# Proving $E\subseteq\mathbb{R}^n$ satisfies Heine-Borel Property if and only if its FIP

How would do we prove $$F_\alpha \ne \emptyset$$? I am not sure how to fully prove the problem, so can I please receive help? Thank you.

$$\def\R{{\mathbb R}} \def\Z{{\mathbb Z}} \def\N{{\mathbb N}}$$ Prove $$E\subseteq\R^n$$ satisfies the Heine-Borel Property if and only if it satisfies the Finite Intersection Property such that given any collection of closed sets $$\{F_\alpha\}_{\alpha\in\mathcal{I}}$$ such that for every finite set $$\{\alpha_1,\,\alpha_2,\, \dots,\,\alpha_n\}\subseteq\mathcal{I}$$, $$\displaystyle{E\cap \bigcap_{i=1}^n F_{\alpha_i}\ne\emptyset}$$, then $$\displaystyle{E\cap \bigcap_{\alpha\in\mathcal{I}} F_{\alpha}\ne\emptyset}$$.

$$\textbf{Proof:}$$ Suppose $$E$$ is compact and $$\{F_\alpha\}_{\alpha\in \mathcal{I}}$$ is a family of closed sets of $$E$$ having the property, for every finite set $$\{\alpha_1,\,\alpha_2,\, \dots,\,\alpha_n\}\subseteq\mathcal{I}$$, $$\displaystyle{E\cap \bigcap_{i=1}^n F_{\alpha_i}\ne\emptyset}$$. To show that $$\displaystyle{E\cap \bigcap_{\alpha\in\mathcal{I}} F_{\alpha}\ne\emptyset}$$, assume $$\displaystyle{\bigcap_{\alpha\in \mathcal{I}} F_\alpha = \emptyset}.$$ Then, $$\displaystyle{\bigcup_{\alpha\in\mathcal{I}} (E-F_\alpha) = E}.$$

Since, $$F_\alpha$$ is closed in $$E$$ for all $$\alpha \in \mathcal{I}$$, therefore, $$(E-F_\alpha)$$ are open in $$E$$ for all $$\alpha \in \mathcal{I}.$$ Therefore, $$\{(E-F_\alpha : \alpha \in \mathcal{I}\}$$ is an open cover of $$E$$. Since, $$E$$ is compact and satisfies Heine-Borel Property, there exists $$\alpha_1,\,\alpha_2,\, \dots,\,\alpha_n \in \mathcal{I}$$ such that $$(E-F_{\alpha_1}) \cup (E-F_{\alpha_2}) \cup \dots \cup (E-F_{\alpha_n}) = X.$$ Hence, $$E\cap \bigcap_{i=1}^n F_{\alpha_i}\ne\emptyset$$ is a contradiction. Therefore, $$\displaystyle{E\cap \bigcap_{\alpha\in\mathcal{I}} F_{\alpha}\ne\emptyset}$$.

Conversely, let each family of closet sets of $$E$$ have finite intersection property. To show each open cover of $$E$$ has finite subcover, i.e., $$E$$ satisfies the Heine-Borel property. Let $$y$$ be an open covering of $$E$$. Then $$\displaystyle{\bigcup_{G\in y} G = E}$$, which implies $$\displaystyle{\bigcap_{G\in y} (E-G) = \emptyset}$$.

Hence, the family of closed sets $$\{(E-G) : G\in y\}$$ has empty intersection. By hypothesis, $$\{(E-G): G\in y\}$$ cannot have finite intersection property. Therefore, there exists $$G_1, G_2, \dots, G_n \in y$$ such that $$(E-G_1) \cap \dots \cap (E-G_n) = \emptyset$$. Thus, implying $$\displaystyle{\bigcup_{i=1}^n G_i = E}$$. Therefore, $$\{G_1, G_2, \dots, G_n\}$$ is a finite subcover of $$y$$.

It's most convenient to just work inside $$E$$, so property 1 is:

Whenever $$F_i, i \in I$$ is a family of (relatively) closed subsets of $$E$$ that has the FIP, then $$\bigcap_i F_i \neq \emptyset$$

and Heine-Borel is just

Whenever $$U_i, i \in I$$ is a (relatively) open cover of $$E$$ then we have a finite subcover.

(a relatively closed subset is of the form $$F \cap E$$ with $$F$$ closed in the ambient space, and likewise for relatively open sets).

Suppose the FIP-property holds for $$E$$. Let $$U_i, i \in I$$ be an open cover in $$E$$. Define $$F_i = E - U_i$$ which are closed in $$E$$. $$\bigcap_i F_i = E- \bigcup_i U_i = \emptyset$$ by de Morgan inside $$E$$, so $$F_i, i \in I$$ does not have the FIP, so $$F_{i_1}, \ldots F_{i_n}$$ exist with tempty intersection, which means that the $$U_{i_1}, \ldots, U_{i_n}$$ cover $$E$$. As the cover was arbitrary, Heine-Borel holds for $$E$$.

Suppose Heine-Borel holds for $$E$$. Let $$F_i, i \in I$$ have FIP. Define $$U_i = E- F_i$$, open in $$E$$. No finite subset of the $$U_i$$ covers $$E$$, because the corresponding $$F_i$$ would have empty intersection, which they don't. So $$U_i, i \in I$$ is not a cover of $$E$$, so $$\bigcap_I F_i \neq \emptyset$$ and $$E$$ has the FIP property.

• Thank you for the explanation Henno! – rudinsimons12 Apr 3 '20 at 14:32