# Is $S=\{0, 1, 1/2, 1/3…, 1/n,…\}$ closed set of natural topology of $\mathbb{R}$?

Is $S=\{0, 1, 1/2, 1/3..., 1/n,...\}$ closed set of natural topology of $\mathbb{R}$?

Since $\mathbb{R}-S = (-\infty, 0) \cup (0,1) \cup (1,1/2)...$ I thought $S$ is a closed set of natural topology of $\mathbb{R}$, but then I realized all of these open intervals includes irrational numbers, but since irrational numbers does not subset of rational numbers and those open intervals include irrational numbers, I thought this is not a closed set of $\mathbb{R}$.

If anyone can explain this, it would be great.

• Your logic was fine until you started talking about irrationals (which have nothing to do with this). You've written $\mathbb{R}-S$ as a union of open sets, so $\mathbb{R}-S$ is open. So, it's complement is closed. – Batman Nov 19 '17 at 14:54
• I think that your expression of $\mathbb{R}-S$ as a union of open intervals is not quite right. It should not include $(0, 1)$ for example. Also it is normal to write intervals with their left endpoint before their right endpoint, e.g. $(1/2,1)$ rather than $(1,1/2)$. How about writing $\mathbb{R}-S=(-\infty,0) \cup (1,\infty) \cup (1/2,1) \cup (1/3, 1/2) \cup \ldots$ ? – Simon Nov 21 '17 at 10:55

$\mathbb R-S$ is the union of open sets, and so is open. Thus $S$ is closed, since it’s the complement of an open set.
I’m not sure why you brought up irrationals in your argument, but everything before that is right. Yes, $\mathbb R-S$ contains irrationals, but that doesn’t change anything.
• @GLHF No, why would it? Every interval in $\mathbb R$ that contains more than one number contains both irrational and rational numbers. – Stella Biderman Nov 19 '17 at 14:59