Show that $M$ is compact $Leftrightarrow$ every real continuous positive function has positive infimum
I already proved the first statement:
If $M$ is compact $Rightarrow$ every positive continuous function $f:Mrightarrow mathbb{R}$ has positive infimum.
Now, I need to prove the converse: If $M$ is a metric space such that every positive continuous function $f:Mrightarrow mathbb{R}$ has positive infimum, so $M$ is compact.
I found this question here: $M$ is compact iff $f:Mtomathbb{R}$ has a positive infimum.
But I want to prove this without using pseudocompactness, and this question uses.
Can someone just give me some hints? I really don't want the answer itself.
general-topology metric-spaces
edited Nov 28 '18 at 3:08
gHem
584
asked Nov 28 '18 at 2:53
Mateus Rocha
819117
add a comment |
I already proved the first statement:
If $M$ is compact $Rightarrow$ every positive continuous function $f:Mrightarrow mathbb{R}$ has positive infimum.
Now, I need to prove the converse: If $M$ is a metric space such that every positive continuous function $f:Mrightarrow mathbb{R}$ has positive infimum, so $M$ is compact.
I found this question here: $M$ is compact iff $f:Mtomathbb{R}$ has a positive infimum.
But I want to prove this without using pseudocompactness, and this question uses.
Can someone just give me some hints? I really don't want the answer itself.
general-topology metric-spaces
edited Nov 28 '18 at 3:08
gHem
584
asked Nov 28 '18 at 2:53
Mateus Rocha
819117
Maybe something like the second answer of this question should work changing $inf$ by $sup$ or $1/n$ by $n$ in the definition of $g$: math.stackexchange.com/questions/1244557/…
– Bias of Priene
Nov 28 '18 at 3:29
add a comment |
I already proved the first statement:
If $M$ is compact $Rightarrow$ every positive continuous function $f:Mrightarrow mathbb{R}$ has positive infimum.
Now, I need to prove the converse: If $M$ is a metric space such that every positive continuous function $f:Mrightarrow mathbb{R}$ has positive infimum, so $M$ is compact.
I found this question here: $M$ is compact iff $f:Mtomathbb{R}$ has a positive infimum.
But I want to prove this without using pseudocompactness, and this question uses.
Can someone just give me some hints? I really don't want the answer itself.
general-topology metric-spaces
edited Nov 28 '18 at 3:08
gHem
584
asked Nov 28 '18 at 2:53
Mateus Rocha
819117
I already proved the first statement:
If $M$ is compact $Rightarrow$ every positive continuous function $f:Mrightarrow mathbb{R}$ has positive infimum.
Now, I need to prove the converse: If $M$ is a metric space such that every positive continuous function $f:Mrightarrow mathbb{R}$ has positive infimum, so $M$ is compact.
I found this question here: $M$ is compact iff $f:Mtomathbb{R}$ has a positive infimum.
But I want to prove this without using pseudocompactness, and this question uses.
Can someone just give me some hints? I really don't want the answer itself.
general-topology metric-spaces
general-topology metric-spaces
edited Nov 28 '18 at 3:08
gHem
584
asked Nov 28 '18 at 2:53
Mateus Rocha
819117
edited Nov 28 '18 at 3:08
gHem
584
asked Nov 28 '18 at 2:53
Mateus Rocha
819117
edited Nov 28 '18 at 3:08
gHem
584
edited Nov 28 '18 at 3:08
gHem
584
edited Nov 28 '18 at 3:08
gHem
584
584
asked Nov 28 '18 at 2:53
Mateus Rocha
819117
asked Nov 28 '18 at 2:53
Mateus Rocha
819117
asked Nov 28 '18 at 2:53
Mateus Rocha
819117
819117
Maybe something like the second answer of this question should work changing $inf$ by $sup$ or $1/n$ by $n$ in the definition of $g$: math.stackexchange.com/questions/1244557/…
– Bias of Priene
Nov 28 '18 at 3:29
add a comment |
Maybe something like the second answer of this question should work changing $inf$ by $sup$ or $1/n$ by $n$ in the definition of $g$: math.stackexchange.com/questions/1244557/…
– Bias of Priene
Nov 28 '18 at 3:29
Maybe something like the second answer of this question should work changing $inf$ by $sup$ or $1/n$ by $n$ in the definition of $g$: math.stackexchange.com/questions/1244557/…
– Bias of Priene
Nov 28 '18 at 3:29
Maybe something like the second answer of this question should work changing $inf$ by $sup$ or $1/n$ by $n$ in the definition of $g$: math.stackexchange.com/questions/1244557/…
– Bias of Priene
Nov 28 '18 at 3:29
add a comment |
1 Answer
1
active
oldest
votes
If there is a sequence ${x_n}$ with no convergent subsequence the $E={x_1,x_2,cdots}$ is a closed set. Define $f:Eto (0,1) $ by $f(x_n)=frac 1 n$. Then $f$ is continuous. By (one form of) Tietze Extension Theorem there exists a continuous function $F:X to (0,1)$ such that $F=f$ on $E$. This continuous function does not have a positive infimum. [I can provide more information on Tietze's Theorem if needed].
