Help in understanding proof of Heine-Borel Theorem from Simmons

Question

I am reading G. F. Simmons' Introduction to Topology and Modern Analysis, and in Section 21 (Compactness), on page 114, he proves the Heine-Borel Theorem (Theorem G). I have some trouble in understanding the proof. It would be great if someone could help me out.

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A subbasic open cover is an open cover whose sets are all in some given open subbase. A class of sets is called a closed subbase if the class of all complements of its sets is an open subbase. The elements of a closed subbase are called subbasic closed sets.

Theorem A. Any closed subspace of a compact space is compact.

Theorem F. A topological space is compact if every subbasic open cover has a finite cover, or equivalently, if every class of subbasic closed sets with the finite intersection property has non-empty intersection.

Now, for the proof of the Heine-Borel Theorem:

Theorem G (the Heine-Borel Theorem). Every closed and bounded subspace of the real line is compact.

Proof. A closed and bounded subspace of the real line is a closed subspace of the real line is a closed subspace of some closed interval $[a,b]$, and by Theorem A it suffices to show that $[a,b]$ is compact.

If $a = b$, this is clear, so we may assume that $a < b$. By Sec. 18, we know that the class of all intervals of the form $[a,d)$ and $(c,b]$, were $c$ and $d$ are any real numbers such that $a < c < b$ and $a < d < b$, is an open subbase for $[a,b]$; therefore the class of all $[a,c]$'s and $[d,b]$'s is a closed subbase. Let $\mathbf{S} = \{ [a,c_i], [d_j,b] \}$ be a class of these subbasic closed sets with the finite intersection property. It suffices by Theorem F to show that the intersection of all sets in $\mathbf{S}$ is non-empty. . . .

I don't understand how it suffices to show that the intersection of all sets in $\mathbf{S}$ is non-empty. Here, $\mathbf{S}$ is a class of subbasic closed sets (with the finite intersection property) that is chosen from a specific closed subbase, namely,

the class of all $[a,c]$'s and $[d,b]$'s . . . where $c$ and $d$ are any real numbers such that $a < c < b$ and $a < d < b$.

So, my question is:

• Is every closed subbase of $[a,b]$ of this form? In other words, if $\mathbf{\tilde{S}}$ is any closed subbase with the finite intersection property, can we show that there is some closed subbase $\mathbf{S}$ of the above form such that $\mathbf{S} \subseteq \mathbf{\tilde{S}}$?

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The proof is concluded by assuming that the intersection of all sets in $\mathbf{S}$ is empty and thereby deriving a contradiction to the hypothesis that $\mathbf{S}$ has the finite intersection property.

Answer 1

He uses (though this is not really needed) the equivalence between covers and families of sets with the FIP (finite intersection property).

Le t$\mathcal{S}$ be the open subbase and define $\mathcal{S}' =\{X\setminus S: S \in \mathcal{S}\}$, the set of complements (a closed subbase). Then the following are equivalent:

1. $X$ is compact (every open cover has a finite subcover)
2. Every open cover from $\mathcal{S}$ has a finite subcover.
3. Every subfamily of $\mathcal{S}'$ with the FIP has non-empty intersection.

The equivalence between 1. and 2. is what Simmons calls theorem $\mathbf{F}$, presumably (it's called the Alexander subbase lemma, more commonly). In the proof he uses the equivalence between the 2. and 3.

The proof of that last equivalence is the same as for the usual compactness equivalence with FIP closed families having non-empty intersection:

suppose 2. holds. Let $\mathcal{F} \subseteq \mathcal{S}'$ be a family with the FIP. If $\bigcap \mathcal{F'} = \emptyset$, $\mathcal{U} := \{X\setminus F: F \in \mathcal{S}'\}$ is an open cover of $X$ (cover because the intersection is supposedly empty) and is a subfamily of $\mathcal{S}$ (its members are the complement of the complement of a member of $\mathcal{S}$...) and so has a finite subcover by 2. But then the corresponding members of $\mathcal{F}$ would contradict the FIP property. So $\bigcap \mathcal{F} \neq \emptyset$. The implication 3. to 2. is similar again, using complements.

My remark here would be that Simmons could just use the open subbase and stick to the fomulation of his $\textbf{F}$ theorem:

Suppose $\{[a, c_i) : i \in I\} \cup \{(d_j, b]: j \in J\}$ form a subbasic cover of $[a,b]$ ;we need at least one of each type to cover both $a$ and $b$. The set $C'=\{c_i : i \in I\}$ has a supremum $p$, say. This must be covered ,but cannot be covered by any set $[a,c_i)$, as all $c_i \le p$, so $p \in (d_j, b]$ for some $d_j$. As $d_j < p$, $d_j$ is not an upper bound for $C'$ so there is some $c_i$ with $d_j < c_i < p$. But then $[a,c_i) ,(d_j ,b]$ are a finite subcover of the original subbasic cover.

So I see no need to go to complements and FIP families at all. If you have Alexander subbase lemma, use it!

Answer 2

Expanding upon the comments by @DanielFischer above.

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The trouble that I faced was that I misunderstood the statement of Theorem F, which could have been worded slightly better. In the book, after defining the terms basic open cover and subbasic open cover, he proves the following result.

Theorem E. A topological space is compact if every basic open cover has a finite subcover.

The trouble actually started here. This theorem could be better stated as follows.

Theorem E (improved). Let $X$ be a topological space with open base $\mathscr{B}$. If every basic open cover whose sets come from $\mathscr{B}$ has a finite subcover, then $X$ is compact.

In particular, one has to note that we are working with one (fixed) open base. I misunderstood this theorem as saying that, "If, for each open base for $X$ we have that every basic open cover with respect to that base has a finite subcover, then $X$ is compact." This is certainly not the case, and upon a moment's reflection we see that the last statement is no different from "If every open cover has a finite subcover then $X$ is compact." Which is just the definition of compactness, so it would not even make it a theorem.

I had the same misunderstanding in Theorem F, which can be improved as follows.

Theorem F (improved). Let $X$ be a topological space with open subbase $\mathscr{S} = \{ S_j \}$. If every subbasic open cover whose sets come from $\mathscr{S}$ has a finite subcover, then $X$ is compact. Equivalently, if every class of subbasic closed sets with the finite intersection property whose sets come from $\{ {S_j}' \}$ has non-empty intersection, then $X$ is compact.