I'm sure this has been asked before, but the proofs I've seen use the fact R is connected or continuous functions is some way. I'm trying to prove it with the things that have only been presented in the book so far (Mathematical Analysis by Apostol).(adsbygoogle = window.adsbygoogle || []).push({});

So, let A be a subset of R which is both open and closed. Assume A is non-empty. I want to end up showing A = R.

Now A is open so by the Representation theorem for open sets in R, A is the union of a countable collection of disjoint component intervals of A. Denote these intervals by [itex] I_{k} = (a_{k},b_{k}) [/itex]

But both [itex]a_{k}[/itex] and [itex]b_{k}[/itex] are accumulation points of A, and so must belong to A (for A is closed).

I'm not really sure how to proceed, or if I'm even using the right path. Would I then say, since A is open there is an r>0 such that [itex] (b_{k}-r,b_{k}+r)[/itex] is in A. Then use the representation again on this, and reiterate to conclude the whole real line is in A?

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# Closed and Open sets in R (or 'clopen')

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