# If $B$ is a basis for a topology $T$, does $B$ necessarily generate $T$?

Let $$(X,\tau)$$ be some topological space. Munkres defines a basis $$\mathcal{B}$$ of $$\tau$$ as a collection of subsets of $$X$$ such that:

$$\mathcal{B}$$ covers $$X$$, and given $$B_1, B_2 \in \mathcal{B}$$, for all $$x \in B_1 \cap B_2$$, there exists some $$B_3 \in \mathcal{B}$$ such that $$x \in B_3 \subset B_1 \cap B_2$$.

Given a collection of subsets $$\mathcal{B}$$ with the above properties, Munkres defines the topology generated by $$\mathcal{B}$$ as all sets $$U$$ such that for all $$x \in U$$, there exists some $$B \in \mathcal{B}$$ such that $$x \in B \subset U$$.

My question is, given some topological space $$(X,\tau)$$ with basis $$\mathcal{B}$$, is $$\tau$$ necessarily the topology generated by $$\mathcal{B}$$? I thought they were different concepts, but in the proof of certain lemmas, they seem to imply one another.

Thanks!

• I presume that if the exact words used are "let $(X,\tau)$ be a topological space with basis $\mathcal{B}$", then yes, $\tau$ is implied to be the one generated by $\mathcal{B}$. Though of course this is more sloppy phrasing than anything. – J_P Jun 16 '19 at 20:39
• As commented and answered by others already, there is a difference between being a basis (for some topology), and given a topology $\tau$ being a basis for that particular topology. For example the set of all singletons in the real line is a basis (for the discrete topology) but it is not a basis (for the usual topology). – Mirko Jun 16 '19 at 21:51

Quoting from Munkres:

If $$X$$ is a set, a basis for a topology on $$X$$ is a collection $$\mathcal{B}$$ of subsets of $$X$$ (called basis elements) such that

1. For each $$x\in X$$, there is at least one basis element $$B$$ containing $$x$$.
2. If $$x$$ belongs to the intersection of two basis elements $$B_1$$ and $$B_2$$, then there is a basis element $$B_3$$ containing $$x$$ such that $$B_3\subset B_1\cap B_2$$.

This definition is not the definition of "$$\mathcal{B}$$ is a basis of the topology $$\tau$$" (note that no topology $$\tau$$ appears in the definition!). Instead, it's the definition of "$$\mathcal{B}$$ is a basis for some topology on $$X$$". Which topology? The topology generated by $$\mathcal{B}$$ (which is the next part of the definition in Munkres).

So when you read "$$\mathcal{B}$$ is a basis for $$\tau$$", it means that $$\mathcal{B}$$ is a basis, and $$\tau$$ is the topology generated by the basis $$\mathcal{B}$$.

• So if I have a collection of subsets B that generate a topology, then the collection B satisfies the first two conditions given? – ChrisWong Jun 17 '19 at 11:00
• @ChrisWong Yes. Since the whole set is in the topology, the generating set must satisfy (1), and since the topology is closed under binary unions, the generating set must satisfy (2). – Alex Kruckman Jun 17 '19 at 12:58

There is a difference between being a basis and a basis for a particular topology. A basis is a collection of subsets of a set $$T$$ satisfying certain properties. It generates a topology as you indicate, and we call it a basis for the topology it generates. It's possible that a proof begins "let $$(T,\tau)$$ be a topological space and $$\mathcal{B}$$..." then there is an implied "for the topology $$\tau$$".