- #1
cuak2000
- 8
- 0
Hey, everyone. I'm trying to prove the following:
[tex] f_n [/tex] and [tex] f_n [/tex] are real-valued function in [tex] \Omega [/tex]
[tex] \{\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 \}
[/tex]
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:
[tex] \exists k \in \mathbb{N} \quad \forall N \quad \exists n \geq N \quad | f_n(\omega) \nrightarrow f(\omega) | \geq 1/k
[/tex]
On the other hand, the last part of R is
[tex] \bigcap^{\infty}_{N=1} \bigcup^{\infty}_{n=1}
\{ \omega : | f_n(\omega) - f(\omega) | \geq 1/k \} [/tex]
which basically takes all the [tex] \omega [/tex] that [tex] \forall N [/tex] have
at least one [tex] n \geq N [/tex] 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
[tex] f_n [/tex] and [tex] f_n [/tex] are real-valued function in [tex] \Omega [/tex]
[tex] \{\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 \}
[/tex]
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:
[tex] \exists k \in \mathbb{N} \quad \forall N \quad \exists n \geq N \quad | f_n(\omega) \nrightarrow f(\omega) | \geq 1/k
[/tex]
On the other hand, the last part of R is
[tex] \bigcap^{\infty}_{N=1} \bigcup^{\infty}_{n=1}
\{ \omega : | f_n(\omega) - f(\omega) | \geq 1/k \} [/tex]
which basically takes all the [tex] \omega [/tex] that [tex] \forall N [/tex] have
at least one [tex] n \geq N [/tex] 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
Last edited: