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I go through a proof of the following.

Let $(\ell_1,d)$ be the metric space of all sequences $x = (\xi_i)_{i \in \mathbb{N}}$ with $\sum_{i=1}^{\infty} |\xi_i| < \infty$ and the metric $$ d(x,y) = \sum_{i=1}^{\infty} |\xi_i - \eta_i|, \qquad x = (\xi_i), y = (\eta_i). $$

Theorem: A subset $M$ of $l_1$ is totally bounded (pre-compact) iff

(i) There is a $K > 0$ with $\sum_{i=1}^{\infty} |\xi_i| \le K$ for all $x = (\xi_i) \in M$;

(ii) $\forall \varepsilon > 0 \exists n_0 \in \mathbb{N} \forall x = (\xi_i) \in M: \sum_{i=n_0}^{\infty} |\xi_i| < \varepsilon$.

The proof goes like this, let $M$ be a subset such that (i) and (ii) hold and let $x = (x_i)$ be a sequences in $\ell_1$ (i.e. a sequence of sequences), we show that it has a sub-sequence $x' = (x'_i)$ with $x'_i = (\xi^{(i)}_j)$ which is a Cauchy-Sequence. So let $\varepsilon > 0$ be given, then select $n_0$ such as in (ii), then for $n,m > n_0$ \begin{align*} d(x_n, x_m) &= \sum_{i=1}^{\infty} | \xi^{(n)}_i - \xi^{(m)}_j | \le \sum_{i=1}^{\infty} | \xi^{(n)}_i | - | \xi^{(m)}_j | \le 2\varepsilon \end{align*} (end of proof)

I dont understand the last steps, why is this sum smaller than $2\varepsilon$?.

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2 Answers 2

up vote 2 down vote accepted

Fix $N$ which works for $\varepsilon/2$: this gives $\sum_{i\geq N}|\xi_i^{(n)}-\xi_i^{(m)}|\leq \varepsilon$. Then use the fact that the sequence of real numbers $\{\xi_k^{(m)}\}_m$ are Cauchy (up to a subsequence) for each $1\leq k\leq N-1$: this gives for each $k$, $N_k$ such that $|\xi_k^{(m)}-\xi_k^{(n)}|\leq \frac\varepsilon{2N}$. Then take $n_0:=\max\{N,\max_{1\leq k\leq N-1}N_k\}$.

For the other direction, we take $\varepsilon>0$, and $n_0$ in (ii) which works for $\varepsilon/2$. Then we use pre-compactness of $[-K,K]^{n_0}$ to get pre-compactness of $M$.

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thanks, btw is the other direction trivial or why is it not mentioned in the proof? –  Stefan Oct 19 '12 at 17:02

I think that the correct thing is \begin{eqnarray} d(x_n, x_m) &=& \sum_{i=1}^{\infty} | \xi^{(n)}_i - \xi^{(m)}_j | \\ &\le &\sum_{i=1}^{\infty} | \xi^{(n)}_i | + | \xi^{(m)}_j |\\ &=& \sum_{i=1}^{\infty} | \xi^{(n)}_i | + \sum_{i=1}^{\infty} | \xi^{(m)}_j | \\ &\le& \varepsilon + \varepsilon . \\ &\le& 2\varepsilon \end{eqnarray}

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