Compactness and connectedness in $Bbb R^3$












1












$begingroup$



Consider the set $$A=left{ begin{pmatrix} x\y\zend{pmatrix} in Bbb R^3: z=x^2+y^2+1right} subset Bbb R^3$$



Prove of disprove: $A$ is connected and compact




The set $A$ is unbounded, since $(sqrt{n},sqrt{n},2n+1)^tin A$ with $$d(textbf{0},(sqrt{n},sqrt{n},2n+1))=sqrt{2n+(2n+1)^2}>n$$ where $textbf{0}$ denotes the origin in $Bbb R^3$ and $n in Bbb N$. So $A$ is not compact.



Is this correct ? How about connectedness? Can I have a hint?






general-topology compactness connectedness







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You're correct, it's not compact as not bounded.
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:01










  • $begingroup$
    For connectedness, try to visualize what this shape looks like. It's a bit like a truncated cone (like the tip was chopped off). It looks connected to me. Do you remember how to prove connectedness?
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:02






  • 1




    $begingroup$
    For connectedness, try writing your set as the image of a function from $Bbb{R}^2$ to $Bbb{R}^3$, and use the fact that the image of a connected set under a continuous map is connected.
    $endgroup$
    – greelious
    Jan 24 at 4:03










  • $begingroup$
    Connectedness comes from the ability to connect two arbitrary points on that surface with a continuous path. This is a surface of revolution, so you should be able to parametrize it in polar coordinate. That should make it super easy to then connect 2 points by a path.
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:04






  • 1




    $begingroup$
    @greelious: Is this function(?) works ? $$f:(x,y) mapsto (x,y,x^2+y^2+1)$$
    $endgroup$
    – Chinnapparaj R
    Jan 24 at 4:11


















1












$begingroup$



Consider the set $$A=left{ begin{pmatrix} x\y\zend{pmatrix} in Bbb R^3: z=x^2+y^2+1right} subset Bbb R^3$$



Prove of disprove: $A$ is connected and compact




The set $A$ is unbounded, since $(sqrt{n},sqrt{n},2n+1)^tin A$ with $$d(textbf{0},(sqrt{n},sqrt{n},2n+1))=sqrt{2n+(2n+1)^2}>n$$ where $textbf{0}$ denotes the origin in $Bbb R^3$ and $n in Bbb N$. So $A$ is not compact.



Is this correct ? How about connectedness? Can I have a hint?






general-topology compactness connectedness







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You're correct, it's not compact as not bounded.
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:01










  • $begingroup$
    For connectedness, try to visualize what this shape looks like. It's a bit like a truncated cone (like the tip was chopped off). It looks connected to me. Do you remember how to prove connectedness?
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:02






  • 1




    $begingroup$
    For connectedness, try writing your set as the image of a function from $Bbb{R}^2$ to $Bbb{R}^3$, and use the fact that the image of a connected set under a continuous map is connected.
    $endgroup$
    – greelious
    Jan 24 at 4:03










  • $begingroup$
    Connectedness comes from the ability to connect two arbitrary points on that surface with a continuous path. This is a surface of revolution, so you should be able to parametrize it in polar coordinate. That should make it super easy to then connect 2 points by a path.
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:04






  • 1




    $begingroup$
    @greelious: Is this function(?) works ? $$f:(x,y) mapsto (x,y,x^2+y^2+1)$$
    $endgroup$
    – Chinnapparaj R
    Jan 24 at 4:11
















1












1








1





$begingroup$



Consider the set $$A=left{ begin{pmatrix} x\y\zend{pmatrix} in Bbb R^3: z=x^2+y^2+1right} subset Bbb R^3$$



Prove of disprove: $A$ is connected and compact




The set $A$ is unbounded, since $(sqrt{n},sqrt{n},2n+1)^tin A$ with $$d(textbf{0},(sqrt{n},sqrt{n},2n+1))=sqrt{2n+(2n+1)^2}>n$$ where $textbf{0}$ denotes the origin in $Bbb R^3$ and $n in Bbb N$. So $A$ is not compact.



