Is $S = (x,y,z) in mathbbR^3: sqrt(x^2+y^2) leq z leq 1$ bounded, compact and/or path connected? The 2019 Stack Overflow Developer Survey Results Are InOpen and connected in $R^n$ revisedProve set is bounded and nonempty?A set is compact if and only if every continouos function is bounded on the set??construct a path between $(-1,0)$ and $(0,2)$How can I prove that any ball in $mathbbR^n$ is connected?Constructing bounded funcions that converge uniformly on compact sets.$fcolonmathbbR^nrightarrow mathbbR^m$ is continuous $Asubset mathbbR^n$ is bounded is $f(A)$ bounded?Proving every rectifiable path (bounded in some sense) is integrablePath-Connected SubsetsDividing a closed connected region into multiple subregions
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Is $S = (x,y,z) in mathbbR^3: sqrt(x^2+y^2) leq z leq 1$ bounded, compact and/or path connected?
The 2019 Stack Overflow Developer Survey Results Are InOpen and connected in $R^n$ revisedProve set is bounded and nonempty?A set is compact if and only if every continouos function is bounded on the set??construct a path between $(-1,0)$ and $(0,2)$How can I prove that any ball in $mathbbR^n$ is connected?Constructing bounded funcions that converge uniformly on compact sets.$fcolonmathbbR^nrightarrow mathbbR^m$ is continuous $Asubset mathbbR^n$ is bounded is $f(A)$ bounded?Proving every rectifiable path (bounded in some sense) is integrablePath-Connected SubsetsDividing a closed connected region into multiple subregions
$begingroup$
I know if it is bounded then it is definitely compact. So how to prove it is bounded or path-connected?
So far I have only done these in 2 dimensions so I'm not sure where to start here.
multivariable-calculus
edited Mar 24 at 9:09
Eric Toporek
10910
asked Mar 24 at 6:06
crowncrown
11
$endgroup$
add a comment |
$begingroup$
I know if it is bounded then it is definitely compact. So how to prove it is bounded or path-connected?
So far I have only done these in 2 dimensions so I'm not sure where to start here.
multivariable-calculus
edited Mar 24 at 9:09
Eric Toporek
10910
asked Mar 24 at 6:06
crowncrown
11
$endgroup$
add a comment |
$begingroup$
I know if it is bounded then it is definitely compact. So how to prove it is bounded or path-connected?
So far I have only done these in 2 dimensions so I'm not sure where to start here.
multivariable-calculus
edited Mar 24 at 9:09
Eric Toporek
10910
asked Mar 24 at 6:06
crowncrown
11
$endgroup$
I know if it is bounded then it is definitely compact. So how to prove it is bounded or path-connected?
So far I have only done these in 2 dimensions so I'm not sure where to start here.
multivariable-calculus
multivariable-calculus
edited Mar 24 at 9:09
Eric Toporek
10910
asked Mar 24 at 6:06
crowncrown
11
edited Mar 24 at 9:09
Eric Toporek
10910
asked Mar 24 at 6:06
crowncrown
11
edited Mar 24 at 9:09
Eric Toporek
10910
edited Mar 24 at 9:09
Eric Toporek
10910
edited Mar 24 at 9:09
Eric Toporek
10910
10910
asked Mar 24 at 6:06
crowncrown
11
asked Mar 24 at 6:06
crowncrown
11
asked Mar 24 at 6:06
crowncrown
11
11
add a comment |
add a comment |
1 Answer
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$begingroup$
You have $||x,y,z|| leq ||x,y,0|| + |z| leq 2$, so it is bounded.
It’s a cone. A point $(x,y,z)$ can be linked to $0$ by $(tx,ty,tz)$, $tin [0,1]$.
answered Mar 24 at 6:25
PierrePierre
3276
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You have $||x,y,z|| leq ||x,y,0|| + |z| leq 2$, so it is bounded.
It’s a cone. A point $(x,y,z)$ can be linked to $0$ by $(tx,ty,tz)$, $tin [0,1]$.
answered Mar 24 at 6:25
PierrePierre
3276
$endgroup$
add a comment |
$begingroup$
You have $||x,y,z|| leq ||x,y,0|| + |z| leq 2$, so it is bounded.
It’s a cone. A point $(x,y,z)$ can be linked to $0$ by $(tx,ty,tz)$, $tin [0,1]$.
answered Mar 24 at 6:25
PierrePierre
3276
$endgroup$
add a comment |
$begingroup$
You have $||x,y,z|| leq ||x,y,0|| + |z| leq 2$, so it is bounded.
It’s a cone. A point $(x,y,z)$ can be linked to $0$ by $(tx,ty,tz)$, $tin [0,1]$.
answered Mar 24 at 6:25
PierrePierre
3276
$endgroup$
You have $||x,y,z|| leq ||x,y,0|| + |z| leq 2$, so it is bounded.
It’s a cone. A point $(x,y,z)$ can be linked to $0$ by $(tx,ty,tz)$, $tin [0,1]$.
answered Mar 24 at 6:25
PierrePierre
3276
edited Mar 24 at 6:32
answered Mar 24 at 6:25
PierrePierre
3276
answered Mar 24 at 6:25
PierrePierre
3276
answered Mar 24 at 6:25
PierrePierre
3276
3276
add a comment |
add a comment |
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