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I have come across to the following question :

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

I think that the unique smallest topology equals the union of all the $\mathscr{T}_\alpha's$, and the unique largest topology contained in all $\mathscr{T}_\alpha$ equals the intersection of all the $\mathscr{T}_\alpha's$ .

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The union is not a topology, it generates one. –  Blah Apr 11 '11 at 5:10
    
If you show just one of: (i) there is a smallest topology containing all the T (ii) there is a largest topology contained in all the T, then the other statement can be seen as a corollary. It's a little easier to verify (ii). Your idea to use the intersection of the family is the correct one. –  Mike F Apr 11 '11 at 5:23
    
@Mike: Why are these corollaries of each other? Wouldn't that only be the case if there were a duality where the complement of a topology is a topology? –  joriki Apr 11 '11 at 5:32
    
@leopard: A comment I made previously was partly incorrect; I'd misread the question. The correct part of my comment was a hint towards Blah's answer. –  joriki Apr 11 '11 at 5:36
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@joriki: Quite generally if you want to show some poset P which has a greatest and largest element is a complete lattice then you need only check that (i) all subsets of P have a lub, or (ii) all subsets of P have a glb. If, say, (i) holds and you have some subset S of P then consider the set S' of all lower bounds for S. The lub of S' is the glb of S. –  Mike F Apr 11 '11 at 5:51

3 Answers 3

Many texts discuss this question, perhaps leaving some of the verifications to the reader. For instance you could look here.

(I guess that by writing these notes and posting them on the web I am implicitly giving the opinion that facts like these are too basic and foundational to make for good exercises for the student. There is no lack of things to ask a student to solve, so in my opinion as an instructor or a writer you might as well write out the really basic proofs in full. I note that, for instance, Bourbaki has this philosophy: a lot of "his" exercises are not only more challenging than many dreary propositions proved in full in the body of the text but are more interesting as well. This is not a very interesting exercise.)

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+1 Because of "This is not a very interesting exercise." :-D –  a.r. Dec 15 '11 at 14:41

The basic fact to verify is that if $\cal{T}_i$, $i \in I$ is an an indexed collection of topologies on a set $X$, then their intersection $\cap_{i \in I} T = \left\{ A \subset X: \forall i \in I: A \in\cal{T}_i \right\}$ is a topology on $X$ as well. This is straightforward from the topology axioms.

Once this is done, then the largest topology that contains all $\cal{T}_i$ can be defined as the intersection of all topologies $\cal{T}$ that contain $\cup_{i \in I} T_i$ as a subfamily, and there is at least one (the discrete topology) so we take the intersection of a non-empty family of topologies, which is a topology as we saw above. And it is clearly the smallest one that contains all $\cal{T}_i$: if $\cal{T}$ is any such topology it is one of the topologies we take the intersection of, and thus clearly is a subset of $\cal{T}$.

Also, the smallest topology contained in all $\cal{T}_i$ is simply their intersection, and this is clearly the largest topology that is contained in all of them.

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How do you show the uniqueness?? –  Ramana Venkata Oct 26 '11 at 14:04
    
If there are 2 minimal topologies $\mathcal{T}$ and $\mathcal{U}$ that contain all $\mathcal{T}_i$, then $\mathcal{T} \subset \mathcal{U}$, as the former is minimal among all such topologies, and the reverse also holds, reasoning for $\mathcal{U}$, hence equality. –  Henno Brandsma Oct 26 '11 at 19:41

Foreachx∈AthereexistsanopensetUsuchthatU⊆A.Sotakeeachxi∈Aforsomeindex IandweknowthereexistsanopensetUisuchthatUi⊆A.Theunionofanycollectionofopen setsisalsoopensoi∈IUiisopen.Also,eachxiisini∈IUisoA⊆i∈IUiand,clearly,as eachUi⊆Awefindi∈IUi⊆A.SoA=i∈IUiisopen

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What happened to your space bar? –  Calvin Lin Jun 10 '13 at 5:20

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