Results 1 to 6 of 6

Thread: Open sets in a separable metric space

  1. #1
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2

    Open sets in a separable metric space

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

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

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

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

    Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by tttcomrader View Post
    Suppose that $\displaystyle (X, \rho ) $ is a separable metric space. Then there exists a sequence $\displaystyle \{ x_n \} $ such that it is dense in X. Prove that any open set $\displaystyle U \subset X $ is a union of elements from the set $\displaystyle 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 $\displaystyle \{ x_n \} $.

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

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

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

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

    Thus every element of U lies in a set in $\displaystyle B$ that is contained in U. The union of these sets (for $\displaystyle x\in U$) is therefore the whole of U.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    Quote Originally Posted by tttcomrader View Post
    Suppose that $\displaystyle (X, \rho ) $ is a separable metric space. Then there exists a sequence $\displaystyle \{ x_n \} $ such that it is dense in X. Prove that any open set $\displaystyle U \subset X $ is a union of elements from the set $\displaystyle 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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2
    Quote Originally Posted by Opalg View Post
    Let $\displaystyle x\in U$. Since U is open, there is a neighbourhood of x contained in U. So there exists a rational number r such that $\displaystyle B(x,r)\subset U.$

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

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

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

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

    Pick $\displaystyle y \in B(x,r)$, then I have $\displaystyle \rho (y,x)<r $, is that how I should proceed?
    Last edited by tttcomrader; Sep 24th 2009 at 09:40 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by tttcomrader View Post
    I follow the proof, but how do I know that $\displaystyle B(x_m,r/2)\subset B(x,r) $? Thanks.

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

    Pick $\displaystyle y \in B(x,r)$, then I have $\displaystyle \rho (y,x)<r $, is that how I should proceed?
    To show that $\displaystyle B(x_m,r/2)\subset B(x,r) $, you need to pick $\displaystyle y\in B(x_m,r/2) $. Then (triangle inequality!) $\displaystyle \rho(y,x)\leqslant\rho(y,x_m) + \rho(x_m,x) < \tfrac r2 + \tfrac r2 = r$, and so $\displaystyle y\in B(x,r)$.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Just a remark :
    So since X is separable, I know that every nonempty open subset U contains at least one element of the sequence $\displaystyle \{ x_n \}$.
    This is actually because $\displaystyle \{x_n\}$ is dense in E.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Open sets in a metric space
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Mar 20th 2011, 11:58 AM
  2. Metric spaces, open sets, and closed sets
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Mar 16th 2011, 05:17 PM
  3. open ball metric space
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Nov 1st 2010, 11:39 AM
  4. Replies: 1
    Last Post: Oct 30th 2010, 01:50 PM
  5. Metric Space , Close and open
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Sep 28th 2009, 07:18 PM

Search Tags


/mathhelpforum @mathhelpforum