Let $S$ be a set. For $\tau$ and $\tau'$ be topologies on $S$. Declare $\tau \leq \tau'$ meaning $id: (S,\tau) \rightarrow (S,\tau')$ is continuous. This is a partial order on the set of all topologies on $S$. Prove there exists an infimum for any subset of this poset.
Attempt: I can first prove $\tau \subseteq \tau'$ iff $id: (S,\tau') \rightarrow (S,\tau)$ is continuous.
I need to show a supremum exists in the partial order by $\subseteq$. So, for $\kappa$ a collection of topologies on $S$, I need to show a topology $\tau_{\kappa}$ on $S$ with the following properties:
For $\tau \in \kappa$, then $\tau \subseteq \tau_{\kappa}$
For $\tau'$ a topology on S for which, for each $\tau \in \kappa, \tau \subseteq \tau'$,then $\tau_{\kappa} \subseteq \tau'$.
So if $\tau_{\kappa} = \bigcup\limits_{U \in \kappa}U $. This satisfies 1) but might not be a topology as indicated from Munkres excercise 4a)
How could I formally show there is a supremum?
Note that $\tau \le \tau'$ if and only if, for all $\mathcal{U} \in \tau'$, $\mathcal{U} = \operatorname{id}^{-1} (\mathcal{U}) \in \tau$. That is, $\tau$ is finer than $\tau'$. So, to find an infimum of a subset $T$ of the poset, we need to find the coarsest topology that is finer than every topology in $T$.
Let $B$ be the union of all topologies in $T$. While $B$ is not necessarily a topology, it makes a perfectly good sub-basis. Form the topology $\tau_0$ by taking all arbitrary unions of finite intersections of $B$. Such a topology is clearly finer than all elements of $T$.
Suppose $\tau_1$ is a lower bound of $T$. Then, $\tau_1$ contains every set in $B$. Moreover, it must contain every finite intersection of sets in $B$, and every arbitrary union of such intersections. In other words, $\tau_1 \le \tau_0$, showing that it is the infimum of $T$.