Is this correct ? How about connectedness? Can I have a hint?






general-topology compactness connectedness







share|cite|improve this question









$endgroup$





Consider the set $$A=left{ begin{pmatrix} x\y\zend{pmatrix} in Bbb R^3: z=x^2+y^2+1right} subset Bbb R^3$$



Prove of disprove: $A$ is connected and compact




The set $A$ is unbounded, since $(sqrt{n},sqrt{n},2n+1)^tin A$ with $$d(textbf{0},(sqrt{n},sqrt{n},2n+1))=sqrt{2n+(2n+1)^2}>n$$ where $textbf{0}$ denotes the origin in $Bbb R^3$ and $n in Bbb N$. So $A$ is not compact.



Is this correct ? How about connectedness? Can I have a hint?






general-topology compactness connectedness




general-topology compactness connectedness






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 at 3:57









Chinnapparaj RChinnapparaj R

5,6382928




5,6382928








  • 1




    $begingroup$
    You're correct, it's not compact as not bounded.
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:01










  • $begingroup$
    For connectedness, try to visualize what this shape looks like. It's a bit like a truncated cone (like the tip was chopped off). It looks connected to me. Do you remember how to prove connectedness?
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:02






  • 1




    $begingroup$
    For connectedness, try writing your set as the image of a function from $Bbb{R}^2$ to $Bbb{R}^3$, and use the fact that the image of a connected set under a continuous map is connected.
    $endgroup$
    – greelious
    Jan 24 at 4:03










  • $begingroup$
    Connectedness comes from the ability to connect two arbitrary points on that surface with a continuous path. This is a surface of revolution, so you should be able to parametrize it in polar coordinate. That should make it super easy to then connect 2 points by a path.
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:04






  • 1




    $begingroup$
    @greelious: Is this function(?) works ? $$f:(x,y) mapsto (x,y,x^2+y^2+1)$$
    $endgroup$
    – Chinnapparaj R
    Jan 24 at 4:11
















  • 1




    $begingroup$
    You're correct, it's not compact as not bounded.
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:01










  • $begingroup$
    For connectedness, try to visualize what this shape looks like. It's a bit like a truncated cone (like the tip was chopped off). It looks connected to me. Do you remember how to prove connectedness?
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:02






  • 1




    $begingroup$
    For connectedness, try writing your set as the image of a function from $Bbb{R}^2$ to $Bbb{R}^3$, and use the fact that the image of a connected set under a continuous map is connected.
    $endgroup$
    – greelious
    Jan 24 at 4:03










  • $begingroup$
    Connectedness comes from the ability to connect two arbitrary points on that surface with a continuous path. This is a surface of revolution, so you should be able to parametrize it in polar coordinate. That should make it super easy to then connect 2 points by a path.
    $endgroup$
    – Stefan Lafon
    Jan 24 at 4:04






  • 1




    $begingroup$
    @greelious: Is this function(?) works ? $$f:(x,y) mapsto (x,y,x^2+y^2+1)$$
    $endgroup$
    – Chinnapparaj R
    Jan 24 at 4:11










1




1




$begingroup$
You're correct, it's not compact as not bounded.
$endgroup$
– Stefan Lafon
Jan 24 at 4:01




$begingroup$
You're correct, it's not compact as not bounded.
$endgroup$
– Stefan Lafon
Jan 24 at 4:01












$begingroup$
For connectedness, try to visualize what this shape looks like. It's a bit like a truncated cone (like the tip was chopped off). It looks connected to me. Do you remember how to prove connectedness?
$endgroup$
– Stefan Lafon
Jan 24 at 4:02




$begingroup$
For connectedness, try to visualize what this shape looks like. It's a bit like a truncated cone (like the tip was chopped off). It looks connected to me. Do you remember how to prove connectedness?
$endgroup$
– Stefan Lafon
Jan 24 at 4:02




1




1




$begingroup$
For connectedness, try writing your set as the image of a function from $Bbb{R}^2$ to $Bbb{R}^3$, and use the fact that the image of a connected set under a continuous map is connected.
$endgroup$
– greelious
Jan 24 at 4:03




$begingroup$
For connectedness, try writing your set as the image of a function from $Bbb{R}^2$ to $Bbb{R}^3$, and use the fact that the image of a connected set under a continuous map is connected.
$endgroup$
– greelious
Jan 24 at 4:03












$begingroup$
Connectedness comes from the ability to connect two arbitrary points on that surface with a continuous path. This is a surface of revolution, so you should be able to parametrize it in polar coordinate. That should make it super easy to then connect 2 points by a path.
$endgroup$
– Stefan Lafon
Jan 24 at 4:04




