Compactness and connectedness in $Bbb R^3$
$begingroup$
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
asked Jan 24 at 3:57
Chinnapparaj RChinnapparaj R
5,6382928
$endgroup$
|
show 5 more comments
$begingroup$
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
asked Jan 24 at 3:57
Chinnapparaj RChinnapparaj R
5,6382928
$endgroup$
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
|
show 5 more comments
$begingroup$
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
asked Jan 24 at 3:57
Chinnapparaj RChinnapparaj R
5,6382928
$endgroup$
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
asked Jan 24 at 3:57
Chinnapparaj RChinnapparaj R
5,6382928
asked Jan 24 at 3:57
Chinnapparaj RChinnapparaj R
5,6382928
asked Jan 24 at 3:57
Chinnapparaj RChinnapparaj R
5,6382928
asked Jan 24 at 3:57
Chinnapparaj RChinnapparaj R
5,6382928
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
|
show 5 more comments
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
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 27 at 14:09
Paul FrostPaul Frost
11.4k3934
$endgroup$
$begingroup$
Thanks!..........
$endgroup$
– Chinnapparaj R
Jan 28 at 5:08
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085438%2fcompactness-and-connectedness-in-bbb-r3%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 27 at 14:09
Paul FrostPaul Frost
11.4k3934
$endgroup$
$begingroup$
Thanks!..........
$endgroup$
– Chinnapparaj R
Jan 28 at 5:08
add a comment |
$begingroup$
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.
answered Jan 27 at 14:09
Paul FrostPaul Frost
11.4k3934
$endgroup$
$begingroup$
Thanks!..........
$endgroup$
– Chinnapparaj R
Jan 28 at 5:08
add a comment |
$begingroup$
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.
answered Jan 27 at 14:09
Paul FrostPaul Frost
11.4k3934
$endgroup$
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.
answered Jan 27 at 14:09
Paul FrostPaul Frost
11.4k3934
answered Jan 27 at 14:09
Paul FrostPaul Frost
11.4k3934
answered Jan 27 at 14:09
Paul FrostPaul Frost
11.4k3934
answered Jan 27 at 14:09
Paul FrostPaul Frost
11.4k3934
11.4k3934
$begingroup$
Thanks!..........
$endgroup$
– Chinnapparaj R
Jan 28 at 5:08
add a comment |
$begingroup$
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
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085438%2fcompactness-and-connectedness-in-bbb-r3%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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