Results 1 to 7 of 7

Math Help - Applying Baire's Thm

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    147

    Applying Baire's Thm

    Call a vector X = (x, y) in R2 "resonant" if it satisfies an equation
    of the form ax + by + c = 0 where a, b, c are integers, not all zero.
    Otherwise X is non-resonant. Use Baire theory to show that the set
    of "non-resonant" vectors is dense in R2.

    All I have so far is that a vector is non-resonant if it satisfies x not rational, y not rational or x /= q1 + q2*y, where q1 and q1 are rational.

    Note: Apply Baire's theory is a nice way of saying that we have to show the set is thick; that is, the intersection of countably many open and dense sets.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by southprkfan1 View Post
    Call a vector X = (x, y) in R2 "resonant" if it satisfies an equation
    of the form ax + by + c = 0 where a, b, c are integers, not all zero.
    Otherwise X is non-resonant. Use Baire theory to show that the set
    of "non-resonant" vectors is dense in R2.

    All I have so far is that a vector is non-resonant if it satisfies x not rational, y not rational or x /= q1 + q2*y, where q1 and q1 are rational.

    Note: Apply Baire's theory is a nice way of saying that we have to show the set is thick; that is, the intersection of countably many open and dense sets.
    Do you have to show first that \mathbb{R}^2 is a Baire space or can you just conclude that it is since it's a completely metrizable space?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    Posts
    147
    Quote Originally Posted by Drexel28 View Post
    Do you have to show first that \mathbb{R}^2 is a Baire space or can you just conclude that it is since it's a completely metrizable space?
    No. We can just use the fact that R2 is complete and so all we have to do is show that the required set (that is, of non-resonant vectors) is thick and the conclusion will follow.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by southprkfan1 View Post
    No. We can just use the fact that R2 is complete and so all we have to do is show that the required set (that is, of non-resonant vectors) is thick and the conclusion will follow.
    My first inclination is that clearly if we let K be the set of resonant vectors then clearly \varphi:K\mapsto\mathbb{Z}^3 given by ax+by+c\mapsto (a,b,c) is an injection. Thus, K is countable. So, for every k\in K consider \mathbb{R}^2-k. This is clearly an open dense set and \mathbb{R}^2-K=\bigcap_{k\in K}\left\{\mathbb{R}^2-k\right\}.

    That could be wrong though, I didn't check it.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2010
    Posts
    147
    Quote Originally Posted by Drexel28 View Post
    My first inclination is that clearly if we let K be the set of resonant vectors then clearly \varphi:K\mapsto\mathbb{Z}^3 given by ax+by+c\mapsto (a,b,c) is an injection. Thus, K is countable. So, for every k\in K consider \mathbb{R}^2-k. This is clearly an open dense set and \mathbb{R}^2-K=\bigcap_{k\in K}\left\{\mathbb{R}^2-k\right\}.

    That could be wrong though, I didn't check it.
    Are we sure the function is an injection? I think you may have to add a condition that (a,b,c) have no common factor or something?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by southprkfan1 View Post
    Are we sure the function is an injection? I think you may have to add a condition that (a,b,c) have no common factor or something?
    \phi(ax+by+c)=\phi(a'x+b'y+c')\implies (a,b,c)=(a',b',c')\implies  a=a'\text{ }b=b'\text{ }c=c'\implies ax+by+c=a'x+b'y+c'
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jun 2009
    Posts
    113
    Quote Originally Posted by southprkfan1 View Post
    Call a vector X = (x, y) in R2 "resonant" if it satisfies an equation
    of the form ax + by + c = 0 where a, b, c are integers, not all zero.
    Otherwise X is non-resonant. Use Baire theory to show that the set
    of "non-resonant" vectors is dense in R2.

    All I have so far is that a vector is non-resonant if it satisfies x not rational, y not rational or x /= q1 + q2*y, where q1 and q1 are rational.

    Note: Apply Baire's theory is a nice way of saying that we have to show the set is thick; that is, the intersection of countably many open and dense sets.
    For any (a,b,c)\in \mathbb{Z}^3 , not all 0,you can easily check that V_{a,b,c}:=\{(x,y,z):ax+by+c\neq 0\} is open. If you show that V_{a,b,c} is dense, you conclude because the set V of all non resonant vectors is the countable intersection of all the V_{a,b,c}. To check this density you have to assume ax+by+c=0 and show that there is (x_1,y_1) "as near as desired" from (x,y) such that ax_1+by_1+c\neq 0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Baire space homeomorphism question - again
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: January 25th 2010, 11:57 AM
  2. powers of Baire space are homeomorphic to Baire space
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: December 15th 2009, 03:46 PM
  3. base for the topology of baire space N^N
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: December 12th 2009, 07:49 AM
  4. Baire Category and Residual Sets
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: August 4th 2009, 09:55 AM
  5. Baire Category Theorem?
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: November 22nd 2008, 01:47 AM

Search Tags


/mathhelpforum @mathhelpforum