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If $(X,\mathcal{T})$ is a perfect space and $A$ is either an open set or a dense set in $(X,\mathcal{T})$, then $A$ has no isolated points.

Proof: $A$ - open, then $\mathcal{T}_A = \{ A \cap U : U \in \mathcal{T}\}$. Note that every element of $\mathcal{T}_A$ is open in $\mathcal{T}$ and $\mathcal{T}_A \subseteq \mathcal{T}$. Since $\mathcal{T}$ does not contain isolated points, $\mathcal{T}_A$, as a subset, also does not contain isolated points. End of proof.

Could you help me to prove this for the dense subset?

I'm thinking about counterexample. Consider a topological space $X = \{a, b, c, d, e\}$ and $\mathcal{T} = \{\emptyset, \{a, b, c, d, e\}, \{b, c, d, e\}, \{b, c\}, \{d, e\} \}$.

Then $(X, \mathcal{T})$ is perfect - it does not contain open singleton sets. $A = \{b, d\}$ is dense in $X$. But then $A$ has isolated points $b$ and $d$: $\mathcal{T}_A = \{ A \cap U : U \in \mathcal{T}\} = \{\emptyset, \{b, d\}, \{b\}, \{d\}\}$?

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As your (correct) example shows, the dense part does not hold always.

It does hold if $X$ is $T_1$, which is for most practical spaces: Let $D$ be dense and suppose $\{d\}$ is open in $D$ for some $d \in D$, so that $U \cap D = \{d\}$ for some open $U$ in $X$.

As $X$ has no isolated points, there is some $x \in U$ with $x \neq d$. As $\{d\}$ is closed in $X$ ($T_1$ ensures this), $U \setminus \{d\} = U \cap (X\setminus\{d\})$ is open and non-empty ($x$ is in it) in $X$ while $(U \setminus \{d\}) \cap D = (U \cap D)\setminus \{d\} = \emptyset$. This contradicts $D$ being dense. So $D$ has no isolated points.

Maybe your text assumes by definition that perfect spaces are $T_1$?

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  • $\begingroup$ It does not have any explicit assumptions. Probably they assume that it is a metric space, because the problem is in the chapter dedicated to the metric spaces. $\endgroup$ – Andreo Apr 22 '18 at 23:46
  • $\begingroup$ @Andreo then $T_1$ is certainly true. $\endgroup$ – Henno Brandsma Apr 23 '18 at 3:48

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