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Let $X$ be a topological space and $\mathbb{B}$ a collection of subsets of $X$. The following are equivalent.

(1)

  • $\mathbb{B}$ $\subseteq$ $\tau$

  • Every open subset of $X$ is the union of some collection of elements in $\mathbb{B}$

(2)

  • $X=$ $\bigcup_{B \in \mathbb{B}}B$

  • If $B_1,B_2 \in \mathbb{B}$ and $ x\in B_1\cap B_2$ then there exists $B_3 \in \mathbb{B}$ such that $x\in B_3 \subseteq B_1 \cap B_2$.

Attempt:

(1) $\implies (2)$: Since $X$ is open in any topology, it follows that it is the union of some collection of elements in $\mathbb{B}$. The first condition is satisfied. Let $B_1,B_2 \in \mathbb{B}$ and $x \in B_1 \cap B_2$. Since the intersection is an open subset of $X$, there exists a union of some collection of elements in $\mathbb{B}$ that is contained within. Therefore, atleast one element in $\mathbb{B}$ containing $x$ is contained in the intersection. This verifies the second condition.

How do I show (2) implies (1)?

2 Answers2

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The two are not equivalent.
The first is the definition of a base for a given topology.
The second is the condition for B to be a base for a topology.
Namely the topology generated by B.
With that topology understood, it is now possible to prove 2 implies 1.

William Elliot
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Definition $(1)$ is the definition of a base $\Bbb B$ for a given topology $\tau$ on a set $X$. A topology is known once you know a base for it: we can say, given $\Bbb B$, that

$$\tau=\{\bigcup B': B' \subseteq \Bbb B\}\tag{a}$$

Proof: $\supseteq$ follows from the fact that $\Bbb B \subseteq \tau$ and topologies are closed under arbitrary unions. $\subseteq$ is a reformulation of the second part of definition $(1)$.

Another slightly alternative way to define the topology from a base is inspired by the way metric topologies are defined by their base of open balls:

$$O \in \tau \iff \forall x \in O: \exists B_x \in \Bbb B: x \in B_x \subseteq O\tag{b}$$

Proof: $\Rightarrow$: By the definition $(1)$ of a base $O$ is a union of some family from $\Bbb B$ and if $x \in O$, there must be such a $B_x$ from this union, as claimed. $\Leftarrow$ we can, when the right hand side holds, easily see that $O=\bigcup\{B_x: x \in O\}$: right to left inclusion as all $B_x$ are a subset of $O$, so their union is too, and left to right as each $x$ is covered by its own $B_x$; then note that $O \in \tau$ by the closure under unions axiom of topology again. QED.

So knowing a base is knowing the topology and that's part of the reason that many topological spaces are defined by defining a collection of subsets of $X$ and declaring it to be a base for some uniquely defined topology $\tau$, using $(a)$ or $(b)$ as a guidance.

But you cannot specify just any $\Bbb B$ as such a collection to define a topology that way: that's where the conditions of $(2)$ come in: following $(a)$, the first part of $(b)$ follows from $X \in \tau$ and the second is a reformulation of a consequence the fact that base sets are open and so their intersection must be a union of base elements. So the conditions under $(b)$ are necessary for definition $(a)$ to work. And if $(b)$ holds, define $\tau$ as in $(a)$, $\emptyset \in \tau$ as $B'=\emptyset$ is allowed, $X \in \tau$ we already saw, closedness under all unions is clear, as a union of union is a union. The finite intersection follows from the second part of $(b)$ plus simple set theory:

$$\bigcup B' \cap \bigcup B'' = \bigcup\{B_1 \cap B_2: B_1 \in B', B_2 \in B''\}$$

and the intersections of base elements are in $\tau$.

Henno Brandsma
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