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.

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

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