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Math Help - Closure Problem

  1. #1
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    Closure Problem

    Hello everyone.

    I am trying to prove that if A \subseteq X is dense then for any open T we have T \subseteq Cl(A \cap T). I tried assuming not, and I arrive at the statement t \in A^c and I can't see why that would be a contradiction.

    Does anyone know of a good proof of this? Also, is it necessary that the T be open?
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  2. #2
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    First if I am correct is assuming that Cl is the closure operator, then I am puzzled by your confusion.
    Here are some facts.
    \left[ {\forall A\,,\,A \subseteq Cl(A)} \right], and Cl(A \cap B) \subseteq Cl(A) \cap Cl(B).

    Because A is dense in X you also have Cl(A) = X.

    Put all that together.
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  3. #3
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    Plato, you proved the opposite inclusion. You proved that:

    Cl(A \cap T) \subseteq Cl(T)

    I want to show that T \subseteq Cl(A \cap T) if T is open.

    As an example if A = \mathbb{Q} and T = (0,1) t hen (0, 1) \subseteq Cl(\mathbb{Q} \cap (0,1)) = [0, 1]
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  4. #4
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    Quote Originally Posted by mpetnuch View Post
    Plato, you proved the opposite inclusion.
    No I did not. If T is open then the statement follows as a matter of definitions. If there is a t which is not in Cl(A \cap T) then realize that t is a limit point of A. But some open set O contains t and is a subset of T. Think about that contradiction.

    You asked about T being open. Let T = \left\{ {\sqrt 2 } \right\}\quad \& \quad A = Q. It is clear that the statement is false using this example.
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  5. #5
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    Quote Originally Posted by Plato View Post
    No I did not. If T is open then the statement follows as a matter of definitions. If there is a t which is not in Cl(A \cap T) then realize that t is a limit point of A. But some open set O contains t and is a subset of T. Think about that contradiction.
    I am sorry, but I still don't see it. Since A is dense means that A intersects every non-empty open set. In particular A \cap T \ne \emptyset. I see this is the reason why it fails if T is not open.

    Also since A is dense means that every point is a limit point of A, so in particular t is a limit point of A \forall t \in T. And since T is open, we can find an open neighborhood of T, U, such that t \in U \subseteq T.

    Now, you asked me to assume that t \notin Cl(A \cap T) and consider that open set U defined above. Since t is a limit point of A implies that A \cap U\backslash\{t\}\ne \emptyset. But I still don't see the contradiction.
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  6. #6
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    Because T is open there is an open set containing t which is a subset of T. Any open containing t must contain must contain a point of A distinct from t. Hence the second point must be in (A \cap T).
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