Proving a set is open/closed?

Question

I need to prove that $S = \{1/n : n\in \mathbb{N} \}$ is neither open nor closed.

To prove that this is not open, I began by using contradiction and assumed that the set is open. Since $1/n$ ranges from $(0,1]$, let $y=supS$. By definition, $y≥s$ for all $s \in S$. So, there cannot exist a $\epsilon >0$ such that $(y- \epsilon,y+ \epsilon)$ is contained in $S$.

Is there a better way to word this?

To prove that this is not closed, I would need to prove that the complement is not closed. But how would I do that?

Answer 1

Well, $\sup S=1$. But you are right: $S$ contains no interval of the form $(1-\varepsilon,1+\varepsilon)$, and therefore it os not open.

And it is not closed because$$S^\complement=(-\infty,0]\cup\left(\frac12,1\right)\cup\left(\frac13,\frac12\right)\cup\ldots$$This set is not open, since $0$ belongs to it, but it contains no interval e(-\varepsilon,\varepsilon)$.

Answer 2

I am not sure what you are trying to say here. You start by talking about y= sup(S). Why not just say "1"? And since the set is {1, 1/2, 1/3, …} 1 is in the open interval (3/2,1/2) which contains no other members of S. 1 is not an interior point so S is not open. For the other part, the sequence 1, 1/2, 1/3, … converges to 0 so the closure of S includes 0 which is not in S so S is not closed.