Show that $M$ is compact $Leftrightarrow$ every real continuous positive function has positive infimum












2














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.










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  • 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
















2














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.










share|cite|improve this question
























  • 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














2












2








2







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.










share|cite|improve this question















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






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edited Nov 28 '18 at 3:08









gHem

584




584










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


















  • 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










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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].






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    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].






    share|cite|improve this answer


























      1














      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].






      share|cite|improve this answer
























        1












        1








        1






        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].






        share|cite|improve this answer












        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].







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 5:42









        Kavi Rama Murthy

        50.4k31854




        50.4k31854






























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