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Set A is open relative to Y iff A also contains points of a set B open in X?

  1. Oct 7, 2012 #1
    Problem:

    I'm trying to make sure i understand the following proof:

    Suppose we have the metric spaces (X, d) and (Y, d) with Y < X. Then

    A is open in Y [itex]\Leftrightarrow[/itex] A = B [itex]\bigcap[/itex] Y where B is an open set in X.​

    Here is the proof I have written down:

    ([itex]\Rightarrow[/itex])
    - Assume A open in Y. We need to produce a B s.t. A = B [itex]\bigcap[/itex] Y.
    - For p [itex]\in[/itex] A, there exists a radius about p r > 0 s.t. { q [itex]\in[/itex] Y: d(p, q) <r} < A.
    - Then let B[itex]_{p}[/itex] be the neighborhood of p in X with radius r[itex]_{p}[/itex].
    - So we have constructed a set B open in X s.t. A = B [itex]\bigcap[/itex] Y.


    ([itex]\Leftarrow[/itex])
    - Assume B open in X and A = B [itex]\bigcap[/itex] Y. Pick any p[itex]\in[/itex] A, which implies p is in B.
    - So there exists some r > 0 s.t. { q [itex]\in[/itex] X: d(p, q) <r} < B.
    - Observe that { q [itex]\in[/itex] X: d(p, q) <r} [itex]\bigcap[/itex] Y < B [itex]\bigcap[/itex] Y = A.
    - Thus we make the left hand side equal to { q [itex]\in[/itex] Y: d(p, q) <r}, i.e. neighborhoods in Y.
    - So there exists a neighborhood of p in Y, so it can also be contained in A.
    - Thus we constructed a set A open in Y.

    My points of confusion are...

    1. Are there any stronger arguments to make? I feel like my proof is quite vague and shaky. I welcome any corrections or advice.

    2. Does this theorem only work if Y has the subspace topology, i.e. the same topology which defines X? I suppose I should have been more careful in my notes unless I should assume that 'd' is the same for both X and Y...

    3. Am I right in assuming that A cannot be open in Y by simply being in Y? It seems like we can just say A is open in Y relative to Y's topology. I guess this goes back to #2, since I am confused if X and Y are being defined with different topologies.

    But does this mean we are also proving A is open in X? Again, this relates to my confusions in #2.

    4. The theorem says A must have parts of both B and Y to be open in Y. I think of B as being any possible open set in X. Now if A is simply a subset of B, A is open in X, correct?

    But since we want to prove A is open in a subset of X, namely Y, we need A to be the intersection of B and Y. B obviously isn't stated to be a subset of Y, so why is it that A is open in Y instead of simply being open in X?

    I think I'm getting confused as to why there needs to be an intersection of B and Y. I know that A is a subset of Y, but I can't figure out why else. Like, I feel that the theorem can simply say A is open in Y and that's it... Why is it that A must contain some of B?

    Is the theorem saying that for a subset A to be open in a subspace Y of X, then that subset must also contain some of the set B which is open in X? Again, doesn't that mean A is open in X as well?

    I'm confusing myself. I'd appreciate any clarity, thank you.
     
  2. jcsd
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