# Show $\{-n + 1/n \ : n \in \mathbb{N}\}$ is a closed set

I have the set

$$A = \{-n + 1/n \ : n \in \mathbb{N}\}$$

My attempt

I tried to find some limit point in A, but

$$\lim_n (-n + 1/n) = -\infty$$

Is there anyone to help?

• Can you show that $A$ has only finitely many elements in any bounded set? – Lord Shark the Unknown Oct 13 '18 at 7:03
• @LordSharktheUnknown bounded subset? or $A \cap B$ is finite, and B is a bounded set? – user198044 Oct 13 '18 at 7:04
• I have this suspicion that $A' = \emptyset \subseteq A$ ? If I'm right, then it can be proved if $\forall n \in \mathbb N, -n + \frac1n$ is isolated. – user198044 Oct 13 '18 at 7:04

The complement of $$A$$ is $$\bigcup_{n\in\mathbb N}(-n-1+\frac1{n+1},-n+\frac1n)\bigcup(0,+\infty)$$, which is a union of open intervals, thus open. Therefore $$A$$, as the complement of an open subset, is closed.
If you want to prove that your set is closed using sequences, you can use the fact that, precisely because $$\lim_{n\to\infty}-n+\frac1n=-\infty$$, the only convergent sequences of numbers of the firm $$-n+\frac1n$$ are those that are constant after a certain point. And every such sequence converges to another element of your set, obviously.
The intersection of the set with any compact subset (i.e. closed and bounded) of $$\mathbb{R}$$ is finite hence closed. As metrics spaces are $$k$$-spaces, the set is closed.