A jew's harp exercise!

[fu^2]

Show that [tex]l_1=\{(a_j)_{j\in\mathbb{N}}\mid a_j\in\mathbb{R}, \displaystyle\sum_{j\geq 0}|a_j|<+\infty\}[/tex], with the standard norm, i,e. [tex]||a||=\sum_j|a_j|[/tex] is separable.

[Fioravante Patrone1]

"cirasa":How shameful I'm an ass with English!

Exactly like me, when I was as old as you

[cirasa]

How shameful I'm an ass with English!

[Fioravante Patrone1]

"cirasa":
I'm giving to you a proof in the case [tex]p=1[/tex].

Let suppose that [tex]a_n\geq 0[/tex]. If it's not so, you can substitute [tex]a_n[/tex] with [tex]|a_n|[/tex].
Since [tex]\displaystyle \sum_{n=N}^\infty a_n\stackrel{N\to\infty}{\to}0[/tex], for every [tex]\varepsilon>0[/tex] there exists [tex]N[/tex] s.t. [tex]\displaystyle \sum_{n=N}^\infty a_n<\frac{\varepsilon}{2}[/tex]
For every [tex]n
So we have the succession [tex]\displaystyle q=(q_n)_{n\in\mathbb{N}}[/tex].
Now we can calculate [tex]\displaystyle \|a-q\|_{l^1}[/tex]. We have:
[tex]\displaystyle \|a-q\|_{l^1}=\sum_{n=0}^\infty|a_n-q_n|=\sum_{n=0}^{N-1}|a_n-q_n|+\sum_{n=N}^\infty a_n<\sum_{n=0}^{N-1}\frac{\varepsilon}{2^{n+1}}+\frac{\varepsilon}{2}[/tex]
[tex]\displaystyle <\sum_{n=0}^\infty\frac{\varepsilon}{2^{n+1}}+\frac{\varepsilon}{2}=\varepsilon[/tex]

Finally, we have to count how many successions are formed by rational numbers and are definitively null.
But I am too tired. So I'll wait for some friends that will finish the proof for me.

Ecco la mia versione (ma attenti che io non ho nessun rispetto per le lingue...):


I will provide a proof in the case [tex]p=1[/tex].

Assume that [tex]a_n\geq 0[/tex]. If not, you can replace [tex]a_n[/tex] with [tex]|a_n|[/tex].
Since [tex]\displaystyle \sum_{n=N}^\infty a_n\stackrel{N\to\infty}{\to}0[/tex], for every [tex]\varepsilon>0[/tex] there exists [tex]N[/tex] s.t. [tex]\displaystyle \sum_{n=N}^\infty a_n<\frac{\varepsilon}{2}[/tex]
For every [tex]n
So, we have defined the sequence [tex]\displaystyle q=(q_n)_{n\in\mathbb{N}}[/tex].
Let us evaluate [tex]\displaystyle \|a-q\|_{l^1}[/tex]. We have:
[tex]\displaystyle \|a-q\|_{l^1}=\sum_{n=0}^\infty|a_n-q_n|=\sum_{n=0}^{N-1}|a_n-q_n|+\sum_{n=N}^\infty a_n<\sum_{n=0}^{N-1}\frac{\varepsilon}{2^{n+1}}+\frac{\varepsilon}{2}[/tex]
[tex]\displaystyle <\sum_{n=0}^\infty\frac{\varepsilon}{2^{n+1}}+\frac{\varepsilon}{2}=\varepsilon[/tex]

We are left with the question: how many are the sequences of rational numbers which are eventually null?
Well, I'm too tired, now. So I'll wait for some friends that will finish the proof for me.

[cirasa]

Thank you!
Here I will learn to use my English!

[dissonance]

I'd like to add a generalization. The same result applies for all [tex]L^p(\Omega)[/tex], provided that [tex]1 \le p < \infty[/tex] and that [tex]( \Omega , \mathcal{M} , \mu)[/tex] is separable, i.e. there exist a countable collection [tex]\{ B_n \}_{n \in \mathbb{N}} \subset \mathcal{M}[/tex] of measurables s.t. [tex]\mathcal{M}[/tex] is the smallest [tex]\sigma[/tex]-algebra containing all of the [tex]B_n[/tex]. (cfr. Brezis Analisi funzionale, Teorema IV.13 - the proof is given only in the special case [tex]\Omega = \mathbb{R}^N[/tex]). Can you prove it? I don't know if it will be an easy task, so be warned.

[OT] @ViciousGoblin: Obviously you're right - I've checked in the dictionary and it says: "definitely: sicuramente, certamente", while "eventually:alla fine, infine". Thank you!

