Does “ All continuous functions are bounded ” or “ All continuous functions attain a maximum ” or together imply the domain is compact?

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Let $(X,d)$ be an infinite metric space satisfying H1 or H2 or both.

H1: (All continuous functions on X to $mathbbR$ are bounded.)

If $f: XtomathbbR$ is continuous on $X$, then $f(x)$ is bounded.

H2: (All continuous functions on $X$ to $mathbbR$ attain a maximum.)

If $f: XtomathbbR$ is continuous on $X$, then there exists at least one point $p in X $ such that $f(p) geq f(x)$ for every $x in X. $

Question

Does H1 only or does H2 only imply $X$ is compact?
Do H1 and H2 together imply $X$ is compact?

Note:

I know " If $X$ is compact then all continuous functions on $X$ to $mathbbR$ are bounded " and " If $X$ is compact then all continuous functions on $X$ to $mathbbR$ attain a maximum ". I am just curious about whether or not the converse still hold.

real-analysis general-topology continuity compactness

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edited Jul 25 at 15:24

asked Jul 25 at 15:18

JI-ZHAN HUANG

575

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You mean "converse" not "contrapositive."
  – saulspatz
  Jul 25 at 15:20
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  See math.stackexchange.com/questions/114123/…
  – Wraith1995
  Jul 25 at 15:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  For metric spaces, each of H1 and H2 implies compactness. Here is H1. For general topological spaces (even locally compact Hausdorff spaces), this is not the case, $omega_1$ (the first uncountable ordinal) is an example of a locally compact Hausdorff space with H1 and H2 that isn't compact.
  – Daniel Fischer♦
  Jul 25 at 15:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @Wraith1995 :-)
  – JI-ZHAN HUANG
  Jul 25 at 15:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @DanielFischer :-)
  – JI-ZHAN HUANG
  Jul 25 at 15:36
  

  
  
  
  

 | 
show 4 more comments

up vote
6
down vote

favorite

1

Let $(X,d)$ be an infinite metric space satisfying H1 or H2 or both.

H1: (All continuous functions on X to $mathbbR$ are bounded.)

If $f: XtomathbbR$ is continuous on $X$, then $f(x)$ is bounded.

H2: (All continuous functions on $X$ to $mathbbR$ attain a maximum.)

If $f: XtomathbbR$ is continuous on $X$, then there exists at least one point $p in X $ such that $f(p) geq f(x)$ for every $x in X. $

Question

Does H1 only or does H2 only imply $X$ is compact?
Do H1 and H2 together imply $X$ is compact?

Note:

I know " If $X$ is compact then all continuous functions on $X$ to $mathbbR$ are bounded " and " If $X$ is compact then all continuous functions on $X$ to $mathbbR$ attain a maximum ". I am just curious about whether or not the converse still hold.

real-analysis general-topology continuity compactness

share|cite|improve this question

edited Jul 25 at 15:24

asked Jul 25 at 15:18

JI-ZHAN HUANG

575

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You mean "converse" not "contrapositive."
  – saulspatz
  Jul 25 at 15:20
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  See math.stackexchange.com/questions/114123/…
  – Wraith1995
  Jul 25 at 15:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  For metric spaces, each of H1 and H2 implies compactness. Here is H1. For general topological spaces (even locally compact Hausdorff spaces), this is not the case, $omega_1$ (the first uncountable ordinal) is an example of a locally compact Hausdorff space with H1 and H2 that isn't compact.
  – Daniel Fischer♦
  Jul 25 at 15:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @Wraith1995 :-)
  – JI-ZHAN HUANG
  Jul 25 at 15:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @DanielFischer :-)
  – JI-ZHAN HUANG
  Jul 25 at 15:36
  

  
  
  
  

 | 
show 4 more comments

up vote
6
down vote

favorite

1

up vote
6
down vote

favorite

1

1

Let $(X,d)$ be an infinite metric space satisfying H1 or H2 or both.

