# Math Help - Applying Baire's Thm

1. ## 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.

2. Originally Posted by southprkfan1
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?

3. Originally Posted by Drexel28
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.

4. Originally Posted by southprkfan1
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.

5. Originally Posted by Drexel28
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?

6. Originally Posted by southprkfan1
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?
$\phi(ax+by+c)=\phi(a'x+b'y+c')\implies (a,b,c)=(a',b',c')\implies$ $a=a'\text{ }b=b'\text{ }c=c'\implies ax+by+c=a'x+b'y+c'$

7. Originally Posted by southprkfan1
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 $ax_1+by_1+c\neq 0$.