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Heine-Cantor theorem asserts that every continuous function on a compact metric space is uniformly continuous.

The converse is , if every continous function is uniformly continuous on a metric space X then X is compact. I have got the answer that the converse is false but I am unable to find a counter example.

The other question is if the converse is false then its negation must be true.

Negation- if every continous function is uniformly continuous on a metric space X and X is not compact.

How do we prove this?

Subhrajyoti Nayak
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  • I think this can help you: https://math.stackexchange.com/questions/316143/inverse-of-heine-cantor-theorem – user773458 Feb 16 '21 at 11:38
  • You cannot prove it: there are non-compact spaces where every continuous function is uniformly continuous. But a valid question it certainly is. – Henno Brandsma Feb 16 '21 at 11:45
  • ${0}$ is a counter-example for your last statment. Your logic is flawed. – Kavi Rama Murthy Feb 16 '21 at 11:50
  • This question has been asked many times. Here is a good answer to the question: https://math.stackexchange.com/questions/316143/inverse-of-heine-cantor-theorem?noredirect=1&lq=1 – Kavi Rama Murthy Feb 16 '21 at 11:53
  • @KaviRamaMurthy Then what would be the correct negation. – Subhrajyoti Nayak Feb 16 '21 at 11:56
  • If every continuous function on $X$ is uniformly continuous then $X$ need not be compact. Or you can say: If every continuous function on $X$ is uniformly continuous then $X$ is not necessarily compact. – Kavi Rama Murthy Feb 16 '21 at 12:18
  • @KaviRamaMurthy I made the negation of ' If A then B ' to 'A and not B'. Btw can I conclude from your negation that compact spaces are not the only spaces on which every continuous function is uniformly continous'. – Subhrajyoti Nayak Feb 16 '21 at 14:15

1 Answers1

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Theorem 1 of this old paper gives conditions on a metric space so that all continuous functions on it are uniformly continuous.

It seems to me that $\Bbb Z$ (as a subspace of $\Bbb R$) is an easy example of such a non-compact space. And also $\Bbb Z \cup \{\frac1n: n =1,2,3,\ldots\}$ is one that is non-discrete too. As the theorem 1 quoted above shows, we do need lots of "far away" isolated points for this to happen, and in "nice" metric spaces this will imply compactness.

BTW the negation of the converse is: there exists a non-compact $X$ such that every continuous function on $X$ is uniformly continuous.

Henno Brandsma
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