H1: (All continuous functions on X to $mathbbR$ are bounded.)

If $f: XtomathbbR$ is continuous on $X$, then $f(x)$ is bounded.

H2: (All continuous functions on $X$ to $mathbbR$ attain a maximum.)

If $f: XtomathbbR$ is continuous on $X$, then there exists at least one point $p in X $ such that $f(p) geq f(x)$ for every $x in X. $

Question

Does H1 only or does H2 only imply $X$ is compact?
Do H1 and H2 together imply $X$ is compact?

Note:

I know " If $X$ is compact then all continuous functions on $X$ to $mathbbR$ are bounded " and " If $X$ is compact then all continuous functions on $X$ to $mathbbR$ attain a maximum ". I am just curious about whether or not the converse still hold.

real-analysis general-topology continuity compactness

share|cite|improve this question

edited Jul 25 at 15:24

asked Jul 25 at 15:18

JI-ZHAN HUANG

575

Let $(X,d)$ be an infinite metric space satisfying H1 or H2 or both.

H1: (All continuous functions on X to $mathbbR$ are bounded.)

If $f: XtomathbbR$ is continuous on $X$, then $f(x)$ is bounded.

H2: (All continuous functions on $X$ to $mathbbR$ attain a maximum.)

If $f: XtomathbbR$ is continuous on $X$, then there exists at least one point $p in X $ such that $f(p) geq f(x)$ for every $x in X. $

Question

Does H1 only or does H2 only imply $X$ is compact?
Do H1 and H2 together imply $X$ is compact?

Note:

I know " If $X$ is compact then all continuous functions on $X$ to $mathbbR$ are bounded " and " If $X$ is compact then all continuous functions on $X$ to $mathbbR$ attain a maximum ". I am just curious about whether or not the converse still hold.

real-analysis general-topology continuity compactness

share|cite|improve this question

edited Jul 25 at 15:24

asked Jul 25 at 15:18

JI-ZHAN HUANG

575

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 15:24

edited Jul 25 at 15:24

edited Jul 25 at 15:24

asked Jul 25 at 15:18

JI-ZHAN HUANG

575

asked Jul 25 at 15:18

JI-ZHAN HUANG

575

asked Jul 25 at 15:18

JI-ZHAN HUANG

575

575

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You mean "converse" not "contrapositive."
  – saulspatz
  Jul 25 at 15:20
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  See math.stackexchange.com/questions/114123/…
  – Wraith1995
  Jul 25 at 15:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  For metric spaces, each of H1 and H2 implies compactness. Here is H1. For general topological spaces (even locally compact Hausdorff spaces), this is not the case, $omega_1$ (the first uncountable ordinal) is an example of a locally compact Hausdorff space with H1 and H2 that isn't compact.
  – Daniel Fischer♦
  Jul 25 at 15:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @Wraith1995 :-)
  – JI-ZHAN HUANG
  Jul 25 at 15:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @DanielFischer :-)
  – JI-ZHAN HUANG
  Jul 25 at 15:36
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You mean "converse" not "contrapositive."
  – saulspatz
  Jul 25 at 15:20
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  See math.stackexchange.com/questions/114123/…
  – Wraith1995
  Jul 25 at 15:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  For metric spaces, each of H1 and H2 implies compactness. Here is H1. For general topological spaces (even locally compact Hausdorff spaces), this is not the case, $omega_1$ (the first uncountable ordinal) is an example of a locally compact Hausdorff space with H1 and H2 that isn't compact.
  – Daniel Fischer♦
  Jul 25 at 15:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @Wraith1995 :-)
  – JI-ZHAN HUANG
  Jul 25 at 15:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you @DanielFischer :-)
  – JI-ZHAN HUANG
  Jul 25 at 15:36
  

  
  
  
  

