Originally Posted by

**Gamma** See I think part of the propblem is your fractions are screwed up, that first set in the union is going to contain everything else in your union, so all that is unnecessary, and i think we can all agree $\displaystyle B \not = (-1, \frac{1}{2})$. Recall $\displaystyle \frac{1}{n}\rightarrow 0$.

I see what you are trying to do though and see the confusion.

By definition, a set B is open iff for every x in B, there is an open set containing x and contained in B.

But consider that point 0 that is in B. There is no open interval (these are the basis elements for the real line in the usual topology) which contains 0 and does not intersect the sequence $\displaystyle \{\frac{1}{n}\}$ (in fact there are an infinite number of points of this intersection for every open set containing 0 by definition of $\displaystyle \frac{1}{n}\rightarrow 0$).

If instead 0 were removed like the teachers correction states (call it B') what you were trying to do is spot on.

$\displaystyle B'=(-1,0)\bigcup_{n=1}^\infty(\frac{1}{n+1},\frac{1}{n} )$

but the way it stands as he originally stated, 0 is in B, but not in B'.