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Subspace topology question.

  1. Nov 19, 2007 #1
    Well it's not homewrok cause i dont need to hand this question in, this is why i decided to put it here. (that, and there isn't a topology forum per se, perhaps it's suited to point set topology so the set theory forum may suit it).

    Now to the question:
    Show that if Y is a subspace of X, and A is a subset of Y, then the subspace topoology on A as a subspace of Y is the same as the subspace topology on A as a subspace of X.
    (^-stands for intersection).
    well so i wrote the specific topologies in hand:
    T is a topology on X and T' a topology on Y.
    T_Y subspace topology on Y from T, i.e: [tex]T_Y[/tex]={ Y^U|U\in T }
    [tex]T_{A_Y}[/tex]={ A^U|U\in T' } is the subspace topology on A as a subspace of Y, and [tex]T_{A_X}[/tex]={ A^U| U\in T }
    is the subspace on A as a subspace of X, now i need to prove that the latter topologies are the same, i.e are equal.
    and so iv'e begun:
    let V be in T_{A_Y} then there exists U in T' such that V=A^U but A is a subset of Y, then V is a subset of Y^U, so because Y and U are in T', also Y^U in T', now i need to prove that U is in T (the topology of X), now there's a lemma that ithink i need to use here but not certain how to apply it, the lemma states that:
    Let Y be a subspace of X, if U is open in Y and Y is open in X then U is open in X.
    now i obvioulsy want to show that U is open in X, so i need Y to be open in X, not sure how to do it, any hints?
    p.s
    also hints for the second part of the proof would be nice but i havent yet tried to prove the second part.
     
    Last edited: Nov 19, 2007
  2. jcsd
  3. Nov 19, 2007 #2
    Suppose O is open in the topology on A induced from Y. Then 0 = A n V for some V open in Y. Since V is open Y we have V = Y n B for some B open in X. Then 0 = A n V = A n Y n B = A n B because A is a subset of Y , hence O is open in the topology induced on A from X.

    Conversely, suppose O is open in the subspace topology on A induced from X. Then O = A n B for some open B in X. Since A is a subset of Y, O = A n B = A n (Y n B) and Y n B is open in Y since B is open in X. Thus O is open in the topology induced on A from Y.

    So the open sets in the topology induced on A from Y are exactly the same as the open sets in the topology induced on A from X. So these topologies coincide.

    Notice two things. I didn't use any theorems, only what it means to be open in each respective topology, and, I didn't introduce any symbols for the topologies, which makes things easier and cleaner imho. Hope this helps.
     
    Last edited: Nov 19, 2007
  4. Nov 20, 2007 #3
    I don't quite understand.
    why because V is open in Y we have V=YnB for some B open in X, the fact that it's open in the subspace topology means that it's open in every other topology on Y, cause i think that this is what is assumed here, cause the above assertion is valid when V is in T_Y although from our assumption it's in T'.
     
  5. Nov 20, 2007 #4

    morphism

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    U is open in the subspace A of Y
    iff U=A[itex]\cap[/itex]V where V is open in the subspace Y of X
    iff U=A[itex]\cap[/itex](Y[itex]\cap[/itex]G)=A[itex]\cap[/itex]G where G is open in X (the second equality holds because A[itex]\cap[/itex]Y=A).
     
    Last edited: Nov 20, 2007
  6. Nov 20, 2007 #5
    OK morphism repeating it doesnt explain it.

    as I said U is in T_A_Y iff there exists V in T' such that U=AnV where T' is the topology on Y, the definition of subspace topology is quite clear every intersection of an open set in the topology on Y with A is in this topology.
    now, if V is in T' does it means that it's in T_Y?
    I mean does a set has only one topology on it ( i think not).
     
  7. Nov 20, 2007 #6

    morphism

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    It's stated that the topology on Y is the subspace topology it gets from X...
     
  8. Nov 21, 2007 #7
    you mean that by saying Y is a subspace of X, it's actually saying that the subspace topology is the topology on Y.
    yes, I see your point, thanks.
     
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