3

3

You mean "converse" not "contrapositive."
– saulspatz
Jul 25 at 15:20

You mean "converse" not "contrapositive."
– saulspatz
Jul 25 at 15:20

6

6

See math.stackexchange.com/questions/114123/…
– Wraith1995
Jul 25 at 15:20

See math.stackexchange.com/questions/114123/…
– Wraith1995
Jul 25 at 15:20

2

2

For metric spaces, each of H1 and H2 implies compactness. Here is H1. For general topological spaces (even locally compact Hausdorff spaces), this is not the case, $omega_1$ (the first uncountable ordinal) is an example of a locally compact Hausdorff space with H1 and H2 that isn't compact.
– Daniel Fischer♦
Jul 25 at 15:33

For metric spaces, each of H1 and H2 implies compactness. Here is H1. For general topological spaces (even locally compact Hausdorff spaces), this is not the case, $omega_1$ (the first uncountable ordinal) is an example of a locally compact Hausdorff space with H1 and H2 that isn't compact.
– Daniel Fischer♦
Jul 25 at 15:33

Thank you @Wraith1995 :-)
– JI-ZHAN HUANG
Jul 25 at 15:35

Thank you @Wraith1995 :-)
– JI-ZHAN HUANG
Jul 25 at 15:35

Thank you @DanielFischer :-)
– JI-ZHAN HUANG
Jul 25 at 15:36

Thank you @DanielFischer :-)
– JI-ZHAN HUANG
Jul 25 at 15:36

 | 
show 4 more comments

2 Answers
2

active

oldest

votes

up vote
3
down vote

Properties H1 and H2 are defined for arbitrary toplogical spaces $X$. If $X$ satisfies H1, it is called pseudocompact. See https://en.wikipedia.org/wiki/Pseudocompact_space.

Clearly H2 implies H1. The converse holds for Tychonoff-spaces (= completely regular spaces). See Theorem 27 in

Hewitt, Edwin. "Rings of real-valued continuous functions. I." Transactions of the American Mathematical Society 64.1 (1948): 45-99

Theorem 30 in this paper states that a normal space is pseudocompact if and only if it is compact.

This gives a complete answer to your question.

share|cite|improve this answer

edited Jul 26 at 12:58

answered Jul 25 at 17:02

Paul Frost

3,613420

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's not clear to me that H2 implies H1. Couldn't $f$ be only bounded above? Would you be willing to clarify that?
  – Todd Wilcox
  Jul 25 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ToddWilcox The function $-f$ is continuous as well, hence attains its maximum, which is (minus) minimum for $f$.
  – A.Γ.
  Jul 25 at 18:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @A.Γ. Are we taking it as given that $X$ is compact? If $X$ is not compact then it seems that H2 does not imply H1. For example, if $X=mathbbR$ and $f$ is a parabola opening downwards, then it has a maximum and no minimum. Ohhh. .Ohh.. I see. All continuous functions. So if $f$ is continuous then $-f$ is also continuous and must have a maximum. My brain is apparently not responding to caffeine today.
  – Todd Wilcox
  Jul 25 at 18:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ToddWilcox If $f$ is continuous, then also $lvert f rvert$ is continuous and must attain a maximum. Thus $f$ is bounded.
  – Paul Frost
  Jul 25 at 18:53
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

H1 $Longrightarrow (X$ is compact).

Suppose $X$ is not compact. Then there is a sequence $x_n in X$ with no subsequence that converges in $X$. (For a metric space, "compact" and "sequentially compact" are equivalent.) Taking a subsequence, we may assume WLOG that all $x_n$ are distinct. The set $E= x_n : n in mathbb N$ is closed in $X$. The set $E$ has the discrete topology. The unbounded function $x_n mapsto n$ is therefore continuous on $E$. A continuous real-valued function on a closed subset of a metric space extends to a continuous real-valued function on the whole space. That will be an unbounded continuous function on $X$.

