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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 $x \in S$.

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  • $\begingroup$ Definition 2 does not imply definition 1. $\endgroup$
    – jMdA
    Feb 6, 2021 at 11:03
  • $\begingroup$ @jMdA Thank you, I just saw your answer. $\endgroup$
    – user56202
    Feb 6, 2021 at 15:43

3 Answers 3

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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 $\tau$.

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  • $\begingroup$ 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). $\endgroup$ Feb 6, 2021 at 2:13
  • $\begingroup$ I see your point. Thank you for the clarification. $\endgroup$
    – user56202
    Feb 6, 2021 at 2:17
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    $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Feb 6, 2021 at 3:30
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    $\begingroup$ 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 $x \in B_i \subset U$. $\endgroup$
    – Lee Mosher
    Feb 6, 2021 at 3:32
  • $\begingroup$ 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 $\endgroup$ Feb 6, 2021 at 14:24
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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.

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

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  • $\begingroup$ Oops I mean to say $S = U$. But yeah I confused myself and blundered in the second part. $\endgroup$
    – user56202
    Feb 6, 2021 at 2:15
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    $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Feb 6, 2021 at 4:23

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