IVT and connectedness on $mathbb{R}^{n}$
$begingroup$
Let $f: mathbb{R}^{n} to mathbb{R}$ be a continuous function and $n geq 2$. Moreover, suppose that there is $c in mathbb{R}$ such that $f^{-1}({c}) = {x in mathbb{R}^{n} mid f(x) = c}$ is bounded.
(a) Prove that ${x in mathbb{R}^{n} mid f(x) geq c }$ or ${x in mathbb{R}^{n} mid f(x) leq c }$ is bounded.
(b) Prove that $f$ has a maximum or minimum in $mathbb{R}^{n}$.
My attempt.
(a)Suppose that both are unbounded. Just take $r > 0$ such that $B_{r}(0) supset f^{-1}({c})$. Since $n geq 2$, $mathbb{R}^{n}-B_{r}(0)$ is connected. But $f$ is continuous, then apply the IVT for conclude that $f^{-1}({c})$ is unbounded, a contradiction.
(b) Suppose, WLOG, that $F = {x in mathbb{R}^{n} mid f(x) geq c }$ is bounded. So, $F$ has a maximum or minimum, namely $f(p)$. If $f(p)$ is a maximum, its easy. If $f(p)$ is a minimum, then we should be $f(p) = c$, since $f(mathbb{R}^{n})$ is connected. So...?
I dont know how to finish the proof. Can someone help me?
real-analysis general-topology connectedness
edited Jan 2 at 5:27
Chris Custer
14.2k3827
asked Jan 2 at 5:19
Lucas CorrêaLucas Corrêa
1,5561421
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb{R}^{n} to mathbb{R}$ be a continuous function and $n geq 2$. Moreover, suppose that there is $c in mathbb{R}$ such that $f^{-1}({c}) = {x in mathbb{R}^{n} mid f(x) = c}$ is bounded.
(a) Prove that ${x in mathbb{R}^{n} mid f(x) geq c }$ or ${x in mathbb{R}^{n} mid f(x) leq c }$ is bounded.
(b) Prove that $f$ has a maximum or minimum in $mathbb{R}^{n}$.
My attempt.
(a)Suppose that both are unbounded. Just take $r > 0$ such that $B_{r}(0) supset f^{-1}({c})$. Since $n geq 2$, $mathbb{R}^{n}-B_{r}(0)$ is connected. But $f$ is continuous, then apply the IVT for conclude that $f^{-1}({c})$ is unbounded, a contradiction.
(b) Suppose, WLOG, that $F = {x in mathbb{R}^{n} mid f(x) geq c }$ is bounded. So, $F$ has a maximum or minimum, namely $f(p)$. If $f(p)$ is a maximum, its easy. If $f(p)$ is a minimum, then we should be $f(p) = c$, since $f(mathbb{R}^{n})$ is connected. So...?
I dont know how to finish the proof. Can someone help me?
real-analysis general-topology connectedness
edited Jan 2 at 5:27
Chris Custer
14.2k3827
asked Jan 2 at 5:19
Lucas CorrêaLucas Corrêa
1,5561421
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb{R}^{n} to mathbb{R}$ be a continuous function and $n geq 2$. Moreover, suppose that there is $c in mathbb{R}$ such that $f^{-1}({c}) = {x in mathbb{R}^{n} mid f(x) = c}$ is bounded.
(a) Prove that ${x in mathbb{R}^{n} mid f(x) geq c }$ or ${x in mathbb{R}^{n} mid f(x) leq c }$ is bounded.
(b) Prove that $f$ has a maximum or minimum in $mathbb{R}^{n}$.
My attempt.
(a)Suppose that both are unbounded. Just take $r > 0$ such that $B_{r}(0) supset f^{-1}({c})$. Since $n geq 2$, $mathbb{R}^{n}-B_{r}(0)$ is connected. But $f$ is continuous, then apply the IVT for conclude that $f^{-1}({c})$ is unbounded, a contradiction.
(b) Suppose, WLOG, that $F = {x in mathbb{R}^{n} mid f(x) geq c }$ is bounded. So, $F$ has a maximum or minimum, namely $f(p)$. If $f(p)$ is a maximum, its easy. If $f(p)$ is a minimum, then we should be $f(p) = c$, since $f(mathbb{R}^{n})$ is connected. So...?
I dont know how to finish the proof. Can someone help me?
real-analysis general-topology connectedness
edited Jan 2 at 5:27
Chris Custer
14.2k3827
asked Jan 2 at 5:19
Lucas CorrêaLucas Corrêa
1,5561421
$endgroup$
Let $f: mathbb{R}^{n} to mathbb{R}$ be a continuous function and $n geq 2$. Moreover, suppose that there is $c in mathbb{R}$ such that $f^{-1}({c}) = {x in mathbb{R}^{n} mid f(x) = c}$ is bounded.