As noted, H2 $Longrightarrow$ H1 is easy, so we also get
H2 $Longrightarrow (X$ is compact).

[I would guess this was known for metric spaces long before Hewitt's great paper in 1948; and the substance of Hewitt's paper was exploring what happens in non-meric spaces.]

share|cite|improve this answer

answered Jul 26 at 13:45

GEdgar

58.4k264163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree with your guess concerning metric spaces. In fact, proofs are fairly simple in this case. However, Hewitt identified pseudocompactness as a property interesting in its own right and clarified its status among the large variety of compactness concepts.
  – Paul Frost
  Jul 26 at 14:18
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Properties H1 and H2 are defined for arbitrary toplogical spaces $X$. If $X$ satisfies H1, it is called pseudocompact. See https://en.wikipedia.org/wiki/Pseudocompact_space.

Clearly H2 implies H1. The converse holds for Tychonoff-spaces (= completely regular spaces). See Theorem 27 in

Hewitt, Edwin. "Rings of real-valued continuous functions. I." Transactions of the American Mathematical Society 64.1 (1948): 45-99

Theorem 30 in this paper states that a normal space is pseudocompact if and only if it is compact.

This gives a complete answer to your question.

share|cite|improve this answer

edited Jul 26 at 12:58

answered Jul 25 at 17:02

Paul Frost

3,613420

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's not clear to me that H2 implies H1. Couldn't $f$ be only bounded above? Would you be willing to clarify that?
  – Todd Wilcox
  Jul 25 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ToddWilcox The function $-f$ is continuous as well, hence attains its maximum, which is (minus) minimum for $f$.
  – A.Γ.
  Jul 25 at 18:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @A.Γ. Are we taking it as given that $X$ is compact? If $X$ is not compact then it seems that H2 does not imply H1. For example, if $X=mathbbR$ and $f$ is a parabola opening downwards, then it has a maximum and no minimum. Ohhh. .Ohh.. I see. All continuous functions. So if $f$ is continuous then $-f$ is also continuous and must have a maximum. My brain is apparently not responding to caffeine today.
  – Todd Wilcox
  Jul 25 at 18:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ToddWilcox If $f$ is continuous, then also $lvert f rvert$ is continuous and must attain a maximum. Thus $f$ is bounded.
  – Paul Frost
  Jul 25 at 18:53
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

Properties H1 and H2 are defined for arbitrary toplogical spaces $X$. If $X$ satisfies H1, it is called pseudocompact. See https://en.wikipedia.org/wiki/Pseudocompact_space.

Clearly H2 implies H1. The converse holds for Tychonoff-spaces (= completely regular spaces). See Theorem 27 in

Hewitt, Edwin. "Rings of real-valued continuous functions. I." Transactions of the American Mathematical Society 64.1 (1948): 45-99

Theorem 30 in this paper states that a normal space is pseudocompact if and only if it is compact.

This gives a complete answer to your question.

share|cite|improve this answer

edited Jul 26 at 12:58

answered Jul 25 at 17:02

Paul Frost

3,613420

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's not clear to me that H2 implies H1. Couldn't $f$ be only bounded above? Would you be willing to clarify that?
  – Todd Wilcox
  Jul 25 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ToddWilcox The function $-f$ is continuous as well, hence attains its maximum, which is (minus) minimum for $f$.
  – A.Γ.
  Jul 25 at 18:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @A.Γ. Are we taking it as given that $X$ is compact? If $X$ is not compact then it seems that H2 does not imply H1. For example, if $X=mathbbR$ and $f$ is a parabola opening downwards, then it has a maximum and no minimum. Ohhh. .Ohh.. I see. All continuous functions. So if $f$ is continuous then $-f$ is also continuous and must have a maximum. My brain is apparently not responding to caffeine today.
  – Todd Wilcox
  Jul 25 at 18:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ToddWilcox If $f$ is continuous, then also $lvert f rvert$ is continuous and must attain a maximum. Thus $f$ is bounded.
  – Paul Frost
  Jul 25 at 18:53
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Properties H1 and H2 are defined for arbitrary toplogical spaces $X$. If $X$ satisfies H1, it is called pseudocompact. See https://en.wikipedia.org/wiki/Pseudocompact_space.

