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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**