Dgdrxrt

A word that means fill it to the required quantity

Slides for 30 min~1 hr Skype tenure track application interview

What is the most efficient way to store a numeric range?

Did Scotland spend $250,000 for the slogan "Welcome to Scotland"?

What do I do when my TA workload is more than expected?

Can a rogue use sneak attack with weapons that have the thrown property even if they are not thrown?

"as much details as you can remember"

Correct punctuation for showing a character's confusion

Keeping a retro style to sci-fi spaceships?

Is it okay to consider publishing in my first year of PhD?

How can I define good in a religion that claims no moral authority?

Are turbopumps lubricated?

If I can cast sorceries at instant speed, can I use sorcery-speed activated abilities at instant speed?

ODD NUMBER in Cognitive Linguistics of WILLIAM CROFT and D. ALAN CRUSE

What could be the right powersource for 15 seconds lifespan disposable giant chainsaw?

Mathematics of imaging the black hole

How to charge AirPods to keep battery healthy?

Is it possible for absolutely everyone to attain enlightenment?

How do PCB vias affect signal quality?

Deal with toxic manager when you can't quit

How do I free up internal storage if I don't have any apps downloaded?

Accepted by European university, rejected by all American ones I applied to? Possible reasons?

Match Roman Numerals

Is it safe to harvest rainwater that fell on solar panels?

Existence of limit for every adherence point in a bounded set $U$ implies $f(U)$ is bounded

0

$begingroup$

Let $U subseteq R^n$ be a bounded set, $f:U rightarrow mathbbR$ and
suppose that for every $x_0 in overline U$, $displaystylelim_x rightarrow x_0f(x)$ exist.
Prove $f(U)$ is bounded.

By the hypothesis, if $x_0 in overline U$ , then for every $epsilon >0$ there exists $delta>0$ such that if $left | x-x_0 right |<delta$ then $left | f(x)-L right |<epsilon$. I don't really see how I can obtain a constant $M>0$ such that $left | w-z right |<M$ for every $w,z in f(U)$. If $f$ was uniformly continuous I can see why $f(U)$ would be bounded but with the limit hypothesis only I'm stuck. Any guideline is welcome.

real-analysis multivariable-calculus

share|cite|improve this question

edited Mar 30 at 21:45

Aloizio Macedo♦

23.8k24088

asked Mar 30 at 21:36

ipreferpiipreferpi

388

$endgroup$

add a comment | 

0

$begingroup$

Let $U subseteq R^n$ be a bounded set, $f:U rightarrow mathbbR$ and
suppose that for every $x_0 in overline U$, $displaystylelim_x rightarrow x_0f(x)$ exist.
Prove $f(U)$ is bounded.

By the hypothesis, if $x_0 in overline U$ , then for every $epsilon >0$ there exists $delta>0$ such that if $left | x-x_0 right |<delta$ then $left | f(x)-L right |<epsilon$. I don't really see how I can obtain a constant $M>0$ such that $left | w-z right |<M$ for every $w,z in f(U)$. If $f$ was uniformly continuous I can see why $f(U)$ would be bounded but with the limit hypothesis only I'm stuck. Any guideline is welcome.

real-analysis multivariable-calculus

share|cite|improve this question

edited Mar 30 at 21:45

Aloizio Macedo♦

23.8k24088

asked Mar 30 at 21:36

ipreferpiipreferpi

388

$endgroup$

add a comment | 

$begingroup$

Let $U subseteq R^n$ be a bounded set, $f:U rightarrow mathbbR$ and
suppose that for every $x_0 in overline U$, $displaystylelim_x rightarrow x_0f(x)$ exist.
Prove $f(U)$ is bounded.

By the hypothesis, if $x_0 in overline U$ , then for every $epsilon >0$ there exists $delta>0$ such that if $left | x-x_0 right |<delta$ then $left | f(x)-L right |<epsilon$. I don't really see how I can obtain a constant $M>0$ such that $left | w-z right |<M$ for every $w,z in f(U)$. If $f$ was uniformly continuous I can see why $f(U)$ would be bounded but with the limit hypothesis only I'm stuck. Any guideline is welcome.

