Proof for continuity of $h(x) = begin{cases} x, & xinmathbb{Q} \ -x, & xnotinmathbb{Q} end{cases}$












4












$begingroup$


I have the given function and I have to find out, where it is continuous and discontinuous...



My guess is, that the function is discontinuous on all $mathbb{R}setminus{0}$ and continuous in $0$. Could you look over my proof and give me a feedback?



Proof:



For $xinmathbb{Q}$ look at the sequence $(q_n)$, which converges to $x$. The sequence $(h(q_n))$ is then equal to the sequence $(q_n)$ and therefore converges to $x$, so we can conclude: $lim_{n->infty}h(q_n))=h(lim_{n->infty}q_n)=h(x)=x$



For $xnotinmathbb{Q}$ look at the sequence $(r_n)$, which converges to $x$. The sequence $(h(r_n))$ is then equal to the sequence $-(r_n)$ and therefore converges to $-x$, so we can conclude: $lim_{n->infty}h(r_n))=xneq-x=h(lim_{n->infty}-r_n)$



Therefore $h$ can't be continuous (I don't know the exact theorem name, but the subsequences of a function has to have the limit $h(x_0)$, if $h(x)$ is continuous in $x_0$



So the function is only continuous in $x=0$. Proof:



Let $epsilon>0$ and $delta =epsilon$, then we have:
$forall epsilon>0 exists delta > 0 forall xin mathbb{R} : |x-x_0|=|x-0|<delta =epsilon Rightarrow |h(x)-h(x_0)|=|h(x)-x_0|=|h(x)|=|x|<epsilon$



We can conclude, that the function is continuous in 0 and discontinuous for every $mathbb{R} setminus 0$










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:09












  • $begingroup$
    By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
    $endgroup$
    – Bekay
    Dec 8 '18 at 19:21












  • $begingroup$
    I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:28












  • $begingroup$
    Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
    $endgroup$
    – Bekay
    Dec 8 '18 at 19:38










  • $begingroup$
    It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:43


















4












$begingroup$


I have the given function and I have to find out, where it is continuous and discontinuous...



My guess is, that the function is discontinuous on all $mathbb{R}setminus{0}$ and continuous in $0$. Could you look over my proof and give me a feedback?



Proof:



For $xinmathbb{Q}$ look at the sequence $(q_n)$, which converges to $x$. The sequence $(h(q_n))$ is then equal to the sequence $(q_n)$ and therefore converges to $x$, so we can conclude: $lim_{n->infty}h(q_n))=h(lim_{n->infty}q_n)=h(x)=x$



For $xnotinmathbb{Q}$ look at the sequence $(r_n)$, which converges to $x$. The sequence $(h(r_n))$ is then equal to the sequence $-(r_n)$ and therefore converges to $-x$, so we can conclude: $lim_{n->infty}h(r_n))=xneq-x=h(lim_{n->infty}-r_n)$



Therefore $h$ can't be continuous (I don't know the exact theorem name, but the subsequences of a function has to have the limit $h(x_0)$, if $h(x)$ is continuous in $x_0$



So the function is only continuous in $x=0$. Proof:



Let $epsilon>0$ and $delta =epsilon$, then we have:
$forall epsilon>0 exists delta > 0 forall xin mathbb{R} : |x-x_0|=|x-0|<delta =epsilon Rightarrow |h(x)-h(x_0)|=|h(x)-x_0|=|h(x)|=|x|<epsilon$



We can conclude, that the function is continuous in 0 and discontinuous for every $mathbb{R} setminus 0$










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:09












  • $begingroup$
    By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
    $endgroup$
    – Bekay
    Dec 8 '18 at 19:21












  • $begingroup$
    I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:28












  • $begingroup$
    Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
    $endgroup$
    – Bekay
    Dec 8 '18 at 19:38










  • $begingroup$
    It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:43
















4












4








4


1



$begingroup$


I have the given function and I have to find out, where it is continuous and discontinuous...



My guess is, that the function is discontinuous on all $mathbb{R}setminus{0}$ and continuous in $0$. Could you look over my proof and give me a feedback?



