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I'm doing this exercise:

Let $\{T_\alpha\}$ be a family of topologies on $X$. Show that there is a unique smallest topology on $X$ containing all the collections $T_\alpha$, and a unique largest topology contained in all $T_\alpha$.

I have proved everything except the unique part.. I just can't get my head around what is meant with unique here. Which may sounds silly.

I have proved that the intersection is a topology. And if you are a topology that is also contained in every $T_\alpha$, than you surely are contained in the intersection, so you are not larger.

But I don't see from what it follows that this intersection is the unique largest topology contained in all $T_\alpha$. One part of my head say it is trivial, the other part gets confused. Like it is redundant to talk about unique in this context.

The same for proving the uniqueness of the smallest topology.

Edit Should I read topology $A$ larger than topology $B$ as, $A$ has more elements than $B$ ? I thought that, because the author uses the word finer for $A \supset B$.

Kasper
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  • Can you prove than an intersection of topologies is a topology, and similarly a union? – Ian Coley Mar 01 '14 at 15:48
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    I can proof it for an intersection. The idea is just that as X, the empty set, arbitrary unions, finite intersections are in every topology $T_\alpha$, it must also be in the intersection. For union, it is not always true, but you can take the union as a subbasis. – Kasper Mar 01 '14 at 15:50
  • @IanColey A union of topologies is not necessarily a topology. – Junglemath May 29 '20 at 19:12

2 Answers2

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If a topology is largest, it's unique automatically. Suppose $\tau_0$ and $\tau_1$ were both largest topologies. Then $\tau_0\supseteq\tau_1$ and $\tau_1\supseteq\tau_0$, hence $\tau_0 = \tau_1$.

Maximal, on the contrary would not have to be unique.

user2345215
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    I think the author of the book means with "larger", having more elements. As he uses "finer" for containing the other. – Kasper Mar 01 '14 at 15:53
  • @Kasper Are you sure? What does that even mean? Like the number of open subsets? He surely doesn't mean that. – user2345215 Mar 01 '14 at 15:56
  • Maybe, you are right :p But why wouldn't he mean that ? – Kasper Mar 01 '14 at 15:58
  • @Kasper I think the reason he uses largest instead of finest to differentiate between largest and maximal elements (under the inclusion ordering), because "not fineable" sounds clumsy for a maximal element, but I'm speculating :) – user2345215 Mar 01 '14 at 16:01
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The set of all topologies contained in $\{T_{\alpha}\}$ has a partial ordering so it has a sense of maximal element. You want to show that if $\tau$ is maximal with respect to this partial ordering then $\tau \subset \tau_0$ where $\tau_0$ is the topology you mentioned. Since $\tau$ was maximal you will have $\tau=\tau_0$.

SomeEE
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