as mentioned in your post? Later seems very strong requirement!
$endgroup$
– Mike V.D.C.
Mar 27 at 4:02
$begingroup$
I am asking about if precomposition with every continuous function from your base spaces are continuous, not just an inclusion map.
$endgroup$
– Liquid
Mar 27 at 4:16
$begingroup$
I am asking about if precomposition with every continuous function from your base spaces are continuous, not just an inclusion map.
$endgroup$
– Liquid
Mar 27 at 4:16
$begingroup$
You should add your proof for the path connected case.
$endgroup$
– Paul Frost
Mar 27 at 16:52
$begingroup$
You should add your proof for the path connected case.
$endgroup$
– Paul Frost
Mar 27 at 16:52
$begingroup$
I added it but while I was writing it up I realized it only really works if the paths in $Xtimes Y$ can be sufficiently short.
$endgroup$
– Liquid
Mar 27 at 18:17
$begingroup$
I added it but while I was writing it up I realized it only really works if the paths in $Xtimes Y$ can be sufficiently short.
$endgroup$
– Liquid
Mar 27 at 18:17
$begingroup$
@Liquid Your function constructed by "connecting paths between $z_n$ and $z_n+1$" need not be continuous at $0$. I assume that you chunk $[0,1]$ into $[1/(n+1), 1/n]$ intervals and take path at each one? Assume that the glueing is continuous. Let $lambda_n:[1/(n+1), 1/n]to X$ be such paths. Note that any sequence $x_n$ such that $x_nin[1/(n+1), 1/n]$ converges to $0$. And thus the sequence $lambda_n(x_n)$ has to converge to $z_infty$ by continuity. With that property you can easily constuct a counterexample: paths with mid-points converging somewhere else.
$endgroup$
– freakish
Mar 27 at 22:01
$begingroup$
@Liquid Your function constructed by "connecting paths between $z_n$ and $z_n+1$" need not be continuous at $0$. I assume that you chunk $[0,1]$ into $[1/(n+1), 1/n]$ intervals and take path at each one? Assume that the glueing is continuous. Let $lambda_n:[1/(n+1), 1/n]to X$ be such paths. Note that any sequence $x_n$ such that $x_nin[1/(n+1), 1/n]$ converges to $0$. And thus the sequence $lambda_n(x_n)$ has to converge to $z_infty$ by continuity. With that property you can easily constuct a counterexample: paths with mid-points converging somewhere else.
$endgroup$
– freakish
Mar 27 at 22:01
|
show 4 more comments
0
active
oldest
votes
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164063%2fis-a-function-x-times-y-to-z-continuous-iff-precomposition-with-continuous-f%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164063%2fis-a-function-x-times-y-to-z-continuous-iff-precomposition-with-continuous-f%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
it is not clear what exactly you are pre-composing? Is it the inclusion map? OR do you want the statement to hold for all continuous functions $Xto Xtimes Y$, as mentioned in your post? Later seems very strong requirement!
$endgroup$
– Mike V.D.C.
Mar 27 at 4:02
$begingroup$
I am asking about if precomposition with every continuous function from your base spaces are continuous, not just an inclusion map.
$endgroup$
– Liquid
Mar 27 at 4:16
$begingroup$
You should add your proof for the path connected case.
$endgroup$
– Paul Frost
Mar 27 at 16:52
$begingroup$
I added it but while I was writing it up I realized it only really works if the paths in $Xtimes Y$ can be sufficiently short.
$endgroup$
– Liquid
Mar 27 at 18:17
$begingroup$
@Liquid Your function constructed by "connecting paths between $z_n$ and $z_n+1$" need not be continuous at $0$. I assume that you chunk $[0,1]$ into $[1/(n+1), 1/n]$ intervals and take path at each one? Assume that the glueing is continuous. Let $lambda_n:[1/(n+1), 1/n]to X$ be such paths. Note that any sequence $x_n$ such that $x_nin[1/(n+1), 1/n]$ converges to $0$. And thus the sequence $lambda_n(x_n)$ has to converge to $z_infty$ by continuity. With that property you can easily constuct a counterexample: paths with mid-points converging somewhere else.
$endgroup$
– freakish
Mar 27 at 22:01