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Simple topology questions

  1. Feb 29, 2008 #1
    1. The problem statement, all variables and given/known data

    1. given a set X and a collection of subsets S, prove there exists a smallest topology containing S
    2. Prove, on R, the topology containing all intervals of the from [a,b) is a topology finer than the euclidean topology, and that the topologies containing the intervals of the form [a,b) and (a,b], respectively, are not comparable.

    3. The attempt at a solution

    1. You can say there exists one, the discrete topology. But the smallest one?

    2. I tried to write an element (a,b) as the union or intersection of intervals of the form [a,b). You could say [tex](a,b)={\bigcup^{\infty}}_{i=1}[a_{i},b), a_{i}=a+\frac{1}{i}[/tex]

    or can you also write:

    [tex](a,b) = [\frac{b+a}{2},a)\cup[\frac{b+a}{2},b)[/tex] ?

    but then still, why would the latter two be incomparible?
     
    Last edited: Feb 29, 2008
  2. jcsd
  3. Feb 29, 2008 #2
    well obviously, one topology has a minimum while the other doesnt and vice versa about the maximum, i guess this is why you can't compare between them.
     
  4. Feb 29, 2008 #3

    morphism

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    1. Certainly any topology that contains S will contain finite intersections of elements of S, and unions of these finite intersections, etc. So how about you try to construct a basis for the smallest topology containing S?
     
  5. Feb 29, 2008 #4

    NateTG

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    For #1, you could also show that the intersection of topologies is a topology.

    For part 2:
    It's implicitly assumed (and I'm guessing actually listed) that the half-open intervals have the condition that [itex](a<b)[/itex] so [itex][\frac{a+b}{2},a)[/itex] is not necessarily open.

    To show that they're incomparable, you have to show that there are open sets in each that are not open in the other. (You might try something like [itex][0,1)[/itex] or [itex][1,0)[/itex]...)
     
  6. Feb 29, 2008 #5
    Nate:

    1. I am doing this, but I'm having problems showing T2). You don't know the intersection of an arbitrary number of opens is open. If it would, you could take a topology that contains S and take it's interesection with all other topologies on X containing S, leaving the smallest. But I think the problem is there may be infinite topologies containing S, and you won't be able to intersect them.

    2. after the first comment I suspected it was listed too, but it was not. But I imagine you're right. Do you agree with my other suggestion of the infinite union? And I think the last part is clear now, thanks.

    morphism:

    1. we're doing the chapter on topology bases next week, so I must say I haven't gotten a real idea of what a basis is yet, but I'm studying topology right now, so I imagine I'll find out tonight and then I will look back to the exercise and try do it your way.
     
    Last edited: Feb 29, 2008
  7. Mar 1, 2008 #6
    anyone?
     
  8. Mar 1, 2008 #7

    morphism

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    Nate was suggesting you intersect two topologies together, not their sets. So a set will belong to the intersection of two topologies if it is open in both of them.
     
  9. Mar 1, 2008 #8
    If you know what a subbasis is, then let S U {X} be a subbasis for the topology. Or, consider the intersection of all topologies containing S.
     
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