(a) Prove that ${x in mathbb{R}^{n} mid f(x) geq c }$ or ${x in mathbb{R}^{n} mid f(x) leq c }$ is bounded.
(b) Prove that $f$ has a maximum or minimum in $mathbb{R}^{n}$.
My attempt.
(a)Suppose that both are unbounded. Just take $r > 0$ such that $B_{r}(0) supset f^{-1}({c})$. Since $n geq 2$, $mathbb{R}^{n}-B_{r}(0)$ is connected. But $f$ is continuous, then apply the IVT for conclude that $f^{-1}({c})$ is unbounded, a contradiction.
(b) Suppose, WLOG, that $F = {x in mathbb{R}^{n} mid f(x) geq c }$ is bounded. So, $F$ has a maximum or minimum, namely $f(p)$. If $f(p)$ is a maximum, its easy. If $f(p)$ is a minimum, then we should be $f(p) = c$, since $f(mathbb{R}^{n})$ is connected. So...?
I dont know how to finish the proof. Can someone help me?
real-analysis general-topology connectedness
real-analysis general-topology connectedness
edited Jan 2 at 5:27
Chris Custer
14.2k3827
asked Jan 2 at 5:19
Lucas CorrêaLucas Corrêa
1,5561421
edited Jan 2 at 5:27
Chris Custer
14.2k3827
asked Jan 2 at 5:19
Lucas CorrêaLucas Corrêa
1,5561421
edited Jan 2 at 5:27
Chris Custer
14.2k3827
edited Jan 2 at 5:27
Chris Custer
14.2k3827
edited Jan 2 at 5:27
Chris Custer
14.2k3827
14.2k3827
asked Jan 2 at 5:19
Lucas CorrêaLucas Corrêa
1,5561421
asked Jan 2 at 5:19
Lucas CorrêaLucas Corrêa
1,5561421
asked Jan 2 at 5:19
Lucas CorrêaLucas Corrêa
1,5561421
1,5561421
add a comment |
add a comment |
1 Answer
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$begingroup$
Because $F$ is bounded, it is compact, so $f$ attains a maximum on $F$, which is a global maximum of $f$.
answered Jan 2 at 5:36
SmileyCraftSmileyCraft
3,761519
$endgroup$
$begingroup$
Oh, sure! $f$ attains a maximum and a minimum. Thank you!
$endgroup$
– Lucas Corrêa
Jan 2 at 5:38
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
Because $F$ is bounded, it is compact, so $f$ attains a maximum on $F$, which is a global maximum of $f$.
answered Jan 2 at 5:36
SmileyCraftSmileyCraft
3,761519
$endgroup$
$begingroup$
Oh, sure! $f$ attains a maximum and a minimum. Thank you!
$endgroup$
– Lucas Corrêa
Jan 2 at 5:38
add a comment |
$begingroup$
Because $F$ is bounded, it is compact, so $f$ attains a maximum on $F$, which is a global maximum of $f$.
answered Jan 2 at 5:36
SmileyCraftSmileyCraft
3,761519
$endgroup$
$begingroup$
Oh, sure! $f$ attains a maximum and a minimum. Thank you!
$endgroup$
– Lucas Corrêa
Jan 2 at 5:38
add a comment |
$begingroup$
Because $F$ is bounded, it is compact, so $f$ attains a maximum on $F$, which is a global maximum of $f$.
answered Jan 2 at 5:36
SmileyCraftSmileyCraft
3,761519
$endgroup$
Because $F$ is bounded, it is compact, so $f$ attains a maximum on $F$, which is a global maximum of $f$.
answered Jan 2 at 5:36
SmileyCraftSmileyCraft
3,761519
answered Jan 2 at 5:36
SmileyCraftSmileyCraft
3,761519
answered Jan 2 at 5:36
SmileyCraftSmileyCraft
3,761519
answered Jan 2 at 5:36
SmileyCraftSmileyCraft
3,761519
3,761519
$begingroup$
Oh, sure! $f$ attains a maximum and a minimum. Thank you!
$endgroup$
– Lucas Corrêa
Jan 2 at 5:38
add a comment |
$begingroup$
Oh, sure! $f$ attains a maximum and a minimum. Thank you!
$endgroup$
– Lucas Corrêa
Jan 2 at 5:38
$begingroup$
Oh, sure! $f$ attains a maximum and a minimum. Thank you!
$endgroup$
– Lucas Corrêa
Jan 2 at 5:38
$begingroup$
Oh, sure! $f$ attains a maximum and a minimum. Thank you!
$endgroup$
– Lucas Corrêa
Jan 2 at 5:38
add a comment |
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