Union of untersections

**wilhelm** · January 7th 2013, 09:48 PM

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.

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

Thank you.

**chiro** · January 8th 2013, 12:27 AM

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).

**Plato** · January 8th 2013, 07:44 AM

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

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

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

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

Recall that with intersection we get only the common part.

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

**wilhelm** · January 8th 2013, 11:33 AM

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

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