# What are the closure and interior of this subset of $\Bbb R$?

Consider in $(\mathbb{R} ,\tau_e)$ ($\tau_e$ is the Euclidean topology) the following subset:

$$X=\{x \in \mathbb{R}:x= \frac{p}{10^q},\; p,q \in \mathbb{Z}\}.$$

Decide if:

(i) $X$ is open in $(\mathbb{R} ,\tau_e)$ and find its interior, and

(ii) $X$ is closed in $(\mathbb{R} ,\tau_e)$ and find its closure.

$X$ is the set formed by rational non-periodic numbers... I think it is not open because every interval in the Euclidean topology contains a periodic rational number, so $\operatorname{Int}(X)=\emptyset$.

To see if $X$ is closed: $C_{\mathbb{R}}(X)=$ periodic rational numbers $\cup$ $\mathbb{R} \setminus \mathbb{Q}$... and is this open? I don't think but I can't show it... so $\operatorname{cl}(X)=X$.

Surely $X$ is not open as $X$ is only countable. Hence its interior is empty.
$X$ is also not closed because it does not contain all its limit points. You can find a sequence of points in $X$ converging to $\pi$, say: $$3/1,\,31/10,\,314/100,\,3141/1000,\dots$$
For the closure, note that any dyadic rational number of the form $m/2^n$ for $m,n\in\Bbb Z$ can be expressed as an element of $X$ since $2^n = 10^n/5^n$, so $m/2^n = 5^nm/10^n$. Since the dyadic rational numbers $D$ are dense in $\Bbb R$, and $D\subset X$, we have $\operatorname{cl}(X) = \Bbb R$, i.e., the closure of $X$ is $\Bbb R$.
• Thanks for the answer. Is my motivation for the "interior" right? I didn't know that property of the dyadic ration numbers... can I show that $C_{\mathbb{R}}(X)$ is not open? (I can't do this..) – VoB Jul 13 '17 at 17:34
• @feddy What do you mean by $C_\Bbb{R}(X)$? I'm not familiar with the notation. I am not sure how you could prove that every interval in $\Bbb{R}$ contains a periodic rational number, but if you could, and supposing that $X$ truly does consist of non-periodic rationals, then you would have shown that the interior is empty, yes. – Alex Ortiz Jul 13 '17 at 17:36
• With that I mean the complementar of $X$ in $\mathbb{R}$. The fact is that I don't know how to prove it. And for the closure... is there a way to show that $C_{\mathbb{R}}(X)=$ is an open set? – VoB Jul 13 '17 at 17:39
• @feddy Ok, I understand now. There is no way to show that $C_\Bbb{R}(X)$ is open because this would imply $X$ is closed, which it is not. – Alex Ortiz Jul 13 '17 at 19:24