proof on neighborhoods and accumulation points

prove that: S is a subset of R. if every neighborhood of x contains at least one point of S other than x itself, then every neighborhood of x contains infinitely many points of S.

i want to make sure my reasoning is valid. for the => direction, i assume that there exists some neighborhood of x that contains only finitely many points of S. then i say that there exists some neighborhood of x inside this neighborhood such that it contains no points of S other than x by picking it small enough, which is a contradiction.

the <= direction seems clear

also, neighborhoods as defined in this context are the open sets of points "a" such that |x-a| < e for any e.

am i missing any subtleties in my argument? thanks.

Re: proof on neighborhoods and accumulation points

Quote:

Originally Posted by

**oblixps** prove that: S is a subset of R. if every neighborhood of x contains at least one point of S other than x itself, then every neighborhood of x contains infinitely many points of S.

This may fly right over your head. But this is true if the space is a Hausdorff space, a $\displaystyle T_2$-space. Every two points are separated by disjoint open sets. Any metric space is a Hausdorff space.

If you understand that definition then the proof is trivial.