# Non-convergence written with sets

1. Aug 10, 2010

### cuak2000

Hey, everyone. I'm trying to prove the following:

$$f_n$$ and $$f_n$$ are real-valued function in $$\Omega$$

$$\{\omega: f_n(\omega) \nrightarrow f(\omega) \} = \\ \bigcup^{\infty}_{k=1} \bigcap^{\infty}_{N=1} \bigcup^{\infty}_{n=1} \{ \omega : | f_n(\omega) - f(\omega) | \geq 1/k \}$$

I am convinced by the proof I've made up, but it isn't formal, so I would appreciate if you could help me give it more formality.
Let's call the left side of the equality L and the right side R.
L can be written:

$$\exists k \in \mathbb{N} \quad \forall N \quad \exists n \geq N \quad | f_n(\omega) \nrightarrow f(\omega) | \geq 1/k$$

On the other hand, the last part of R is
$$\bigcap^{\infty}_{N=1} \bigcup^{\infty}_{n=1} \{ \omega : | f_n(\omega) - f(\omega) | \geq 1/k \}$$

which basically takes all the $$\omega$$ that $$\forall N$$ have
at least one $$n \geq N$$ that makes the absolute difference bigger than 1/k
If you take the union for all k, then you have the definition for being in L.

cd

Last edited: Aug 10, 2010
2. Sep 26, 2010

### Landau

There's something wrong with your R. You take an intersection over N=1 to infty, but the index N does not appear in the collection over which the intersection is taken. You probably meant

$$\bigcup^{\infty}_{k=1} \bigcap^{\infty}_{N=1} \bigcup^{\infty}_{n=N} \{ \omega : | f_n(\omega) - f(\omega) | \geq 1/k \}$$

and then you are already done, since the condition for omega to belong to this one and to to LHS are the same, namely

$$\exists k \in \mathbb{N} \quad \forall N \quad \exists n \geq N \quad | f_n(\omega) \nrightarrow f(\omega) | \geq 1/k$$