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I encountered the fact that ' to prove function $f$ on metric space $X$ is continuous, it is enough to prove $f$ is continuous on every compact subspace $X$'.

I thought this fact is obvious because ' In metric space, notion of continuous is equivalent to sequential continuous, so it is enough to prove $f$ is continuous at every point of $X$, but every one point in $X$ is compact, so the fact follows.

Is this correct?

P.S.

Some websites reads like this, 'If $x_n\to x$, then the set $\{x\} \cup \{ x_n : n \in \mathbb{N}\}$ is compact', so the fact follows.But I think above explanation is enough.

Pont
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1 Answers1

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No that is not sufficient.

The issue with your argument is that indeed every point is compact. But to deal with continuity of a map $f$ at a point $p$, you need to have a neighborhood of $p$ where $f$ is defined. This is not the case for a single point $p$.

mathcounterexamples.net
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  • Isn't continuity a pointwise argument? In other word, ’$f$ is continuous on X’ is equivalent to '$f$ is continuous on every point of $X$'. – Pont Jul 25 '21 at 11:26
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    @Nekojiru yes, but continuity of $f$ at $x$ and continuity of $f\vert_{{x}}$ are not the same thing. In fact, the latter is always true, while the former might not be. – Vercassivelaunos Jul 25 '21 at 11:51
  • Oh, your explanation is very clear, thank you. Could you tell me why we can conclude that $f$ is continuous at x if we could prove ' if xn converges to x, xn→x, then, {xn}∪{x} is compact ? – Pont Jul 25 '21 at 14:00
  • @Nekojiru See this answer for more details. – mathcounterexamples.net Jul 25 '21 at 15:47