Hi! I have this two related questions:(adsbygoogle = window.adsbygoogle || []).push({});

(1) I was thinking that [itex]\mathbb{Q}[/itex] as a subset of [itex] \mathbb{R} [/itex] is a closed set (all its points are boundary points).

But when I think of [itex]\mathbb{Q}[/itex] not like a subset, but like a topological space (with the inherited subspace topology), are all it's points still boundary points? I think not, as there ir no "exterior" points now. So are the only clopen sets of [itex]\mathbb{Q}[/itex] the empty set and the full [itex]\mathbb{Q}[/itex] set now?

(2) The other question: I saw this example in the book Counterexamples in Analysis:

A function continuous on a closed interval and not bounded there (and therefor, not uniformly continous there) (Note: the domain is a subset of [itex] \mathbb{Q} [/itex])

[itex] f(x) = \frac{1}{x^2 - 2} \ \ \ 0 \leq x \leq 2 [/itex]

It confuses me because of the theorem that a continuous function over a compact set is uniformly continuous. I dont know in what part of the proof it was used that the compact set has to be also a complete space.

Thanks.

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# Boundary points of subsets when viewed with the subset topology

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