Return to Answer

Let $\ f(m)\ $ be the number of the $m$-th partition intervals which intersect $\ A.\ $

Theorem 1  There exists $\ A\ $ which is not dense in any non-trivial interval whenever

$$ \lim_{m=\infty} \frac {f(m)}m\ =\ 0 $$

Proof  The required $\ A\ $ can be given as follows:

$$ A\ :=\\ \ \left\{ \frac {2\cdot k-1}{2\cdot m}: \ m\in\Bbb N\ \ \text{and} \ \ k\in\Bbb Z\ \ \text{and}\ \ 0<k\le f(m)\,\right\} $$

End of Proof

On the other hand,

Theorem 2  If $\ \lim_{m=\infty} f(m) = \infty\ $ then there exists respective $\ A\ $ which is dense in $\ [0;1]$.

Proof  The required $\ A\ $ can be given by:

$$ A\ :=\\ \ \left\{ \frac{2\cdot k-1}{2\cdot f(m)}:\ m\in\Bbb N \ \ \text{and}\ \ f(m)>0\ \ \text{and}\ \ k\in\Bbb Z \ \ \text{and}\ \ 0<k\le f(m) \right\} $$

End of Proof

Let the number of the $m$-th partition intervals which intersect $\ A\ $ be $\ \ge f(m).\ $ Then

Theorem  There exists $\ A\ $ which is not dense in any non-trivial interval whenever

$$ \lim_{m=\infty} \frac {f(m)}m\ =\ 0 $$

Proof  The required $\ A\ $ can be given as follows:

$$ A\ :=\\ \ \left\{ \frac {2\cdot k-1}{2\cdot m}: \ m\in\Bbb N\ \ \text{and} \ \ k\in\Bbb Z\ \ \text{and}\ \ 0<k\le f(m)\,\right\} $$

End of Proof

Let $\ f(m)\ $ be the number of the $m$-th partition intervals which intersect $\ A.\ $

Theorem 1  There exists $\ A\ $ which is not dense in any non-trivial interval whenever

$$ \lim_{m=\infty} \frac {f(m)}m\ =\ 0 $$

Proof  The required $\ A\ $ can be given as follows:

$$ A\ :=\\ \ \left\{ \frac {2\cdot k-1}{2\cdot m}: \ m\in\Bbb N\ \ \text{and} \ \ k\in\Bbb Z\ \ \text{and}\ \ 0<k\le f(m)\,\right\} $$

End of Proof ce On the other hand,

Theorem 2  If $\ \lim_{m=\infty} f(m) = \infty\ $ then there exists respective $\ A\ $ which is dense in $\ [0;1]$.

Proof  The required $\ A\ $ can be given by:

$$ A\ :=\ \left\{ \frac{2\cdot k-1}{2\cdot f(m)}:\ m\in\Bbb N \ \ \text{and}\ \ f(m)>0\ \ \text{and}\ \ k\in\Bbb Z \ \ \text{and}\ \ 0<k\le f(m) \right\} $$

End of Proof

Let $\ f(m)\ $ be the number of the $m$-th partition intervals which intersect $\ A.\ $

Theorem 1  There exists $\ A\ $ which is not dense in any non-trivial interval whenever

$$ \lim_{m=\infty} \frac {f(m)}m\ =\ 0 $$

Proof  The required $\ A\ $ can be given as follows:

$$ A\ :=\\ \ \left\{ \frac {2\cdot k-1}{2\cdot m}: \ m\in\Bbb N\ \ \text{and} \ \ k\in\Bbb Z\ \ \text{and}\ \ 0<k\le f(m)\,\right\} $$

End of Proof

On the other hand,

Theorem 2  If $\ \lim_{m=\infty} f(m) = \infty\ $ then there exists respective $\ A\ $ which is dense in $\ [0;1]$.

Proof  The required $\ A\ $ can be given by:

$$ A\ :=\\ \ \left\{ \frac{2\cdot k-1}{2\cdot f(m)}:\ m\in\Bbb N \ \ \text{and}\ \ f(m)>0\ \ \text{and}\ \ k\in\Bbb Z \ \ \text{and}\ \ 0<k\le f(m) \right\} $$

End of Proof

It is not possible, and even quite a bit more is not possible.

Let $\ f(m)\ $ be the number of the $m$-th partition intervals which intersect $\ A.\ $

Theorem  There exists $\ A\ $ which is not dense in any non-trivial interval whenever

$$ \lim_{m=\infty} \frac {f(m)}m\ =\ 0 $$

Proof  The required $\ A\ $ can be given as follows:

$$ A\ :=\\ \ \left\{ \frac {2\cdot k-1}{2\cdot m}: \ m\in\Bbb N\ \ \text{and} \ \ k\in\Bbb Z\ \ \text{and}\ \ 0<k\le f(m)\,\right\} $$

End of Proof

Let $\ f(m)\ $ be the number of the $m$-th partition intervals which intersect $\ A.\ $

Theorem 1  There exists $\ A\ $ which is not dense in any non-trivial interval whenever

$$ \lim_{m=\infty} \frac {f(m)}m\ =\ 0 $$

Proof  The required $\ A\ $ can be given as follows:

$$ A\ :=\\ \ \left\{ \frac {2\cdot k-1}{2\cdot m}: \ m\in\Bbb N\ \ \text{and} \ \ k\in\Bbb Z\ \ \text{and}\ \ 0<k\le f(m)\,\right\} $$

End of Proof ce On the other hand,

Theorem 2  If $\ \lim_{m=\infty} f(m) = \infty\ $ then there exists respective $\ A\ $ which is dense in $\ [0;1]$.

Proof  The required $\ A\ $ can be given by:

$$ A\ :=\ \left\{ \frac{2\cdot k-1}{2\cdot f(m)}:\ m\in\Bbb N \ \ \text{and}\ \ f(m)>0\ \ \text{and}\ \ k\in\Bbb Z \ \ \text{and}\ \ 0<k\le f(m) \right\} $$

End of Proof