# Is set of all real numbers dense in itself?

According to Wikipedia, "In mathematics, a subset of a topological space is said to be dense-in-itself if it contains no isolated points."

I think $$R$$ is dense in itself because $$R$$ contains all its limit points as any non-empty open set in $$R$$ will have a neighbourhood of $$x$$ (a limit point of that set), will contain elements of $$R$$ other than itself. Maybe I'm wrong, please be kind. This is just what I thought of as reason for $$R$$ being dense in itself. Is this the right reason? Is $$R$$ dense in itself? Thanks in advance.

You;re right; if you want to be formal about it:

If $$x\in \Bbb R$$ were an isolated point, then there'd be an $$r>0$$ such that $$(x-r,x+r) \subseteq \{x\}$$, but this is nonsense, as $$x + \frac{r}{2} \in (x-r,x+r)$$ but $$x+\frac{r}{2} \notin \{x\}$$. This contradiction shows that $$\Bbb R$$ has no isolated points.

This argument works (in adapted form) for any ordered space $$X$$ with a dense order ($$\forall x,y \in X: (x < y) \to (\exists z\in X: x < z ). So also for $$\Bbb Q$$ e.g.

• Thanks for the reply. I can't accept it because I already accepted the one before. But thanks a lot Commented Nov 9, 2019 at 18:08
• Ohh turns out I can, so I accepted your answer. Thank you. Commented Nov 10, 2019 at 1:57

Your argument is correct indeed. And it also works for $$\mathbb Q$$. And for $$\mathbb R^n$$.

• Never mind. I got why it works for n-dimensional Real space. Thanks though. Commented Oct 2, 2019 at 12:08
• As it stands, your argument consists of a chain of two implications, one false, the other rather incoherent, but the intended meaning is false. However the basic idea is right. Just delete "because R contains all its limit points" and tidy up the rest. (Something like: as every neighbourhood of any point x contains points of R other than x.) Commented Oct 2, 2019 at 12:39
• Okay cool. Thanks David Commented Oct 2, 2019 at 13:35