# Find the closure of ${1,\frac12,\frac13,\cdots, \frac1n,\cdots}$ in $\mathbb{R}$ with usual topology

My understanding is this:

Let A=$\left\{1,\frac12,\frac13,\cdots, \frac1n,\cdots\right\}$

$0$ is a limit point of A, since any open set containing $0$ in $\mathbb{R},U$ contains a point of A different from $0$. But $0 \notin A$, so $A$ is not closed.

So $\bar{A} \neq A$.

My guess is $\bar{A}=A\cup A'$=$\left\{0,1,\frac12,\frac13,\cdots, \frac1n,\cdots\right\}$

Is this correct?

• Yes, you are correct. – Javi Apr 2 '18 at 12:23
• Just a tip - you need to say that $0$ is the only limit point outside of the set, otherwise you are correct – asdf Apr 2 '18 at 12:25
• Your guess is correct. However the proof is incomplete. Indeed, $0$ is a limit point of $A$. What you are missing is that there are no other limit points. – freakish Apr 2 '18 at 12:25

Your guess is correct, but you should also check that no other number is in $A'$.
For $x>1$, take $\delta=|x-1|/2$, hence $x\in B(x,\delta)$, whose intersection with $A$ is empty.
For $x\in (0,1)\setminus A$, let $n$ such that $\frac{1}{n+1}<x<\frac{1}{n}$. Take $\delta=\min\{|x-\frac{1}{n+1}|,|x-\frac{1}{n}|\}/2$. Then, again $x\in B(x,\delta)$, with empty intersection with $A$.
For $x<0$, similar to the case $x>1$, but taking $\delta=|x-0|/2=|x|/2$.