Clearly H2 implies H1. The converse holds for Tychonoff-spaces (= completely regular spaces). See Theorem 27 in

Hewitt, Edwin. "Rings of real-valued continuous functions. I." Transactions of the American Mathematical Society 64.1 (1948): 45-99

Theorem 30 in this paper states that a normal space is pseudocompact if and only if it is compact.

This gives a complete answer to your question.

share|cite|improve this answer

edited Jul 26 at 12:58

answered Jul 25 at 17:02

Paul Frost

3,613420

Properties H1 and H2 are defined for arbitrary toplogical spaces $X$. If $X$ satisfies H1, it is called pseudocompact. See https://en.wikipedia.org/wiki/Pseudocompact_space.

Clearly H2 implies H1. The converse holds for Tychonoff-spaces (= completely regular spaces). See Theorem 27 in

Hewitt, Edwin. "Rings of real-valued continuous functions. I." Transactions of the American Mathematical Society 64.1 (1948): 45-99

Theorem 30 in this paper states that a normal space is pseudocompact if and only if it is compact.

This gives a complete answer to your question.

share|cite|improve this answer

edited Jul 26 at 12:58

answered Jul 25 at 17:02

Paul Frost

3,613420

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 12:58

edited Jul 26 at 12:58

edited Jul 26 at 12:58

answered Jul 25 at 17:02

Paul Frost

3,613420

answered Jul 25 at 17:02

Paul Frost

3,613420

answered Jul 25 at 17:02

Paul Frost

3,613420

3,613420

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's not clear to me that H2 implies H1. Couldn't $f$ be only bounded above? Would you be willing to clarify that?
  – Todd Wilcox
  Jul 25 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ToddWilcox The function $-f$ is continuous as well, hence attains its maximum, which is (minus) minimum for $f$.
  – A.Γ.
  Jul 25 at 18:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @A.Γ. Are we taking it as given that $X$ is compact? If $X$ is not compact then it seems that H2 does not imply H1. For example, if $X=mathbbR$ and $f$ is a parabola opening downwards, then it has a maximum and no minimum. Ohhh. .Ohh.. I see. All continuous functions. So if $f$ is continuous then $-f$ is also continuous and must have a maximum. My brain is apparently not responding to caffeine today.
  – Todd Wilcox
  Jul 25 at 18:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ToddWilcox If $f$ is continuous, then also $lvert f rvert$ is continuous and must attain a maximum. Thus $f$ is bounded.
  – Paul Frost
  Jul 25 at 18:53
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's not clear to me that H2 implies H1. Couldn't $f$ be only bounded above? Would you be willing to clarify that?
  – Todd Wilcox
  Jul 25 at 18:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ToddWilcox The function $-f$ is continuous as well, hence attains its maximum, which is (minus) minimum for $f$.
  – A.Γ.
  Jul 25 at 18:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @A.Γ. Are we taking it as given that $X$ is compact? If $X$ is not compact then it seems that H2 does not imply H1. For example, if $X=mathbbR$ and $f$ is a parabola opening downwards, then it has a maximum and no minimum. Ohhh. .Ohh.. I see. All continuous functions. So if $f$ is continuous then $-f$ is also continuous and must have a maximum. My brain is apparently not responding to caffeine today.
  – Todd Wilcox
  Jul 25 at 18:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ToddWilcox If $f$ is continuous, then also $lvert f rvert$ is continuous and must attain a maximum. Thus $f$ is bounded.
  – Paul Frost
  Jul 25 at 18:53
  
  

  
  
  
  