Proof:



For $xinmathbb{Q}$ look at the sequence $(q_n)$, which converges to $x$. The sequence $(h(q_n))$ is then equal to the sequence $(q_n)$ and therefore converges to $x$, so we can conclude: $lim_{n->infty}h(q_n))=h(lim_{n->infty}q_n)=h(x)=x$



For $xnotinmathbb{Q}$ look at the sequence $(r_n)$, which converges to $x$. The sequence $(h(r_n))$ is then equal to the sequence $-(r_n)$ and therefore converges to $-x$, so we can conclude: $lim_{n->infty}h(r_n))=xneq-x=h(lim_{n->infty}-r_n)$



Therefore $h$ can't be continuous (I don't know the exact theorem name, but the subsequences of a function has to have the limit $h(x_0)$, if $h(x)$ is continuous in $x_0$



So the function is only continuous in $x=0$. Proof:



Let $epsilon>0$ and $delta =epsilon$, then we have:
$forall epsilon>0 exists delta > 0 forall xin mathbb{R} : |x-x_0|=|x-0|<delta =epsilon Rightarrow |h(x)-h(x_0)|=|h(x)-x_0|=|h(x)|=|x|<epsilon$



We can conclude, that the function is continuous in 0 and discontinuous for every $mathbb{R} setminus 0$










share|cite|improve this question











$endgroup$




I have the given function and I have to find out, where it is continuous and discontinuous...



My guess is, that the function is discontinuous on all $mathbb{R}setminus{0}$ and continuous in $0$. Could you look over my proof and give me a feedback?



Proof:



For $xinmathbb{Q}$ look at the sequence $(q_n)$, which converges to $x$. The sequence $(h(q_n))$ is then equal to the sequence $(q_n)$ and therefore converges to $x$, so we can conclude: $lim_{n->infty}h(q_n))=h(lim_{n->infty}q_n)=h(x)=x$



For $xnotinmathbb{Q}$ look at the sequence $(r_n)$, which converges to $x$. The sequence $(h(r_n))$ is then equal to the sequence $-(r_n)$ and therefore converges to $-x$, so we can conclude: $lim_{n->infty}h(r_n))=xneq-x=h(lim_{n->infty}-r_n)$



Therefore $h$ can't be continuous (I don't know the exact theorem name, but the subsequences of a function has to have the limit $h(x_0)$, if $h(x)$ is continuous in $x_0$



So the function is only continuous in $x=0$. Proof:



Let $epsilon>0$ and $delta =epsilon$, then we have:
$forall epsilon>0 exists delta > 0 forall xin mathbb{R} : |x-x_0|=|x-0|<delta =epsilon Rightarrow |h(x)-h(x_0)|=|h(x)-x_0|=|h(x)|=|x|<epsilon$



We can conclude, that the function is continuous in 0 and discontinuous for every $mathbb{R} setminus 0$







real-analysis calculus continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 19:37







Bekay

















asked Dec 8 '18 at 18:59









BekayBekay

212




212








  • 1




    $begingroup$
    Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:09












  • $begingroup$
    By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
    $endgroup$
    – Bekay
    Dec 8 '18 at 19:21












  • $begingroup$
    I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:28












  • $begingroup$
    Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
    $endgroup$
    – Bekay
    Dec 8 '18 at 19:38










  • $begingroup$
    It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:43
















  • 1




    $begingroup$
    Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:09












  • $begingroup$
    By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
    $endgroup$
    – Bekay
    Dec 8 '18 at 19:21












  • $begingroup$
    I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:28












  • $begingroup$
    Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
    $endgroup$
    – Bekay
    Dec 8 '18 at 19:38










  • $begingroup$
    It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
    $endgroup$
    – Tki Deneb
    Dec 8 '18 at 19:43










1




1




$begingroup$
Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:09






$begingroup$
Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:09














$begingroup$
By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
$endgroup$
– Bekay
Dec 8 '18 at 19:21






$begingroup$
By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
$endgroup$
– Bekay
Dec 8 '18 at 19:21














$begingroup$
I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:28






$begingroup$
I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:28














$begingroup$
Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
$endgroup$
– Bekay
Dec 8 '18 at 19:38




$begingroup$
Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
$endgroup$
– Bekay
Dec 8 '18 at 19:38












$begingroup$
It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:43






$begingroup$
It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:43












0






active

oldest

votes











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031482%2fproof-for-continuity-of-hx-begincases-x-x-in-mathbbq-x-x-not%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
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031482%2fproof-for-continuity-of-hx-begincases-x-x-in-mathbbq-x-x-not%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Aardman Animations

Are they similar matrix

“minimization” problem in Euclidean space related to orthonormal basis