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Math Help - more topology

  1. #1
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    more topology

    This problem has five parts, but I think I can get two of them. So

    Let B be the collection of subsets of R in the form [a,b) for all a < b, the empty set, and the whole set included.

    Let T be the topology on R determined by taking all arbitrary unions of sets in B. Show the sequence {1+ (1/n), n= 1,2, ...} converges to one in the topology T.

    I thought it was obvious that this sequence converges to 1 because 1/n converges to 0, but then I got confused by the next part.

    Show that the sequence {1 - (1/n), n=1,2,...} does NOT converge to 1 in the topology T. I'm assuming this may have something to do with [a,b) where b is not included, but I"m not sure.

    Also, considering the product topology R x R determined by T. Show that the set of points {(x,-x) | x is in R} is discrete in the relative topology.

    I'm not sure what discrete in the relative topology means. Does it have anything to do with discrete metrics?

    Thanks
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  2. #2
    Moo
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    Hello,

    Let B be the collection of subsets of R in the form [a,b) for all a < b, the empty set, and the whole set included.

    Let T be the topology on R determined by taking all arbitrary unions of sets in B. Show the sequence {1+ (1/n), n= 1,2, ...} converges to one in the topology T.

    I thought it was obvious that this sequence converges to 1 because 1/n converges to 0, but then I got confused by the next part.

    Show that the sequence {1 - (1/n), n=1,2,...} does NOT converge to 1 in the topology T. I'm assuming this may have something to do with [a,b) where b is not included, but I"m not sure.
    No, it has to do with the definition of convergence in a topology
    Here is the general definition :
    Let (x_n)_n \in X be a sequence of points in X.
    (x_n) converges to x if for any open neighbourhood \mathcal{U} of x :
    \exists N_x \in \mathbb{N}, ~ \forall n \geq N, x_n \in \mathcal{U}


    So for the first one, you're approaching 1 by superior values. It is easy to prove that for any open set containing 1 (which is a neighbourhood), and particulary for the open sets in the form [1,a), a>1, the definition is verified.

    For the second one, you're approaching 1 by inferior values. It is sufficient to take a counterexample :
    [1,a) is an open set, but in no way there is an N such that for all n>N, x_n=1-\frac 1n is in this open set, because it is <1.
    hence it doesn't converge to 1.




    For the second part, to prove a set is discrete, show that any of its element and subset is an open set in the relative topology. I'm sorry I cannot help much here as I don't understand what the set exactly is
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  3. #3
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    Quote Originally Posted by EricaMae View Post
    Also, considering the product topology R x R determined by T. Show that the set of points {(x,-x) | x is in R} is discrete in the relative topology.

    I'm not sure what discrete in the relative topology means. Does it have anything to do with discrete metrics?
    The relative topology on a subspace simply means the topology where the open sets are the open sets in the whole space, intersected with the subspace.

    In \mathbb{R}^2 with the product half-open interval topology, the set \{(x,y):a\leqslant x<a+1,\ -a\leqslant y<-a+1\} is an open set containing the point (a,-a). It's easy to check that this open set does not contain any point of the form (x,-x) apart from (a,-a). That shows that the subspace \{(x,-x):x\in\mathbb{R}\} is discrete for the relative topology.
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