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What shown below is a reference from "Elementos de Topología General" by Fidel Cassarubias Segura and Ángel Tamariz Mascarúa

Lemma

If $X$ is continuum and if $A$ is closed in $X$ then $C\cap\partial A\neq\varnothing$ for any component $C$ of $A$.

Proof. Let be $x_0\in C$ and so we consider the collection $\mathcal{F}$ of clopen sets of $A$ that contain $x_0$. So we know that $C=\bigcap\mathcal{F}$. Therefore we suppose that $C\cap\partial A=\varnothing$. So since $\mathcal{F}$ has the finite intersection property and since $\partial A$ is compact then there exist $F\in\mathcal{F}$ such that $F\cap\partial A=\varnothing$. So we choose an open set $U$ of $X$ such that $U\cap A=F$. So if $F\cap\partial A=\varnothing$ then $F=U\cap A^°$ and so $F$ is open in $X$; however $F$ is closed in $X$ and so if $X$ is connected it follows that $F=X$ and so $\partial A=\varnothing$ that is impossible.

For sake of completeness here is the original text of the proof: I hope mine was a good translation.

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Well I don't understand why there exist $F\in\mathcal{F}$ such that $F\cap\partial A=\varnothing$. So could someone explain this to me, please?

Henno Brandsma
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Antonio Maria Di Mauro
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1 Answers1

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We are assuming that $C\cap\operatorname{bdry}A=\varnothing$. Suppose that $F\cap\operatorname{bdry}A\ne\varnothing$ for each $F\in\mathscr{F}$. Then $\mathscr{F}_0=\{F\cap\operatorname{bdry}A:F\in\mathscr{F}\}$ would be a family of non-empty closed subsets of the compact set $\operatorname{bdry}A$ having the finite intersection property, so $\bigcap\mathscr{F}_0$ would be non-empty. But then

$$\varnothing=C\cap\operatorname{bdry}A=\left(\bigcap\mathscr{F}\right)\cap\operatorname{bdry}A=\bigcap\mathscr{F}_0\ne\varnothing\;,$$

which is absurd.

Brian M. Scott
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  • Okay, but unfortunately I don't understand why $\mathscr{F}_0$ has the finite intersection property: so could you explain better, please? – Antonio Maria Di Mauro Apr 20 '20 at 21:18
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    @AntonioMariaDiMauro $\mathcal{F}$ is even closed under finite intersections, so $\forall F \in \mathcal{F}: F \cap \partial A \neq \emptyset$ is equivalent to FIP of $\mathcal{F}$ restricted to $\partial A$ – Henno Brandsma Apr 20 '20 at 21:26
  • @HennoBrandsma Why? I don't see this. – Antonio Maria Di Mauro Apr 20 '20 at 21:27
  • @AntonioMariaDiMauro the definition of FIP, plus the fact that a finite intersection of clopens containing $x_0$ is again a clopen containing $x_0$. – Henno Brandsma Apr 20 '20 at 21:28
  • Okay, so we know that $\mathscr{F}$ has FIP, since $x_0\in F$ for any $F\in\mathscr{F}$; then any finite intersection of clopens containing $x_0$ is again a clopen containig $x_0$ and so for any finite subcollection $\mathscr{G}$ of $\mathscr{F}$ it follows that $\bigcap\mathscr{G}\in\mathscr{F}$ and so $\bigcap\mathscr{G}\cap\partial A\neq\varnothing$, right? – Antonio Maria Di Mauro Apr 20 '20 at 21:30
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    @AntonioMariaDiMauro Suppose $F_1 \cap \partial A, \ldots F_n \cap \partial A$ are finitely many sets in $\mathscr{F}0$. Then $\bigcap{i=1}^n F_i \in \mathscr{F}$ so by assumption $\emptyset \neq (\bigcap_{i=1}^n F_i) \cap \partial A = \bigcap_{i=1}^n (F_i \cap \partial A)$. – Henno Brandsma Apr 20 '20 at 21:33
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    @AntonioMariaDiMauro: (Sorry: I was away for a bit.) Let me expand what Henno said a bit. Suppose that $\mathscr{H}\subseteq\mathscr{F}$ is finite; each $F\in\mathscr{H}$ is a clopen set in $A$ containing $x_0$, so $\bigcap\mathscr{H}$ is a clopen set in $A$ containing $x_0$. Thus, $\bigcap\mathscr{H}\in\mathscr{F}$. If $F\cap\operatorname{bdry}A\ne\varnothing$ for each $F\in\mathscr{F}$, then $\bigcap\mathscr{H}\cap\operatorname{bdry}A\ne\varnothing$. – Brian M. Scott Apr 20 '20 at 21:34
  • Okay, I understood. Thanks to both!!! – Antonio Maria Di Mauro Apr 20 '20 at 21:37
  • @AntonioMariaDiMauro: You’re welcome! – Brian M. Scott Apr 20 '20 at 21:38
  • @AntonioMariaDiMauro onwards to the next question... – Henno Brandsma Apr 20 '20 at 21:41
  • @HennoBrandsma Surely! – Antonio Maria Di Mauro Apr 20 '20 at 21:45