Find a subset of the real numbers

[mathmari]

Hey!

I have to find an open and dense subset of the real numbers with arbitrarily small measure.

Since the set of the rational numbers is dense, could we use a subset of the rationals?? (Wondering)

How could I find such a subset, that the measure is arbitrarily small?? (Wondering)

 

[mathmari]

Could I take the subset $$A_n= \left ( q_n-\frac{\epsilon}{2}, q_n+\frac{\epsilon}{2} \right ), \epsilon>0$$ ?? (Wondering)

Then $A= \cup_n A_n$.

$$\mu(A)=\mu \left ( \cup_n A_n \right ) \leq \sum_n \mu (A_n)=\sum_n \mu \left ( \left ( q_n-\frac{\epsilon}{2}, q_n+\frac{\epsilon}{2} \right ) \right )=\sum_n \epsilon$$

 

[Euge]

Could I take the subset $$A_n= \left ( q_n-\frac{\epsilon}{2}, q_n+\frac{\epsilon}{2} \right ), \epsilon>0$$ ?? (Wondering)

Then $A= \cup_n A_n$.

$$\mu(A)=\mu \left ( \cup_n A_n \right ) \leq \sum_n \mu (A_n)=\sum_n \mu \left ( \left ( q_n-\frac{\epsilon}{2}, q_n+\frac{\epsilon}{2} \right ) \right )=\sum_n \epsilon$$

Is $\{q_n\}_{n=1}^\infty$ an enumeration of the rationals? If so, then you are on the right track. You should change the $\epsilon/2$ in your $A_n$ to $\epsilon/2^{n+2}$. Then your $A$ is open and dense, with $\mu(A) < \epsilon$.

 

[mathmari]

Is $\{q_n\}_{n=1}^\infty$ an enumeration of the rationals?

What do you mean by "an enumeration of rationals"?? (Wondering)

You should change the $\epsilon/2$ in your $A_n$ to $\epsilon/2^{n+2}$. Then your $A$ is open and dense, with $\mu(A) < \epsilon$.

Yes, you're right! (Yes)

By changing the $\epsilon/2$ to $\epsilon/2^{n+2}$ we get the sum $$\sum_n \frac{\epsilon}{2^{n+1}}=\frac{\epsilon}{2} \sum_n \left ( \frac{1}{2} \right )^n=\frac{\epsilon}{2}2=\epsilon$$ (Mmm)

 

[Euge]

What do you mean by "an enumeration of rationals"?? (Wondering)

It means that the map $f : \Bbb N \to \Bbb Q$ given by $f(n) = q_n$ for all $n\in \Bbb N$, is a bijection.

By changing the $\epsilon/2$ to $\epsilon/2^{n+2}$ we get the sum $$\sum_n \frac{\epsilon}{2^{n+1}}=\frac{\epsilon}{2} \sum_n \left ( \frac{1}{2} \right )^n=\frac{\epsilon}{2}2=\epsilon$$ (Mmm)

You're very close, but since we're summing over all $n\ge 1$,

\(\displaystyle \sum_n \frac{\epsilon}{2^{n+1}} = \frac{\epsilon}{2} < \epsilon.\)

 

[mathmari]

It means that the map $f : \Bbb N \to \Bbb Q$ given by $f(n) = q_n$ for all $n\in \Bbb N$, is a bijection.

I understand!

So, do I have to mention it before I take this subset?? (Wondering)

You're very close, but since we're summing over all $n\ge 1$,

\(\displaystyle \sum_n \frac{\epsilon}{2^{n+1}} = \frac{\epsilon}{2} < \epsilon.\)

Oh, I see! (flower)

 

[Euge]

I understand!

So, do I have to mention it before I take this subset?? (Wondering)

You just need to mention in the beginning that $\{q_n\}$ is an enumeration of the rationals.

 

[mathmari]

You just need to mention in the beginning that $\{q_n\}$ is an enumeration of the rationals.

Ok! Thanks a lot! (Happy)