# Proof that $\{x \}$ is nowhere dense if and only if $x$ is not an isolated point of $X$.

The following is a proof of the result in the title of the question.
I am writing it for various reasons, in particular to check if the proof is indeed correct, and to get a feedback on my writing skills.

Proposition: $\{x \}$ is nowhere dense if and only if $x$ is not an isolated point of $X$.

Proof:
$[\Rightarrow]$ Assume that int(cl($\{ x \})) = \varnothing$. Thus, $X \neq \{ x \}$ and the result trivially holds.
$[\Leftarrow]$ Assume that $x$ is not an isolated point of $X$. Thus, for every $\epsilon > 0$, $B_\epsilon (x) \cap X \neq \varnothing$. Hence, $X$ is not a singleton. To prove that int(cl($\{ x \})) = \varnothing$, notice that it is the same to prove cl($X \setminus \{ x \}) = X$. Thus, let $y \in M$ be arbitrary. We have to prove that, for every $\epsilon>0$, $B_\epsilon (x) \cap (X \setminus \{ x \}) \neq \varnothing$, but this holds trivially if $X$ is not a singleton. Hence, the result is established. QED

Any feedback is most welcome!

• You seem confused about the meaning of the terms involved. Nothing in the question or the argument should be about whether or not $X$ is a singleton, that is irrelevant to the problem. – Andrés E. Caicedo Dec 30 '14 at 15:50
• @AndresCaicedo: First of all, thanks for your feedback. Thus, is the "proof" completely wrong? Could you please tell me what is a wrong inferential step? Just because to me they looked - and still look - ok, thus I cannot do anything particularly useful with the feedback as it stands. :) – Kolmin Dec 30 '14 at 15:54
• Yes, both directions of the proof are completely wrong. Begin by looking up the right definition of "isolated point" and make sure to include it explicitly. – Andrés E. Caicedo Dec 30 '14 at 15:57
• My line of reasoning was that $x$ is not an isolated point, thus it is a fortiori a limit point. Indeed, the definition of limit point says that $x \in X$ is a limit point of a set $A \subseteq X$ iff for every $\epsilon >0$, $(B_\epsilon (x)\setminus \{x \} ) \cap A \neq \varnothing$. Then, an isolated point of a set $A \subseteq X$ is a point $x \in A$ that it is not a limit point of $A$. – Kolmin Dec 30 '14 at 16:05
• In this case, taking the previous definitions, $A = X$ and the result to be proven should be int(cl($\{x\}) = \varnothing \Longleftrightarrow \forall \epsilon > 0 ( B_\epsilon (x) \setminus \{x \} ) \cap X \neq \varnothing)$. – Kolmin Dec 30 '14 at 16:07

## 1 Answer

I’ll give a fairly detailed commentary on your argument.

$[\Rightarrow]$ Assume that int(cl($\{ x \})) = \varnothing$. Thus, $X \neq \{ x \}$ and the result trivially holds.

It’s true that if $\operatorname{int}\operatorname{cl}\{x\}=\varnothing$, then $X\ne\{x\}$, but this by no means implies that $x$ is not an isolated point. For a simple counterexample, note that $\Bbb Z\ne\{0\}$, but $0$ certainly is an isolated point of $\Bbb Z$. The simplest approach here is to prove the contrapositive: assume that $x$ is an isolated point, and show that $\{x\}$ is not nowhere dense. This actually is trivial: if $x$ is isolated, then $\{x\}$ is an open subset of $\operatorname{cl}\{x\}$, so $\operatorname{int}\operatorname{cl}\{x\}\ne\varnothing$.

$[\Leftarrow]$ Assume that $x$ is not an isolated point of $X$. Thus, for every $\epsilon > 0$, $B_\epsilon (x) \cap X \neq \varnothing$.

$B_\epsilon(x)\cap X\ne\varnothing$ for each $\epsilon>0$ even if $x$ is an isolated point of $X$. What you want here is that $B_\epsilon(x)\cap\big(X\setminus\{x\}\big)\ne\varnothing$ for each $\epsilon>0$.

Hence, $X$ is not a singleton. To prove that int(cl($\{ x \})) = \varnothing$, notice that it is the same to prove cl($X \setminus \{ x \}) = X$. Thus, let $y \in M$ be arbitrary.

I assume that $M$ is a typo for $X$.

We have to prove that, for every $\epsilon>0$, $B_\epsilon (x) \cap (X \setminus \{ x \}) \neq \varnothing$,

No, you have to show that $B_\epsilon(y)\cap\big(X\setminus\{x\}\big)\ne\varnothing$ for each $\epsilon>0$.

but this holds trivially if $X$ is not a singleton.

This is sufficiently lacking in detail that I can’t tell whether you have in mind a correct argument or not; you need to expand it a bit. Something like this would do:

If $y\in X\setminus\{x\}$, then obviously $y\in B_\epsilon(y)\cap\big(X\setminus\{x\}\big)\ne\varnothing$ for each $\epsilon>0$, and by hypothesis $B_\epsilon(x)\cap\big(X\setminus\{x\}\big)\ne\varnothing$ for each $\epsilon>0$, so $B_\epsilon(y)\cap\big(X\setminus\{x\}\big)\ne\varnothing$ for each $\epsilon>0$, and $\operatorname{cl}\big(X\setminus\{x\}\big)=X$, as desired.

• First of all something mundane: let me say that this answer is absolutely brilliant, and I was seriously thinking to put a bounty on this question because it is on a topic that is fairly important, and that it is obvious I don't understand properly. – Kolmin Dec 31 '14 at 18:53
• Regarding the answer, one point here, and two questions shortly after, to be sure I see what's going on. Yes, indeed $M$ is a typo, and also the $x$ in $B \in (x) \cap (X \setminus \{x\}) \neq \varnothing$ is a typo (true, far more dangerous). – Kolmin Dec 31 '14 at 19:01
• @Kolmin: By definition a point $x$ is isolated if and only if $\{x\}$ is an open set. In the metric setting this means that $x$ is an isolated point if and only if there is an $\epsilon>0$ such that $B_\epsilon(x)=\{x\}$. – Brian M. Scott Dec 31 '14 at 19:05
• @Kolmin: Yes, with the changes that I suggested it becomes a perfectly fine proof of the implication in that direction. – Brian M. Scott Dec 31 '14 at 19:07
• @Kolmin: That’s right: a singleton can be open or not, depending on the space. For example, every singleton in $\Bbb Z$ is open, but no singleton in $\Bbb R$ is open. In the subspace $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ of $\Bbb R$, every singleton except $\{0\}$ is open. – Brian M. Scott Dec 31 '14 at 19:12