$begingroup$
Connectedness comes from the ability to connect two arbitrary points on that surface with a continuous path. This is a surface of revolution, so you should be able to parametrize it in polar coordinate. That should make it super easy to then connect 2 points by a path.
$endgroup$
– Stefan Lafon
Jan 24 at 4:04




1




1




$begingroup$
@greelious: Is this function(?) works ? $$f:(x,y) mapsto (x,y,x^2+y^2+1)$$
$endgroup$
– Chinnapparaj R
Jan 24 at 4:11






$begingroup$
@greelious: Is this function(?) works ? $$f:(x,y) mapsto (x,y,x^2+y^2+1)$$
$endgroup$
– Chinnapparaj R
Jan 24 at 4:11












1 Answer
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The continuous map $f : mathbb{R}^2 to mathbb{R}^3, f(x,y) = (x,y,x^2+y^2+1)$, establishes a homeomorphism between $mathbb{R}^2$ and $A = f(mathbb{R}^2)$. If $p : mathbb{R}^3 to mathbb{R}^2, p(x,y,z) = (x,y)$ denotes the (continuous!) projection, then $p mid_A : A to mathbb{R}^2$ is the inverse homeomorphism. This shows that $A$ is connected but not compact.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks!..........
    $endgroup$
    – Chinnapparaj R
    Jan 28 at 5:08











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

The continuous map $f : mathbb{R}^2 to mathbb{R}^3, f(x,y) = (x,y,x^2+y^2+1)$, establishes a homeomorphism between $mathbb{R}^2$ and $A = f(mathbb{R}^2)$. If $p : mathbb{R}^3 to mathbb{R}^2, p(x,y,z) = (x,y)$ denotes the (continuous!) projection, then $p mid_A : A to mathbb{R}^2$ is the inverse homeomorphism. This shows that $A$ is connected but not compact.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks!..........
    $endgroup$
    – Chinnapparaj R
    Jan 28 at 5:08
















1












$begingroup$

The continuous map $f : mathbb{R}^2 to mathbb{R}^3, f(x,y) = (x,y,x^2+y^2+1)$, establishes a homeomorphism between $mathbb{R}^2$ and $A = f(mathbb{R}^2)$. If $p : mathbb{R}^3 to mathbb{R}^2, p(x,y,z) = (x,y)$ denotes the (continuous!) projection, then $p mid_A : A to mathbb{R}^2$ is the inverse homeomorphism. This shows that $A$ is connected but not compact.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks!..........
    $endgroup$
    – Chinnapparaj R
    Jan 28 at 5:08














1












1








1





$begingroup$

The continuous map $f : mathbb{R}^2 to mathbb{R}^3, f(x,y) = (x,y,x^2+y^2+1)$, establishes a homeomorphism between $mathbb{R}^2$ and $A = f(mathbb{R}^2)$. If $p : mathbb{R}^3 to mathbb{R}^2, p(x,y,z) = (x,y)$ denotes the (continuous!) projection, then $p mid_A : A to mathbb{R}^2$ is the inverse homeomorphism. This shows that $A$ is connected but not compact.






share|cite|improve this answer









$endgroup$



The continuous map $f : mathbb{R}^2 to mathbb{R}^3, f(x,y) = (x,y,x^2+y^2+1)$, establishes a homeomorphism between $mathbb{R}^2$ and $A = f(mathbb{R}^2)$. If $p : mathbb{R}^3 to mathbb{R}^2, p(x,y,z) = (x,y)$ denotes the (continuous!) projection, then $p mid_A : A to mathbb{R}^2$ is the inverse homeomorphism. This shows that $A$ is connected but not compact.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 27 at 14:09









Paul FrostPaul Frost

11.4k3934




11.4k3934












  • $begingroup$
    Thanks!..........
    $endgroup$
    – Chinnapparaj R
    Jan 28 at 5:08


















  • $begingroup$
    Thanks!..........
    $endgroup$
    – Chinnapparaj R
    Jan 28 at 5:08
















$begingroup$
Thanks!..........
$endgroup$
– Chinnapparaj R
Jan 28 at 5:08




$begingroup$
Thanks!..........
$endgroup$
– Chinnapparaj R
Jan 28 at 5:08


















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