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 $d$: $\mathcal{T}_A = \{ A \cap U : U \in \mathcal{T}\} = \{\emptyset, \{b, d\}, \{b\}, \{d\}\}$?


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 $T_1$?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.