How to find the closure in $l^{\infty}$ of the set
$$M =\{(a_1,a_2,..........) : \textrm{all but finitely many } a_i=0\}?$$
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De donde eres Andres? – May 14 '14 at 02:45
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Google Translate: "Where are you Andres?" :) Don't worry everyone, I've got this – May 14 '14 at 02:46
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Also, that seems to be a personal question... – May 14 '14 at 02:48
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I don't get the joke... – IAmNoOne May 14 '14 at 02:49
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Congratulations to the translation team--nice cooperative effort. Now, to answer it ;) – MPW May 14 '14 at 02:52
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@Nameless "Estaban hablando en español, así que estaba traduciendo para todos." – May 14 '14 at 02:52
1 Answers
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Estoy pensando que "cerradura" es el "closure" del conjunto. Si este es el caso, creo que la respuesta es $\{x = (a_1, a_2, a_3, \ldots): \lim_{n \rightarrow \infty} a_n = 0\}$. Considera $x_n \in M$ and $x_n \rightarrow x= (a_1, a_2, a_3, \ldots)$ en $\| \cdot \|_\infty$
Prove it by showing that any sequence in the closure converges to zero and that any sequence that converges to zero is in the closure.
The key is that if $x$ is in the closure, then there exists a sequence $x_n$ in $M$ such that $\lim_{n \rightarrow \infty} \|x-x_n\| = 0$
Paul Hurst
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"I'm thinking that "lock" is the "closure" of the set. If this is the case, I think the answer is..." the mad translator strikes again – May 14 '14 at 02:54