Definition of a basis for a topology on $X$

Question

These two definitions are from Topology, II Edition, James R. Munkres.

Definition 1.

A topology on a set $X$ is a collection $\mathfrak{I}$ of subsets of $X$ having the following properties:

$(1)$ $\emptyset $ and $X$ are in $\mathfrak{I}.$

$(2)$ The union of the elements of any subcollection of $\mathfrak{I}$ is in $\mathfrak{I}.$

$(3)$ The intersection of the elements of any finite subcollection of $\mathfrak{I}$ is in $\mathfrak{I}.$

If $X$ is a topological space with topology $\mathfrak{I},$ we say that a subset $U$ of $X$ is an open set of $X$ if $U$ belongs to the collection $\mathfrak{I}$

Definition 2

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

• For each $x∈X$, there is at least one basis element $B$ containing $x$

• 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 $B_3⊆B_1∩B_2$.

If $\mathscr{B}$ satisfies these two conditions, then we define the topology $\mathfrak{I}$ generated by $\mathscr{B}$ as follows : A subset $U$ of $X$ is said to be open in $X$( i.e., to be an element of $\mathfrak{I}$ ) if for each $x\in U,$ there is a basis element $B\in \mathscr{B}$ such that $x\in B$ and $B\subset U.$

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Suppose for a set $X$ I have a collection $\mathscr{B}$ satisfying the two properties mentioned in definition 2.

Now consider a subset $S$ of $X$ such that for each $x\in S,$ there is a basis element $B\in \mathscr{B}$ such that $x\in B$ and $B\subset S.$

Form $\mathfrak{I}_1$ such that we choose minimum number of $S_{\alpha}$ required so that $\mathfrak{I}_1$ forms a topology on $X.$

Now if I inject some more such subsets so that now $\mathfrak{I}_2$ is a topology on $X.$

Is such a scenario possible? I don't know , seems plausible to me. If that is the case . . .

does definition 2 say $\mathscr{B}$ is a base which "generates" multiple topologies on a set $X$?

Answer 1

Note that the definition for the topology generated by $\mathscr{B}$ is the set of all subsets $U$ of $X$ with the stated property. It follows that the topology generated is unique.

Answer 2

Given some topology $\mathcal T$ generated by a basis $\mathcal B$ on some set $X$, it is always possible to make a finer topology $\mathcal T_d$, where $\mathcal T \subseteq \mathcal T_d$, by declaring that every set is open. Indeed, when the new basis is the set of all singletons, we obtain the finest possible topology $\mathcal T_d$ known as the discrete topology. However, $\mathcal B$ is only a basis for $\mathcal T$, not $\mathcal T_d$. In general, a basis generates a unique topology, but a given topology can be generated by more than one choice of basis.

Answer 3

There are two ways to approach bases: we start with a topology $(X,\mathcal{T})$ and then a base $\mathcal{B}$ for that topology is a subcollection of $\mathcal{T}$ (so they're open sets) such that every $O \in \mathcal{T}$ can be written as a union of elements from $\mathcal{B}$, so $O = \bigcup \mathcal{B}'$ for some $\mathcal{B}' \subseteq \mathcal{B}$. Note that this means that for every $x \in O$ there is some $B_x \in \mathcal{B}'$ such that $x \in B_x$ (definition of union) and as $B_x \subseteq \bigcup \mathcal{B}'$ we thus have the property that for every open $O$ and every $x \in O$, there is some base element $B$ such that $x \in B \subseteq O$.

This connects to the other way to approach it: we start with a family of subsets of a set $X$ called $\mathcal{B}$ and we want to define the unique topology that $\mathcal{B}$ is a base for: this scenario happens with metric spaces where we use metric balls or ordered spaces where we use open intervals etc. The topology is uniquely determined by the condition that all unions should be in the same topology and we must be able to write all open sets as unions of base elements, and in particular the intersection of two base elements must be a union of base elements, and this determines the second condition that the candidate base family has to fulfill. The first is because $X$ has to be open and thus must be in the set of unions. These two are enough: defining a topology as all unions of subfamilies of $\mathcal{B}$ or equivalently (as we saw before) $O$ is open iff for all $x \in O$ there is some $B_x \in \mathcal{B}$ with $x \in B_x \subseteq O$, we get the uniquely defined topology that our starting family $\mathcal{B}$ is a base for in the first sense.