# Are these two definitions of base for a topology equivalent?

I have some experience with metric spaces, but I never took a dedicated Topology class. I recently encountered the term "base for a topology", and looking on Wikipedia, there seem to be two definitions. The very first line of the article says:

[Definition 1]: In mathematics, a base or basis for the topology $$\tau$$ of a topological space $$(X, \tau)$$ is a family $$B$$ of open subsets of $$X$$ such that every open set is equal to a union of some sub-family of $$B.$$

However, the first thing in the "Definition and basic properties" section is:

[Definition 2] A base is a collection $$B$$ of open subsets of $$X$$ satisfying the following properties:

1. The base elements cover $$X$$.
2. Let $$B_1$$, $$B_2$$ be base elements and let $$I$$ be their intersection. Then for each $$x$$ in $$I$$, there is a base element $$B_3$$ containing $$x$$ such that $$B_3$$ is subset of $$I$$.

I believe I have proved these definitions to be equivalent. But I want to make sure I haven't made any hidden assumptions, because I know that unlike metric spaces, General Topology can be very unintuitive for the outsider.

Def 1 $$\implies$$ Def 2:

Suppose $$B$$ is a base according to definition 1. Take any two base elements $$B_1, B_2$$. Their intersection $$I$$ is an open set, so it is a union of base elements. If $$x$$ is in $$I$$, then it is in the union of those base elements, so it is in at least one of them. Therefore $$x$$ is in a base element which is contained in the intersection.

Def 2 $$\implies$$ Def 1:

Suppose $$B = \{B_i|i \in I\}$$ is a base according to definition $$2$$. Let $$U$$ be an open set of $$X$$. The set $$S = \bigcup_{i \in I}U \cap B_i.$$ is a union of oepn sets, and we claim that $$S = B$$. It is obvious that $$S \subseteq B$$. To see the reverse, let $$x$$ be any element of $$B$$. Since the base covers $$X$$, there is a base element $$b$$ such that $$x \in b$$. Therefore $$x \in U \cap b$$, so $$x \in S$$.

• Definition 2 does not imply definition 1.
– jMdA
Feb 6, 2021 at 11:03
• @jMdA Thank you, I just saw your answer. Feb 6, 2021 at 15:43

These two definitions are not defining the same thing, so they are not "equivalent".

In definition 1, one is given the set $$X$$ and the topology $$\tau$$ on $$X$$. What is being defined is a basis for the topology $$\tau$$.

In definition 2, one is given just the set $$X$$. What is being defined is a basis. Period.

Note the syntactic difference in these two terminologies: definition 1 is relative to a given topology; definition 2 is not relative to anything. (Well... both of them are relative to the given set $$X$$).

These definitions are connected in the following manner:

Theorem: Given a topology $$\tau$$, a basis $$\mathcal B$$ for $$\tau$$ (as in definition 1) is a basis (definition 2).

Theorem: Given a basis $$\mathcal B$$ (as in definition 2), there exists a unique topology $$\tau$$ on $$X$$ such that $$\mathcal B$$ is a basis for $$\tau$$.

• His definition $2$ already has "open sets" in it, so he is assuming that a topology is already present and wishes to show that his collection is a base for it. He just elaborated it a lot more in the first definition (or, whoever wrote that did). Feb 6, 2021 at 2:13
• I see your point. Thank you for the clarification. Feb 6, 2021 at 2:17
• I can see that, but it looks to me like that was inserted inadvertantly, because if one takes that "open set" phrase literally then definition 2 is not equivalent to definition 1. Definition 2 is equivalent to the statement that the topology generated by $\mathcal B$ is coarser than the topology of open sets, but it is not equivalent to the statement that the topology generated by $\mathcal B$ is equal to the topology of open sets. Feb 6, 2021 at 3:30
• What's missing in Definition 2 is the statement that for each open set $U$ and each $x \in U$ there is a basis element $B_i \in \mathcal B$ such that $x \in B_i \subset U$. Feb 6, 2021 at 3:32
• Yeah, I guess in his def 2 he could literally take the basis to just be $\lbrace X \rbrace$. I didn't even notice it, just used to seeing that definition hahaha Feb 6, 2021 at 14:24

I will give a counterexample to Definition 2 implying Definition 1.

Consider any basis $$\mathcal{B}$$ for $$\tau$$, and take any open $$U$$ construct the alternate basis $$\mathcal{B}_U$$ consisting of every set of the form $$B\cup U$$ for $$B$$ any set in $$B\in\mathcal{B}$$.

The collection $$\mathcal{B}_U$$ is still a basis satisfying Definition 2, but not Definition 1.

• What you are suggesting is a counterexample to "Definition 2 implies Definition 1". Feb 6, 2021 at 13:19
• Thank you for catching my mistake.
– jMdA
Feb 6, 2021 at 13:20

EDIT: It's been pointed out that your definition is actually a bit off; the proof below assumes a corrected definition, which I assume was just a typo. Tbh I didn't even notice it haha

Your first part is correct, but the second part has problems. For example, $$S$$ is a subset of $$X$$, but you say that it's equal to $$B$$, a basis, which is a collection of sets. Also, every element in $$S$$ is contained in $$U$$, but $$X$$ is not necessarily contained in $$U$$, so you can't get a base.

The way to do $$2 \implies 1$$ is suppose $$B$$ is a basis according to the second definition, and let $$U$$ be any open set in $$X$$. We want to show that it's the union of elements of $$B$$. By definition $$2$$ of a basis, if $$x \in U = U \cap U$$ then there is an open set $$B_x \in B$$ such that $$x \in B_x \subset U$$. Now $$U = \cup B_x$$, since each $$B_x$$ is contained in $$U$$, but also every point of $$U$$ is in some $$B_x$$. I think you had the right idea, just wrote it down wrong.

• Oops I mean to say $S = U$. But yeah I confused myself and blundered in the second part. Feb 6, 2021 at 2:15
• Definition 2 does not actually say that if $U$ is any open set and if $x \in U$ then there exists $B_x \in B$ such that $x \in B_x \subset U$. That's what is missing if one wanted 2 to be equivalent to 1. Feb 6, 2021 at 4:23