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Math Help - Open sets in a separable metric space

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    Open sets in a separable metric space

    Suppose that (X, \rho ) is a separable metric space. Then there exists a sequence  \{ x_n \} such that it is dense in X. Prove that any open set  U \subset X is a union of elements from the set  B = \{ B(x_n,r) : r \in \mathbb {Q} \} .

    My idea so far.

    So since X is separable, I know that every nonempty open subset U contains at least one element of the sequence  \{ x_n \} .

    So say that  U_1 is an nonempty open subset in X, then there should exist at least one element from the sequence, say  x_m, \ m \in \mathbb {N}, that is contained in U. And since U is open, we can find some  \epsilon _1 > 0 such that  B(x_m, \epsilon _1) \subset U .

    Also, I believe that I can pick a point, say  y \in U , such that  x_m \in B(y, \delta ) ^c for some small  \delta > 0 . Then I should be able to find another point x_i \in B(y, \delta ) since it is open. Then I can get another  B(x_i , \epsilon _2 ) ...

    But how should I proceed to get  U \subset \bigcup B(x_n,r) ?

    Thank you!
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    Quote Originally Posted by tttcomrader View Post
    Suppose that (X, \rho ) is a separable metric space. Then there exists a sequence  \{ x_n \} such that it is dense in X. Prove that any open set  U \subset X is a union of elements from the set  B = \{ B(x_n,r) : r \in \mathbb {Q} \} .

    My idea so far.

    So since X is separable, I know that every nonempty open subset U contains at least one element of the sequence  \{ x_n \} .

    So say that  U_1 is an nonempty open subset in X, then there should exist at least one element from the sequence, say  x_m, \ m \in \mathbb {N}, that is contained in U. And since U is open, we can find some  \epsilon _1 > 0 such that  B(x_m, \epsilon _1) \subset U .

    Also, I believe that I can pick a point, say  y \in U , such that  x_m \in B(y, \delta ) ^c for some small  \delta > 0 . Then I should be able to find another point x_i \in B(y, \delta ) since it is open. Then I can get another  B(x_i , \epsilon _2 ) ...

    But how should I proceed to get  U \subset \bigcup B(x_n,r) ?
    Let x\in U. Since U is open, there is a neighbourhood of x contained in U. So there exists a rational number r such that B(x,r)\subset U.

    Next, the sequence \{x_n\} is dense in X, so there exists m such that \rho(x_m,x)<r/2. Then x\in B(x_m,r/2)\subset B(x,r)\subset U.

    Thus every element of U lies in a set in B that is contained in U. The union of these sets (for x\in U) is therefore the whole of U.
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    Quote Originally Posted by tttcomrader View Post
    Suppose that (X, \rho ) is a separable metric space. Then there exists a sequence  \{ x_n \} such that it is dense in X. Prove that any open set  U \subset X is a union of elements from the set  B = \{ B(x_n,r) : r \in \mathbb {Q} \} .
    I know that you want comment on you have done.
    But frankly I do not follow what you are trying to do.

    But this follows at once from a standard theorem.
    A separable metric space has a countable basis.
    If you can prove that then you will have done this problem.
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    Quote Originally Posted by Opalg View Post
    Let x\in U. Since U is open, there is a neighbourhood of x contained in U. So there exists a rational number r such that B(x,r)\subset U.

    Next, the sequence \{x_n\} is dense in X, so there exists m such that \rho(x_m,x)<r/2. Then x\in B(x_m,r/2)\subset B(x,r)\subset U.

    Thus every element of U lies in a set in B that is contained in U. The union of these sets (for x\in U) is therefore the whole of U.

    I follow the proof, but how do I know that  B(x_m,r/2)\subset B(x,r) ? Thanks.

    So if I pick an element  y \in B(x,r) , I have  \rho (x,y) < r , and since x\in B(x_m, \frac {r}{2} ), I then have  \rho (x,x_m) < \frac {r}{2} <r.

    Pick  y \in B(x,r), then I have  \rho (y,x)<r , is that how I should proceed?
    Last edited by tttcomrader; September 24th 2009 at 10:40 AM.
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  5. #5
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    Quote Originally Posted by tttcomrader View Post
    I follow the proof, but how do I know that  B(x_m,r/2)\subset B(x,r) ? Thanks.

    So if I pick an element  y \in B(x,r) , I have  \rho (x,y) < r , and since x\in B(x_m, \frac {r}{2} ), I then have  \rho (x,x_m) < \frac {r}{2} <r.

    Pick  y \in B(x,r), then I have  \rho (y,x)<r , is that how I should proceed?
    To show that  B(x_m,r/2)\subset B(x,r) , you need to pick y\in B(x_m,r/2) . Then (triangle inequality!) \rho(y,x)\leqslant\rho(y,x_m) + \rho(x_m,x) < \tfrac r2 + \tfrac r2 = r, and so y\in B(x,r).
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    Just a remark :
    So since X is separable, I know that every nonempty open subset U contains at least one element of the sequence \{ x_n \}.
    This is actually because \{x_n\} is dense in E.
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