# Showing that the rationals are not locally compact

Yes, you just restrict the domain. So the continuity follows from Theorem 18.2, page 107...

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Yes, I just looked it up. OK, thanks!

Let [a, b] be a interval of rational numbers. This interval contains at least one irrational number c, and it is closed in our subspace. Then the intervals [a, c-1/n] and [c+1/n, b], for some suitable "starting" n (and hence n+1, n+2, etc.), such that c-1/n and c+1/n lie in the interval [a, b], cover [a, b] but have no finite subcover (I found a proof that [0, 1] is not compact in the rationals, so this should be analogous).

So this contradicts the theorem I was using in post #1, and the rationals are not locally compact.

I hope this works and that I didn't write something stupid again, but there always seem to be some issues when I consider the rationals, so I'd like to clear them up.
I'm confused, is that an open cover, because it seems like all of those sets are closed? Or is this different because we're talking about the closure being compact?