[ViciousGoblin]

"dissonance":[quote="cirasa"]a succession [tex]q=(q_n)_{n\in\mathbb{N}}[/tex] formed by rational numbers that are definitely null

I was wondering how would it possible to rewrite this statement in order that it sounds less as translated from Italian (which is the problem my statements always have - for example, this very statement is brutally translated from Italian ). What about

a rational sequence [tex]q=(q_n)_{n \in \mathbb{N}}[/tex] definitely null

Is is correct to write "definitely null" in the end? Maybe

a definitely null rational sequence [tex]q=(q_n)_{n \in \mathbb{N}}[/tex]

is better? What do you think?[/quote]

I guess "definitely" has a different meaning in english compared to "definitivamente" - I am definitely sure about this. The correct adverb should be "eventually" (while "eventualmente" should translate with "possibly") - I also sometimes write "for $n$ large" (or "for large $n$" ?? - here I am hesitant, but may be both would do) .
If I were to write the above sentence I would use "a sequence of rationals which are eventually zero" .

But don't take me too seriously....

[fu^2]

ok, is a good solution!

The same construction you can do in [tex]C_0=\{a\in l^{\infty} : |a_i|\to 0\}[/tex] and, as you said, in all [tex]l^p[/tex], with [tex]1\neq p <+\infty[/tex]

[dissonance]

"cirasa":I think you're right and my English is horrible!

You say so because you haven't seen mine... This is a good place to learn, though.

[cirasa]

"dissonance":Is is correct to write "definitely null" in the end? Maybe

a definitely null rational sequence [tex]q=(q_n)_{n \in \mathbb{N}}[/tex]

is better? What do you think?

I think you're right and my English is horrible!

[dissonance]

"cirasa":a succession [tex]q=(q_n)_{n\in\mathbb{N}}[/tex] formed by rational numbers that are definitely null

I was wondering how would it possible to rewrite this statement in order that it sounds less as translated from Italian (which is the problem my statements always have - for example, this very statement is brutally translated from Italian ). What about

a rational sequence [tex]q=(q_n)_{n \in \mathbb{N}}[/tex] definitely null

Is is correct to write "definitely null" in the end? Maybe

a definitely null rational sequence [tex]q=(q_n)_{n \in \mathbb{N}}[/tex]

is better? What do you think?

[cirasa]

Sorry, but my English is so bad. I hope I won't do too many grammatical errors.


I'm giving to you a proof in the case [tex]p=1[/tex].

Let suppose that [tex]a_n\geq 0[/tex]. If it's not so, you can substitute [tex]a_n[/tex] with [tex]|a_n|[/tex].
Since [tex]\displaystyle \sum_{n=N}^\infty a_n\stackrel{N\to\infty}{\to}0[/tex], for every [tex]\varepsilon>0[/tex] exists [tex]N[/tex] s.t. [tex]\displaystyle \sum_{n=N}^\infty a_n<\frac{\varepsilon}{2}[/tex]
For every [tex]n
So we have the succession [tex]\displaystyle q=(q_n)_{n\in\mathbb{N}}[/tex].
Now we can calculate [tex]\displaystyle \|a-q\|_{l^1}[/tex]. We have:
[tex]\displaystyle \|a-q\|_{l^1}=\sum_{n=0}^\infty|a_n-q_n|=\sum_{n=0}^{N-1}|a_n-q_n|+\sum_{n=N}^\infty a_n<\sum_{n=0}^{N-1}\frac{\varepsilon}{2^{n+1}}+\frac{\varepsilon}{2}[/tex]
[tex]\displaystyle <\sum_{n=0}^\infty\frac{\varepsilon}{2^{n+1}}+\frac{\varepsilon}{2}=\varepsilon[/tex]

Finally, we have to count how many successions are formed by rational numbers and are definitively null.
But I am too tired. So I'll wait for some friends that will finish the proof for me.

[gugo82]

"cirasa":For every [tex]a=(a_n)_{n\in\mathbb{N}}[/tex] you can approximate [tex]a[/tex] with a succession [tex]q=(q_n)_{n\in\mathbb{N}}[/tex] formed by rational numbers that are definitively null.

We all would like to read a proof of this statement.

[cirasa]

For every [tex]a=(a_n)_{n\in\mathbb{N}}[/tex] you can approximate [tex]a[/tex] with a succession [tex]q=(q_n)_{n\in\mathbb{N}}[/tex] formed by rational numbers that are definitively null.
It's true for every [tex]l^p[/tex]-space ([tex]1\leq p<\infty[/tex])!

It's my first time I write my post in English, I hope you understand it. Bye!