1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Subspace topology question.

  1. Nov 19, 2007 #1

    MathematicalPhysicist

    User Avatar
    Gold Member

    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

    MathematicalPhysicist

    User Avatar
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    MathematicalPhysicist

    User Avatar
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    It's stated that the topology on Y is the subspace topology it gets from X...
     
  8. Nov 21, 2007 #7

    MathematicalPhysicist

    User Avatar
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Subspace topology question.
  1. Topology question (Replies: 29)

  2. Topology question(s) (Replies: 3)

Loading...