real-analysis multivariable-calculus

share|cite|improve this question

edited Mar 30 at 21:45

Aloizio Macedo♦

23.8k24088

asked Mar 30 at 21:36

ipreferpiipreferpi

388

$endgroup$

share|cite|improve this question

edited Mar 30 at 21:45

Aloizio Macedo♦

23.8k24088

asked Mar 30 at 21:36

ipreferpiipreferpi

388

share|cite|improve this question

edited Mar 30 at 21:45

Aloizio Macedo♦

23.8k24088

asked Mar 30 at 21:36

ipreferpiipreferpi

388

edited Mar 30 at 21:45

Aloizio Macedo♦

23.8k24088

edited Mar 30 at 21:45

Aloizio Macedo♦

23.8k24088

asked Mar 30 at 21:36

ipreferpiipreferpi

388

asked Mar 30 at 21:36

ipreferpiipreferpi

388

1 Answer
1

active

oldest

votes

2

$begingroup$

Suppose that $f(U)$ is not bounded. Then there exists a sequence $(x_n)$ in $U$ such that $f(x_n)$ diverges to infinity ($+infty$ or $-infty$).

Now $U$ is bounded, so there exists a subsequence $(x_varphi(n))$ of $(x_n)$ converging to a point $y in overlineU$. By hypothesis, $lim_x rightarrow y f(x)$ exists, so $lim_n rightarrow +infty f(x_varphi(n))$ exists. It is impossible because by construction, $f(x_varphi(n))$ has to diverge to infinity.

share|cite|improve this answer

answered Mar 30 at 21:42

TheSilverDoeTheSilverDoe

5,465216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By contradiction seems the hypothesis works very well. Thanks.
  $endgroup$
  – ipreferpi
  Mar 30 at 21:53
  

  
  
  
  

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168811%2fexistence-of-limit-for-every-adherence-point-in-a-bounded-set-u-implies-fu%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

Suppose that $f(U)$ is not bounded. Then there exists a sequence $(x_n)$ in $U$ such that $f(x_n)$ diverges to infinity ($+infty$ or $-infty$).

Now $U$ is bounded, so there exists a subsequence $(x_varphi(n))$ of $(x_n)$ converging to a point $y in overlineU$. By hypothesis, $lim_x rightarrow y f(x)$ exists, so $lim_n rightarrow +infty f(x_varphi(n))$ exists. It is impossible because by construction, $f(x_varphi(n))$ has to diverge to infinity.

share|cite|improve this answer

answered Mar 30 at 21:42

TheSilverDoeTheSilverDoe

5,465216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By contradiction seems the hypothesis works very well. Thanks.
  $endgroup$
  – ipreferpi
  Mar 30 at 21:53
  

  
  
  
  

add a comment | 

2

$begingroup$

Suppose that $f(U)$ is not bounded. Then there exists a sequence $(x_n)$ in $U$ such that $f(x_n)$ diverges to infinity ($+infty$ or $-infty$).

Now $U$ is bounded, so there exists a subsequence $(x_varphi(n))$ of $(x_n)$ converging to a point $y in overlineU$. By hypothesis, $lim_x rightarrow y f(x)$ exists, so $lim_n rightarrow +infty f(x_varphi(n))$ exists. It is impossible because by construction, $f(x_varphi(n))$ has to diverge to infinity.

share|cite|improve this answer

answered Mar 30 at 21:42

TheSilverDoeTheSilverDoe

5,465216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By contradiction seems the hypothesis works very well. Thanks.
  $endgroup$
  – ipreferpi
  Mar 30 at 21:53
  

  
  
  
  

add a comment | 

$begingroup$

Suppose that $f(U)$ is not bounded. Then there exists a sequence $(x_n)$ in $U$ such that $f(x_n)$ diverges to infinity ($+infty$ or $-infty$).

Now $U$ is bounded, so there exists a subsequence $(x_varphi(n))$ of $(x_n)$ converging to a point $y in overlineU$. By hypothesis, $lim_x rightarrow y f(x)$ exists, so $lim_n rightarrow +infty f(x_varphi(n))$ exists. It is impossible because by construction, $f(x_varphi(n))$ has to diverge to infinity.

share|cite|improve this answer

answered Mar 30 at 21:42

TheSilverDoeTheSilverDoe

5,465216

$endgroup$

share|cite|improve this answer

answered Mar 30 at 21:42

TheSilverDoeTheSilverDoe

5,465216

answered Mar 30 at 21:42

TheSilverDoeTheSilverDoe

5,465216

answered Mar 30 at 21:42

TheSilverDoeTheSilverDoe

5,465216