It's not clear to me that H2 implies H1. Couldn't $f$ be only bounded above? Would you be willing to clarify that?
– Todd Wilcox
Jul 25 at 18:18

It's not clear to me that H2 implies H1. Couldn't $f$ be only bounded above? Would you be willing to clarify that?
– Todd Wilcox
Jul 25 at 18:18

@ToddWilcox The function $-f$ is continuous as well, hence attains its maximum, which is (minus) minimum for $f$.
– A.Γ.
Jul 25 at 18:27

@ToddWilcox The function $-f$ is continuous as well, hence attains its maximum, which is (minus) minimum for $f$.
– A.Γ.
Jul 25 at 18:27

@A.Γ. Are we taking it as given that $X$ is compact? If $X$ is not compact then it seems that H2 does not imply H1. For example, if $X=mathbbR$ and $f$ is a parabola opening downwards, then it has a maximum and no minimum. Ohhh. .Ohh.. I see. All continuous functions. So if $f$ is continuous then $-f$ is also continuous and must have a maximum. My brain is apparently not responding to caffeine today.
– Todd Wilcox
Jul 25 at 18:33

@A.Γ. Are we taking it as given that $X$ is compact? If $X$ is not compact then it seems that H2 does not imply H1. For example, if $X=mathbbR$ and $f$ is a parabola opening downwards, then it has a maximum and no minimum. Ohhh. .Ohh.. I see. All continuous functions. So if $f$ is continuous then $-f$ is also continuous and must have a maximum. My brain is apparently not responding to caffeine today.
– Todd Wilcox
Jul 25 at 18:33

@ToddWilcox If $f$ is continuous, then also $lvert f rvert$ is continuous and must attain a maximum. Thus $f$ is bounded.
– Paul Frost
Jul 25 at 18:53

@ToddWilcox If $f$ is continuous, then also $lvert f rvert$ is continuous and must attain a maximum. Thus $f$ is bounded.
– Paul Frost
Jul 25 at 18:53

add a comment | 

up vote
1
down vote

H1 $Longrightarrow (X$ is compact).

Suppose $X$ is not compact. Then there is a sequence $x_n in X$ with no subsequence that converges in $X$. (For a metric space, "compact" and "sequentially compact" are equivalent.) Taking a subsequence, we may assume WLOG that all $x_n$ are distinct. The set $E= x_n : n in mathbb N$ is closed in $X$. The set $E$ has the discrete topology. The unbounded function $x_n mapsto n$ is therefore continuous on $E$. A continuous real-valued function on a closed subset of a metric space extends to a continuous real-valued function on the whole space. That will be an unbounded continuous function on $X$.

As noted, H2 $Longrightarrow$ H1 is easy, so we also get
H2 $Longrightarrow (X$ is compact).

[I would guess this was known for metric spaces long before Hewitt's great paper in 1948; and the substance of Hewitt's paper was exploring what happens in non-meric spaces.]

share|cite|improve this answer

answered Jul 26 at 13:45

GEdgar

58.4k264163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree with your guess concerning metric spaces. In fact, proofs are fairly simple in this case. However, Hewitt identified pseudocompactness as a property interesting in its own right and clarified its status among the large variety of compactness concepts.
  – Paul Frost
  Jul 26 at 14:18
  

  
  
  
  

add a comment | 

up vote
1
down vote

H1 $Longrightarrow (X$ is compact).

Suppose $X$ is not compact. Then there is a sequence $x_n in X$ with no subsequence that converges in $X$. (For a metric space, "compact" and "sequentially compact" are equivalent.) Taking a subsequence, we may assume WLOG that all $x_n$ are distinct. The set $E= x_n : n in mathbb N$ is closed in $X$. The set $E$ has the discrete topology. The unbounded function $x_n mapsto n$ is therefore continuous on $E$. A continuous real-valued function on a closed subset of a metric space extends to a continuous real-valued function on the whole space. That will be an unbounded continuous function on $X$.

