Re: Union of untersections

Hey wilhelm.

I would consider the smallest end-points in magnitude from the origin for each value of m.

The smallest in the negative is -1 and the smallest on the positive side is unbounded (since n/m -> infinity).

So with this you are going to have (-1,infinity) intuitively with the above reasoning (which may be wrong and if it is please point it out).

Re: Union of untersections

Quote:

Originally Posted by

**wilhelm** Hi. I was wondering if you could help me solve this. I've tried drawing it but it doesn't work, and anyway, I want to be able to solve it in a more elegant manner.

$\displaystyle \bigcap_{m=1}^{+ \infty} \bigcup_{n=1}^{+ \infty} (-\frac {m}{n}, \frac {n}{m})$

$\displaystyle \bigcap_{m=1}^{+ \infty}\left[ \bigcup_{n=1}^{+ \infty} (-\frac {m}{n}, \frac {n}{m})\right]$

Note that I added some [] to emphasize the union is done first. With union we "never loose any elements"

Thus for each $\displaystyle M$ the union yields $\displaystyle (-M,\infty)$.

Recall that with intersection we get only **the common part**.

So what is the common part of all of those unions?

Re: Union of untersections

Is the answer $\displaystyle (-1, + \infty)$ ?