Example 2

Least upper bound of a set of topologies. Every family $(\mathfrak{T}_i)_{i \in I}$ of topologies on a set $X$ has a least upper bound $\mathfrak{T}$ in the ordered set of all topologies on $X$, i.e. there exists a topology on $X$ which is coarsest among all the topologies on $X$ which are finer than each of the $\mathfrak{T}_i$. To see this we may apply Proposition 4, taking $Y_i$ to be the set $X$ with the topology $\mathfrak{T}_i$, and $f_i$ to be the identity map $X \to Y_i$; $\mathfrak{T}$ is the coarsest topology for which all the mappings $f_i$ are continuous.

Let $\mathfrak{S}$ be an arbitrary set of subsets of a set $X$; amongst the topologies $\mathfrak{T}$ on $X$ for which the sets of $\mathfrak{S}$ are open, there is a topology $\mathfrak{T}_0$ which is coarser than all the others and which is called the topology generated by $\mathfrak{S}$. For each set $U \in \mathfrak{S}$ let $\mathfrak{T}_U$ be the topology whose open sets are $\varnothing, U$ and $X$ [it is clear that this set of subsets of $X$ satisfies (O_I) and (O_{II})]; then $\mathfrak{T}_0$ is just the least upper bound of the topologies $\mathfrak{T}_U$. By Proposition 4, if $\mathfrak{B}$ is the set of finite intersections of sets belonging to $\mathfrak{S}$, then $\mathfrak{B}$ is a base of the topology $\mathfrak{T}_0$. We say that $\mathfrak{S}$ is a subbase of $\mathfrak{T}_0$.