answered Nov 28 '18 at 5:42
Kavi Rama Murthy
50.5k31854
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%2f3016637%2fshow-that-m-is-compact-leftrightarrow-every-real-continuous-positive-functi%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
If there is a sequence ${x_n}$ with no convergent subsequence the $E={x_1,x_2,cdots}$ is a closed set. Define $f:Eto (0,1) $ by $f(x_n)=frac 1 n$. Then $f$ is continuous. By (one form of) Tietze Extension Theorem there exists a continuous function $F:X to (0,1)$ such that $F=f$ on $E$. This continuous function does not have a positive infimum. [I can provide more information on Tietze's Theorem if needed].
answered Nov 28 '18 at 5:42
Kavi Rama Murthy
50.5k31854
add a comment |
If there is a sequence ${x_n}$ with no convergent subsequence the $E={x_1,x_2,cdots}$ is a closed set. Define $f:Eto (0,1) $ by $f(x_n)=frac 1 n$. Then $f$ is continuous. By (one form of) Tietze Extension Theorem there exists a continuous function $F:X to (0,1)$ such that $F=f$ on $E$. This continuous function does not have a positive infimum. [I can provide more information on Tietze's Theorem if needed].
answered Nov 28 '18 at 5:42
Kavi Rama Murthy
50.5k31854
add a comment |
If there is a sequence ${x_n}$ with no convergent subsequence the $E={x_1,x_2,cdots}$ is a closed set. Define $f:Eto (0,1) $ by $f(x_n)=frac 1 n$. Then $f$ is continuous. By (one form of) Tietze Extension Theorem there exists a continuous function $F:X to (0,1)$ such that $F=f$ on $E$. This continuous function does not have a positive infimum. [I can provide more information on Tietze's Theorem if needed].
answered Nov 28 '18 at 5:42
Kavi Rama Murthy
50.5k31854
If there is a sequence ${x_n}$ with no convergent subsequence the $E={x_1,x_2,cdots}$ is a closed set. Define $f:Eto (0,1) $ by $f(x_n)=frac 1 n$. Then $f$ is continuous. By (one form of) Tietze Extension Theorem there exists a continuous function $F:X to (0,1)$ such that $F=f$ on $E$. This continuous function does not have a positive infimum. [I can provide more information on Tietze's Theorem if needed].
answered Nov 28 '18 at 5:42
Kavi Rama Murthy
50.5k31854
answered Nov 28 '18 at 5:42
Kavi Rama Murthy
50.5k31854
answered Nov 28 '18 at 5:42
Kavi Rama Murthy
50.5k31854
answered Nov 28 '18 at 5:42
Kavi Rama Murthy
50.5k31854
50.5k31854
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3016637%2fshow-that-m-is-compact-leftrightarrow-every-real-continuous-positive-functi%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
Maybe something like the second answer of this question should work changing $inf$ by $sup$ or $1/n$ by $n$ in the definition of $g$: math.stackexchange.com/questions/1244557/…
– Bias of Priene
Nov 28 '18 at 3:29