Non-convergence written with sets

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

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<br />

On the other hand, the last part of R is
\bigcap^{\infty}_{N=1} \bigcup^{\infty}_{n=1} <br /> \{ \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.
Thanks in advance,

cd

 

[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} <br /> \{ \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<br />