Is the set $B=\{\frac{1}{n}\mid n\in Z^+\}$ open or closed?

Question

Given the set $B=\{\frac{1}{n}\mid n\in Z^+\}$, decide whether it's open or closed. So far, I worked out that it is closed since it contains all its limit points, starting at $1$. These limit points would be the discrete elements in the set itself, no? Thanks for help in advance!

Answer 1

I preassume that $B$ must be looked at as a subset of $\mathbb R$ and that $\mathbb R$ is equipped with its usual (order)topology.

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If $(a,b)$ is an interval with $0\in(a,b)$ then $0<b$ so that - for $n$ large enough - we have $\frac1n\in(0,b)\subseteq(a,b)$.

This makes clear that $0$ is a limitpoint of $B$ and $0\notin B$ justifies the conclusion that $B$ is not closed.

If the set would be open then some $\epsilon>0$ must exist such that $(1-\epsilon,1+\epsilon)\subseteq B$. This is evidently not the case so we conclude that $B$ is not open.

Answer 2

The answer to both is no.

Since $0\not\in B$, we have that $B$ is not closed.

Since you cannot find some $r>0$ such that $B_r(1)\subseteq B$ (because $1+\frac{r}{2}\not\in B$), we have that $B$ is not open in $\mathbb{R}$.