Can a cube of discontinuous function be continuous?
Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.
continuity
add a comment |
Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.
continuity
3
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
2 hours ago
What is your domain? It matters really quite a lot.
– user3482749
2 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
2 hours ago
add a comment |
Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.
continuity
Can a cube (meaning $g(x) = f(x)^3 = f(x) cdot f(x) cdot f(x)$) of discontinuous function $f: D to mathbb{R}$ ($D$ is subset of $mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.
continuity
continuity
edited 2 hours ago
J. Abraham
asked 2 hours ago
J. AbrahamJ. Abraham
463314
463314
3
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
2 hours ago
What is your domain? It matters really quite a lot.
– user3482749
2 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
2 hours ago
add a comment |
3
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
2 hours ago
What is your domain? It matters really quite a lot.
– user3482749
2 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
2 hours ago
3
3
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
2 hours ago
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
2 hours ago
What is your domain? It matters really quite a lot.
– user3482749
2 hours ago
What is your domain? It matters really quite a lot.
– user3482749
2 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
2 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3065701%2fcan-a-cube-of-discontinuous-function-be-continuous%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
add a comment |
Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
add a comment |
Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
Since $phi : mathbb{R} to mathbb{R}, phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $phi circ f$ is continuous.
edited 2 hours ago
answered 2 hours ago
Paul FrostPaul Frost
9,5002631
9,5002631
add a comment |
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3065701%2fcan-a-cube-of-discontinuous-function-be-continuous%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
3
The point is not that $x longmapsto x^3$ is injective, as much as $x longmapsto x^{1/3}$ is continuous.
– Mindlack
2 hours ago
What is your domain? It matters really quite a lot.
– user3482749
2 hours ago
I think the injectivity is very much to the point. $f(x) =sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)cdot g(x)$ continuous.
– MJD
2 hours ago