Clarification regarding basis for a topology

Question

This might be super trivial but I if possible would like some clarification on this topic.

I am reading from Munkres' Topology, 2nd edition, page 78 (if interested).

My question regards what a basis for a topology means. Below is what I am thinking it means; besides its definition I wonder if the following is true:

[$\mathcal{B}$ is a basis for a topology $\tau$] $\Rightarrow \tau$ is the topology generated by $\mathcal{B}$, that is $$\tau=\tau_\mathcal{B}=\{ U \subseteq X: \forall x\in U \exists B\in\mathcal{B}\left( x\in B \subseteq U\right)\}.$$

Essentially what I want is that when we say $\mathcal{B}$ is a basis for a topology, then the topology that $\mathcal{B}$ is a basis for (might just be definition here) is the topology $\tau_\mathcal{B}$ mentioned above.

*Munkres' defines a basis for a topology as:

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 $$(1)\, \text{For each $x \in X$, there is at least one basis element $B$ containing $x$}$$ $$(2)\, \text{If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis}$$ $$\text{element $B_3$ containing $x$ such that $B_3 \subseteq B_1 \cap B_2$.}$$ If $\mathcal{B}$ satisfies these two conditions, then we define the topology $\tau$ generated by $\mathcal{B}$ as follows: A subset $U$ of $X$ is said to be open in $X$ if for each $x \in U$, there is a basis element $B \in \mathcal{B}$ such that $x \in B$ and $B\subseteq U$.

Later in a lemma we prove:

Let $X$ be a set; let $\mathcal{B}$ be a basis for a topology $\tau$ on $X$. Then $\tau$ equals the collection of all unions of elements of $\mathcal{B}.$

In the proof of this lemma, it looks to me like they are saying that $\tau=\tau_\mathcal{B}$ as I defined above so this leads me to think that at this point, when we say $\mathcal{B}$ is a basis for a topology $\tau$ on $X$, we are meaning $\tau=\tau_\mathcal{B}$.

I am still trying to get a sense of what your answers means, Berci. In an older topology book that I used in undergrad, we defined a basis of a topology as being some subset of a topology $\tau$ where each element of $\tau$ could be written as the union of members of $\mathcal{B}$, which I am sure is equivalent to how it is defined in Munkres' text, I am just trying to put some pieces together so that I have some hard statements to prove.

Answer 1

$\mathcal B$ is a basis of topology $\tau$, by definition, if all the open sets (i.e., elements of $\tau$) can be obtained by arbitrary unions of elements of $\mathcal B$.

That is, $U\in\tau \iff U=\bigcup_i B_i$ for some $B_i\in\mathcal B$. So that, it means indeed that for any point $x\in U$, it is $x\in B_i$ for some $i$.

For the other direction, if your condition holds, i.e. for every $x\in U$, there is a $B_x\in\mathcal B$ such that $x\in B_x\subseteq U$, then $U=\bigcup_{x\in U}{B_x}$.

Answer 2

I'm not quite sure what the question is, but hopefully this will clarify.

Given a collection $\mathcal{B}$ of subsets of $X$, there is a least $\mathcal{C} \supseteq \mathcal{B}$ closed under arbitrary unions. There is also a least $\mathcal{D} \supseteq \mathcal{B}$ closed under not only arbitrary unions, but also finite intersections*. We call $\mathcal{D}$ the "topology generated by $\mathcal{B}$," and we might call $\mathcal{C}$ the "generalized topology generated by $\mathcal{B}$." Its always the case that the topology generated by $\mathcal{B}$ includes the generalized topology generated by $\mathcal{B}$. Whenever this is inclusion is an equality, we say that $\mathcal{B}$ is a basis for the topology $\mathcal{D}.$

I got the term "generalized topology" from this article.

This basic analysis applies in many other situations. Given a collection $B$ of elements of a group $G$, there is a least $C \supseteq B$ that is a submonoid of $G$. Call it the submonoid generated by $B$. There is also a least $D \supseteq B$ that is a subgroup, call it the subgroup generated by $B$. Its always the case that $D \supseteq C$. If, furthermore, this is an equality, it would be acceptable to say that $B$ is a (monoidal) basis for $D$.

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*The intersection of the empty collection should be interpreted as equaling $X.$