Existence of limit for every adherence point in a bounded set $U$ implies $f(U)$ is bounded The 2019 Stack Overflow Developer Survey Results Are InExistence of mixed partials in Clairaut's theorem.Every continuous function $f: mathbbR to mathbbR$ is uniformly continuous on every bounded set.Not equicontinuoushow to determine the existence of double limit?Every point in the open set is a limit pointDerivative and uniform continuityConvergent subsequence for every sequence implies complete and totally boundedSet representation of bounded functionsIf omega limit set contains only one point, $x^*$, then $lim_ttoinftyphi(t;x_0)=x^*$Proving a limit diverges if $f$ is not bounded above
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Existence of limit for every adherence point in a bounded set $U$ implies $f(U)$ is bounded
The 2019 Stack Overflow Developer Survey Results Are InExistence of mixed partials in Clairaut's theorem.Every continuous function $f: mathbbR to mathbbR$ is uniformly continuous on every bounded set.Not equicontinuoushow to determine the existence of double limit?Every point in the open set is a limit pointDerivative and uniform continuityConvergent subsequence for every sequence implies complete and totally boundedSet representation of bounded functionsIf omega limit set contains only one point, $x^*$, then $lim_ttoinftyphi(t;x_0)=x^*$Proving a limit diverges if $f$ is not bounded above
$begingroup$
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
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
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
edited Mar 30 at 21:45
Aloizio Macedo♦
23.8k24088
asked Mar 30 at 21:36
ipreferpiipreferpi
388
$endgroup$
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
real-analysis multivariable-calculus
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
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
edited Mar 30 at 21:45
Aloizio Macedo♦
23.8k24088
23.8k24088
asked Mar 30 at 21:36
ipreferpiipreferpi
388
asked Mar 30 at 21:36
ipreferpiipreferpi
388
asked Mar 30 at 21:36
ipreferpiipreferpi
388
388
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
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 |
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1 Answer
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1 Answer
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active
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active
oldest
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$begingroup$
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.
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.
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.
answered Mar 30 at 21:42
TheSilverDoeTheSilverDoe
5,465216
$endgroup$
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.
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
answered Mar 30 at 21:42
TheSilverDoeTheSilverDoe
5,465216
5,465216
$begingroup$
By contradiction seems the hypothesis works very well. Thanks.
$endgroup$
– ipreferpi
Mar 30 at 21:53
add a comment |
$begingroup$
By contradiction seems the hypothesis works very well. Thanks.
$endgroup$
– ipreferpi
Mar 30 at 21:53
$begingroup$
By contradiction seems the hypothesis works very well. Thanks.
$endgroup$
– ipreferpi
Mar 30 at 21:53
$begingroup$
By contradiction seems the hypothesis works very well. Thanks.
$endgroup$
– ipreferpi
Mar 30 at 21:53
add a comment |
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