Let $X$ be a set and $\{\tau_\alpha\}$ be a family of topologies on $X$. Show that there is a unique coarsest topology that is finer than every $\tau_\alpha$.
Clearly $\mathcal{S}:=\bigcup_\alpha \tau_\alpha$ is a subbasis, as $\tau_\alpha$ is a topology on $X$ for every $\alpha$. Moreover, we have observed that $$\mathcal{B}_s:=\left\{\overset{n}{\underset{i=1}{\bigcap}} U_i : U_i \in \tau_\alpha \text{ for some } \alpha \right\}$$ is a basis and generates the topology $\tau_s:=\left\{\bigcup_\beta V_\beta : V_\beta \in \mathcal{B}_s \text{ for all } \beta \right\}$ (Lemma $13.1$).
Let $\tau_\alpha$ be given. Suppose $U \in \tau_\alpha$. Then $U \in \mathcal{B}_s$, and so $U \in \tau_s$. Therefore, $\tau_\alpha \subset \tau_s$ for all $\alpha$. In other words, $\tau_s$ is finer than every $\tau_\alpha$.
Assume there exists a topology $\tau'$ such that $\tau'$ is finer than every $\tau_\alpha$ and $\tau' \subset \tau_s$ ($\tau'$ is coarser than $\tau_s$). Suppose $U \in \tau_s$. Since $U \in \tau_s$ we know that $U$ is the union of elements from $\mathcal{B}_s$. So we may write $U=\bigcup_\beta V_\beta$. Furthermore, for every $\beta$ we have $V_\beta=\bigcap_{i=1}^n U_i$ and each $U_i$ is in $\tau_\alpha \subset \tau'$ for some $\alpha$. So $V_\beta \in \tau'$ for every $\beta$. Since $\tau'$ is a topology $\bigcup_\beta V_\beta=U \in \tau'$. So we have $U \in \tau'$, and $\tau' \supset \tau_s$. Therefore, $\tau' = \tau_s$
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The book we are using (Munkres) shows/tells us how a subbasis can be used to "generate" (taking the finite intersections) a basis and that this collection is indeed a basis but it doesn't
designate a proposition/lemma/thm/corollary for this. Likewise with showing that the topology generated by a basis is a topology. It does give a lemma letting us know what the topology generated by a basis looks like. I might just ask our prof if not
citing anything for the first few lines is ok. -
The second paragraph might be too trivial given the subbasis but I
already typed it up.
Best Answer
It can be simpler, I think. No need to write doen the base or reason from contraposition at the end.
Let $\mathcal{S}=\bigcup_{\alpha \in S} \tau_\alpha$. This defines a subbase for a topology $\tau_s$ on $X$, which means that $\tau_S$ is the smallest topology on $X$ that contains $\mathcal{S}$. (This is an easy lemma proved in Munkres' book as well, this minimality characterises/defines $\tau_S$.)
As $\forall \alpha : \tau_\alpha \subseteq \mathcal{S} \subseteq \tau_S$, $\tau_S$ is finer than all $\tau_\alpha$.
If $\tau$ is any topology topology that is finer then all $\tau_\alpha$, then $\forall \alpha: \tau_\alpha \subseteq \tau$ which immediately implies $\mathcal{S} \subseteq \tau$ and as $\tau_S$ is the minimal topology containing $\mathcal{S}$ by definition, $\tau_S \subseteq \tau$. So $\tau_S$ is the coarsest topology among all such $\tau$. The topology $\tau_S$ is unique with this property, because if $\tau'$ is another topology with the same property, $\tau_S$ and $\tau'$ are coarser than each other by definition, an so equal. (Or, the sup in any poset is unique, which is what we really re-prove here.)