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A base for the closed sets in a topological space $ X $ is a family of closed sets in $ X $, such that every closed set is an intersection of some subfamily.

  1. $ \mathcal{F}$ is a base for the closed sets in $ X $ iff the familily of complements of members of $ \mathcal{F}$ is a base for the open sets.

  2. $ \mathcal{F} $ is a base for the closed sets for some topology on $ X $ iff (a) whenever $ F_{1} $ and $ F_{2} $ belong to $\mathcal{F} $, $ F_{1} \cup F_{2} $ is an intersection of elements of $ \mathcal{F} $, and (b) $ \bigcap_ {F \in \mathcal{F}} {F} = \emptyset $

I have been demonstrating this problem directly using the definition of base and assuming that a base for closed sets in a topological space $ (X, \tau) $ is a family of closed sets in $ (X, \tau) $, however, I have had difficulties reaching the desired result. I have also tried for reduction to the absurd, but the test turns out to be a bit cumbersome.

Curious
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1 Answers1

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Let $\mathcal{F}$ be a base for the closed sets of $X$.

Define $$\mathcal{B} = \{X\setminus F: F \in \mathcal{F}\}$$

Let $O$ be an open subset of $X$. Then $X\setminus O$ is closed, so by assumption there is a subfamily $\mathcal{F}' \subseteq \mathcal{F}$ such that

$$X\setminus O = \bigcap \{F: F \in \mathcal{F}'\}$$

and taking complements on both sides and using de Morgan's law(s):

$$O = X\setminus \left(\bigcap \{F: F \in \mathcal{F}'\}\right)= \\ \bigcup \{X\setminus F: F \in \mathcal{F}'\}$$

showing that $O$ is a union of a subfamily of $\mathcal{B}$, so $\mathcal{B}$ is a base for $X$.

The reverse implication is entirely similar.

part 2, is a reformulation of the fact that $\mathcal{B}$ is a base for a topology on $X$ iff

  • For every $B_1, B_2 \in \mathcal{B}$, $B_1 \cap B_2$ is a union of elements of $\mathcal{B}$ (sometimes put as $$\forall x \in B_1 \cap B_2: \exists B_3 \in \mathcal{B}: x \in B_3 \subseteq B_1 \cap B_2$$

which is the same thing)

  • $\bigcup \mathcal{B} = X$.

The dual statements for the complements after de Morgan are exactly the statements (a) and (b) under 2. in the original question. So it's rather trivial that the complements of a family as under your 2, obeys my two demands and vice versa.

Henno Brandsma
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