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In Introductory Real Analysis by Kolmogorov and Fomin the following definitions are given:

Let A and B be two subsets of a metric space $R$. Then $A$ is said to be dense in $B$ if $B \subset [A]$. In particular, $A$ is said to be everywhere dense (in $R$) if $[A] = R$. A set $A$ is said to be nowhere dense if it is dense in no (open) sphere at all.

Here my question. If $A$ is dense in $B$, should not be $[A]=B$? Further, it is not clear to me the definition of nowhere dense.

Asaf Karagila
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Paolo Secchi
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    If $A$ is a subset of $B$ and the closure is taken in $B$, then you are right. But the authors don't assume that $A$ is a subset of $B$ and they are taking closure in the bigger space $R$. Eg. $\mathbb Q$ is dense in $(0,1)$. – Kavi Rama Murthy Sep 13 '24 at 09:46
  • @geetha290krm: the authors don't assume that $A$ is a subset of $B$ --- Even if $A \subset B,$ we still might have $[A] \neq B.$ Let $R$ be the real numbers and let $A$ be the rational numbers and let $B$ be the algebraic numbers. (Paolo Secchi) For "nowhere dense", maybe this MSE answer will help. – Dave L. Renfro Sep 13 '24 at 10:44
  • @DaveL.Renfro Why not just $B=(0,1)$ and $B= A \cap \mathbb Q$ (in $R$). – Kavi Rama Murthy Sep 13 '24 at 11:26
  • @geetha290krm: Yes, that also works. It seems I included more layers of density than needed. – Dave L. Renfro Sep 13 '24 at 16:12
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    What is the notation $[A]$? Never seen it. – Jakobian Sep 13 '24 at 19:11
  • @Jakobian: I haven't seen this notation either, or at least I don't recall having seen it before. But it's clear from context that $[A]$ means the (topological) closure of $A.$ – Dave L. Renfro Sep 13 '24 at 20:09
  • @DaveL.Renfro Yes. I checked the book and the notation is really there. What an odd choice by the authors. – Jakobian Sep 13 '24 at 20:27
  • @DaveL.Renfro Oddly enough, I see that in precisely one place, when talking about relative compactness, they did write $\overline{M}$. And in all other places they write $[M]$. It's really weird. – Jakobian Sep 13 '24 at 20:37

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