# supremum analog for collection of topologies (proof check)

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, $$\tau' = \tau_s$$

• 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.

• I’m rusty, so forgive me if this is obvious, but why does $U \notin \tau_{\alpha}$ for every $\alpha$ imply that $U \notin \mathcal{B}_S$? Couldn’t there be an open set $U \in \mathcal{B}_S$ that isn’t in any $\tau_{\alpha}$, but is the finite intersection of open sets from different topologies, since your subbasis is $\bigcup_{\alpha} \tau_{\alpha}$?
– Joe
Aug 29 '19 at 2:41
• Assume $U \in \mathcal{B}_s$. So $U=\cap_{i=1}^n U_i$ and each $U_i$ is in $\tau_\alpha$ for some $\alpha$. Since $\tau_\alpha \subset \tau'$ for every $\alpha$, it follows that $U_i \in \tau'$ for every $i=1,\ldots, n$. But $U \notin \tau'$. This contradicts the fact that $\tau'$ is a topology. Perhaps this is something I should include in my argument, as the concern @Joe raised did give me pause to think of this. Aug 29 '19 at 3:58
• It's not a supremum "analog", but the actual supremum of the family $\{\tau_\alpha: \alpha \in S\}$ in the poset of all topologies on $X$ ordered by inclusion. Hence unicity is automatic if the sup exists. Aug 29 '19 at 4:31
• I think my uniqueness argument was completely wrong. I have changed it. Aug 29 '19 at 6:42

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.)
• Are you saying that because I assume $\tau_\alpha \subset \tau'$ for every $\alpha$, and $\tau_s$ is the unique topology generated by the subbase $\mathcal{S}=\cup_\alpha \tau_\alpha$, it then follows that $U \in \tau'$ whenever $U \in \tau_s$ (or more simply put, $\tau' \supset \tau_s$)? Do you think there is something inherently wrong with anything that I have written? My work on this problem basically just consisted of thinking how I might explicitly define (terminology) this topology $\tau_s$. I thought of this basis before anything else, and then I just tried to work from there. Aug 29 '19 at 22:36