Open, closed, compact I have the following question.  I know the answers but am struggling with how to write up the work formally.  Any help would be appreciated.  
Consider $B=\{1-\frac{1}{n} :n=1,2,\cdots\}$.  Is $B$ open? Closed? Compact?  Justify.
I know that it is not open or closed, thus not compact. 
 A: $B$ is not closed, because it has an accumulation point which is not in $B$ (can you see which one)?
$B$ is not open, because $0 \in B$ but there is no interval $(-\epsilon, \epsilon) \subseteq B$.
It is not compact because it has an open cover by intervals of the form $(0, 1-1/n)$, but this cover has no finite subcover.
A: HINT: It may be helpful to write out a few members of $B$:
$$B=\left\{0,\frac12,\frac23,\frac34,\dots\right\}\;.$$


*

*To help you explain why $B$ is not open: is there any $\epsilon>0$ such that $(-\epsilon,\epsilon)\subseteq B$?  

*To help you explain why $B$ is not closed: can you find a limit point of $B$ that is not in $B$?

A: $B\subset \mathbb{R}$ with the usual topology. $B$ is not open since (for example) $0\in B$ is not an interior point since every neighborhood of $0$ intersects $B^c$. $B$ is not closed since $1$ is a limit point not contained in $B$ (a sequence in $B$ converging to 1 is $\{1-1/n\}$). In a metric space, compact sets are closed. $B$ is not closed, hence not compact. (Edit: I took out the bit of B being a metric space itself. That was not relevant.)
