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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.

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    $\begingroup$ 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 $\bigcup_{\alpha} \tau_{\alpha}$? $\endgroup$
    – Joe
    Aug 29 '19 at 2:41
  • $\begingroup$ 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. $\endgroup$
    – M A Pelto
    Aug 29 '19 at 3:58
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    $\begingroup$ 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. $\endgroup$ Aug 29 '19 at 4:31
  • $\begingroup$ I think my uniqueness argument was completely wrong. I have changed it. $\endgroup$
    – M A Pelto
    Aug 29 '19 at 6:42
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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.)

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  • $\begingroup$ I do see now how we/I may use other statements in the text to write an overall more simple argument. But I believe that I fixed my uniqueness argument (to me it seems to have been totally wrong) and I think that I will stick with it, along with everything else I wrote. At least for the sake of trying to use my original work. $\endgroup$
    – M A Pelto
    Aug 29 '19 at 6:46
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    $\begingroup$ @MattAPelto The unicty argument you gave essentially repeats the unicity argument for the topology generated topology from a subbase. It's nicer to just quote and just it than to reprove it every time you use it. $\endgroup$ Aug 29 '19 at 20:50
  • $\begingroup$ 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. $\endgroup$
    – M A Pelto
    Aug 29 '19 at 22:36
  • $\begingroup$ @MattAPelto yes, the first indeed folllows right away. You’re reproving it, so it’s not wrong, just superfluous. $\endgroup$ Aug 30 '19 at 4:19

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