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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.

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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 ($\forall x,y \in X: (x < y) \to (\exists z\in X: x < z <y)$). So also for $\Bbb Q$ e.g.

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  • $\begingroup$ Thanks for the reply. I can't accept it because I already accepted the one before. But thanks a lot $\endgroup$ Commented Nov 9, 2019 at 18:08
  • $\begingroup$ Ohh turns out I can, so I accepted your answer. Thank you. $\endgroup$ Commented Nov 10, 2019 at 1:57
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Your argument is correct indeed. And it also works for $\mathbb Q$. And for $\mathbb R^n$.

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  • $\begingroup$ Never mind. I got why it works for n-dimensional Real space. Thanks though. $\endgroup$ Commented Oct 2, 2019 at 12:08
  • $\begingroup$ 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.) $\endgroup$ Commented Oct 2, 2019 at 12:39
  • $\begingroup$ Okay cool. Thanks David $\endgroup$ Commented Oct 2, 2019 at 13:35

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