How to show $\{1/n : n \in\mathbb Z_+\} \subset \mathbb R$ is neither closed nor open I am getting stuck when I try to think of how would one go about showing that this set below is not closed.
$X = \{1/n : n \in \mathbb Z_+\} \subset \mathbb R$
I know that this set cannot be open, as any finite collection of points in $R$ is not open.
Additionally, what would happen if we perform the operation $X\cup\{0\}$? Would it become a closed set?
 A: $X$ is not closed, precisely because $0$ is a limit point of $X$ that does not lie in $X$. And it's in fact the only such point, adding $0$ to $X$ does make it a closed set. 
And as all intervals (that form a base for the topology of $\Bbb R$) are  uncountable so are all open sets. $X$ being countable (not finite) cannot be open.
A: 1) It's closed if all the limit points of $X$ are points of $X$.  So it's not closed if there is at least one limit point of $X$ that isn't in $X$.  What are the limit points of $X$?  Show one of them isn't in $X$?  (Actually $X$ has exactly one limit point and it is not in $X$.  What is it?
2) $X$ is open if every point of $x$ has a neighborhood that is a subset of $X$.  So it's not open if there is at least one point that doesn't have a neighborhood that is a subset of $X$.  Can you find such a point in $X$?  (Actually all point are like that.)  (Hint.  An open neighborhood of $x$ is the open interval $(x-\epsilon, x+\epsilon)$.  Is there any $\frac 1n \in X$ so that $(\frac 1n-\epsilon,\frac 1n +\epsilon) \not\subset X$?)
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1) If you had to guess what a limit point of $X= \{\frac 1n\}$ is what would it be?
If you guessed $0$ you are right.
Prove $0$ is a limit point:
Let $\epsilon$ be any real, $\epsilon >0$ and and $n$ be any natural where$n > \frac 1{\epsilon}$ then $0 < \frac 1n < \epsilon$ and $d(0,\frac 1n) < \epsilon$ so $\frac 1n\in (0-\epsilon,0+\epsilon)$ and $0$ is a limit point.
And $0\not \in X$ so $X$ is not closed.
What if we hadn't guessed that $0$ was a limit point?  Well, then we'd have our work cut out for us.
Let's try to find a limit point of $X$.  Let $x \in \mathbb R$ and let $r_n = d(x,\frac 1n)$ (assuming $x\ne \frac 1n$).  $0\le r_n$ so $\{r_n\}$ is bounded below by $0$.  So let $e=\inf \{r_n\}$ exists. If $e > 0$ then $(x-e, x+e)$ has not values of $X$ and so $x$ is not a limit point unless $\inf\{r_n\} = \inf|x-\frac 1n| = 0$. Now if $x\ge 1$ and $\frac 1n \ne x$ then $|x-\frac 1n|>\frac 12$.  If $\frac 1{k+1} < x < \frac 1k$ for some $k$ then $|x-\frac 1n|\ge \min(x-\frac 1{k+1},\frac 1k -x)> 0$ and if $x = \frac 1k$ then $|x-\frac 1n| \ge \frac 1k - \frac 1{k+1} > 0$ and if $x < 0$ then $|x-\frac 1n|> |x|>0$.  
So the only possible limit point is $x =0$.
......
2).  This is easier.  Let $\frac 1n \in X$.  Then $\frac 1{n+1}< \frac 1{n} $.
If $\frac 1{n+1} < x < \frac 1n$ then $x\not \in X$.  So there for any possible $\epsilon > 0$ we have $(\frac 1n - \epsilon, \frac 1n + \epsilon)\not \subset X$ because there will always be an $x < \frac 1n$ that is not in $X$.
So $X$ is not open.
