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 at 3:08
gHem
584
asked Nov 28 at 2:53
Mateus Rocha
817117
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 at 3:08
gHem
584
asked Nov 28 at 2:53
Mateus Rocha
817117
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 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 at 3:08
gHem
584
asked Nov 28 at 2:53
Mateus Rocha
817117
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 at 3:08
gHem
584
asked Nov 28 at 2:53
Mateus Rocha
817117
edited Nov 28 at 3:08
gHem
584
asked Nov 28 at 2:53
Mateus Rocha
817117
edited Nov 28 at 3:08
gHem
584
edited Nov 28 at 3:08
gHem
584
edited Nov 28 at 3:08
gHem
584
584
asked Nov 28 at 2:53
Mateus Rocha
817117
asked Nov 28 at 2:53
Mateus Rocha
817117
asked Nov 28 at 2:53
Mateus Rocha
817117
817117
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 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 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 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 at 3:29
add a comment |
1 Answer
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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].
answered Nov 28 at 5:42
Kavi Rama Murthy
48.3k31854
add a comment |
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1 Answer
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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].
answered Nov 28 at 5:42
Kavi Rama Murthy
48.3k31854
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 at 5:42
Kavi Rama Murthy
48.3k31854
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 at 5:42
Kavi Rama Murthy
48.3k31854
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 at 5:42
Kavi Rama Murthy
48.3k31854
answered Nov 28 at 5:42
Kavi Rama Murthy
48.3k31854
answered Nov 28 at 5:42
Kavi Rama Murthy
48.3k31854
answered Nov 28 at 5:42
Kavi Rama Murthy
48.3k31854
48.3k31854
add a comment |
add a comment |
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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 at 3:29