As noted, H2 $Longrightarrow$ H1 is easy, so we also get
H2 $Longrightarrow (X$ is compact).

[I would guess this was known for metric spaces long before Hewitt's great paper in 1948; and the substance of Hewitt's paper was exploring what happens in non-meric spaces.]

share|cite|improve this answer

answered Jul 26 at 13:45

GEdgar

58.4k264163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree with your guess concerning metric spaces. In fact, proofs are fairly simple in this case. However, Hewitt identified pseudocompactness as a property interesting in its own right and clarified its status among the large variety of compactness concepts.
  – Paul Frost
  Jul 26 at 14:18
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

H1 $Longrightarrow (X$ is compact).

Suppose $X$ is not compact. Then there is a sequence $x_n in X$ with no subsequence that converges in $X$. (For a metric space, "compact" and "sequentially compact" are equivalent.) Taking a subsequence, we may assume WLOG that all $x_n$ are distinct. The set $E= x_n : n in mathbb N$ is closed in $X$. The set $E$ has the discrete topology. The unbounded function $x_n mapsto n$ is therefore continuous on $E$. A continuous real-valued function on a closed subset of a metric space extends to a continuous real-valued function on the whole space. That will be an unbounded continuous function on $X$.

As noted, H2 $Longrightarrow$ H1 is easy, so we also get
H2 $Longrightarrow (X$ is compact).

[I would guess this was known for metric spaces long before Hewitt's great paper in 1948; and the substance of Hewitt's paper was exploring what happens in non-meric spaces.]

share|cite|improve this answer

answered Jul 26 at 13:45

GEdgar

58.4k264163

H1 $Longrightarrow (X$ is compact).

Suppose $X$ is not compact. Then there is a sequence $x_n in X$ with no subsequence that converges in $X$. (For a metric space, "compact" and "sequentially compact" are equivalent.) Taking a subsequence, we may assume WLOG that all $x_n$ are distinct. The set $E= x_n : n in mathbb N$ is closed in $X$. The set $E$ has the discrete topology. The unbounded function $x_n mapsto n$ is therefore continuous on $E$. A continuous real-valued function on a closed subset of a metric space extends to a continuous real-valued function on the whole space. That will be an unbounded continuous function on $X$.

As noted, H2 $Longrightarrow$ H1 is easy, so we also get
H2 $Longrightarrow (X$ is compact).

[I would guess this was known for metric spaces long before Hewitt's great paper in 1948; and the substance of Hewitt's paper was exploring what happens in non-meric spaces.]

share|cite|improve this answer

answered Jul 26 at 13:45

GEdgar

58.4k264163

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 13:45

GEdgar

58.4k264163

answered Jul 26 at 13:45

GEdgar

58.4k264163

answered Jul 26 at 13:45

GEdgar

58.4k264163

58.4k264163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree with your guess concerning metric spaces. In fact, proofs are fairly simple in this case. However, Hewitt identified pseudocompactness as a property interesting in its own right and clarified its status among the large variety of compactness concepts.
  – Paul Frost
  Jul 26 at 14:18
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree with your guess concerning metric spaces. In fact, proofs are fairly simple in this case. However, Hewitt identified pseudocompactness as a property interesting in its own right and clarified its status among the large variety of compactness concepts.
  – Paul Frost
  Jul 26 at 14:18
  

  
  
  
  

I agree with your guess concerning metric spaces. In fact, proofs are fairly simple in this case. However, Hewitt identified pseudocompactness as a property interesting in its own right and clarified its status among the large variety of compactness concepts.
– Paul Frost
Jul 26 at 14:18

I agree with your guess concerning metric spaces. In fact, proofs are fairly simple in this case. However, Hewitt identified pseudocompactness as a property interesting in its own right and clarified its status among the large variety of compactness concepts.
– Paul Frost
Jul 26 at 14:18

add a comment | 

 

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