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$c_0 \subset \ell^\infty$ is the subspace of all sequences of scalars converging to zero.

Let $x \in \bar{c_0}$. Then there exists a Cauchy sequence of sequences $(x^k)_{k \in \mathbb{N}}$ in $c_0$, where $x^k = (a_1^k, a_2^k,a_3^k \dots )$ such that $x^k \to x = (a_1, a_2, a_3, \dots) $ under the $\ell^\infty$ norm. We will show $x \in c_0$.

By definition of convergence, for any $\epsilon > 0$, there exists $N_1 \in \mathbb{N}$ such that $$\|x^k - x\|_{\ell^\infty} < \epsilon /3 \quad \forall \, \, k \ge N_1$$

This implies $|a_i^k - a_i| \le \sup_{j \ge 1} \{|a_j^k - a_j|\} < \epsilon/3 \quad \forall \,\, i \ge 1$,

Since $x$ is convergent in $\ell^\infty$, it follows that for all $k \in \mathbb N$ and $\epsilon > 0$, there exists $N_2(\epsilon) \in \mathbb N$ such that $$|a_i^k - a_j^k| < \epsilon/3 \quad \forall \, \, i, j \ge N_2$$

Then, $$|a_i - a_j| \le |a_i - a_i^k| + |a_i^k - a_j^k| + |a_j^k - a_j| < \epsilon \quad \forall \, \, i, j \ge N_2$$

This implies $x = (a_1, a_2, \dots )$ is a Cauchy sequence of numbers in $\ell^\infty$ and so it's convergent.To show $x \in c_0$, we must show $a_i \to 0$.

$|a_i|\le |a_i - a_i^k| + |a_i^k|$

This is where I get stuck. Can I claim $|a_i^k| < 2\epsilon /3$?

Daniel
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bananaspace
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  • $\lim_{k\to \infty} a^k_i =0$ by assumption. – Vincent Boelens Mar 28 '15 at 11:53
  • @bananaspace I can't understand when you say "since $x$ is convergent in $\ell^\infty$", what do you mean with that?, $\ell^\infty$ is composed by bounded sequences, not convergent sequences. I think that you wanted to use "since $x^k$ is a convergent sequence for each $k$", and in such a case, the value of $N_2$ would depend on $k$ too. – Daniel Mar 28 '15 at 19:42

2 Answers2

2

To see that a subspace of a metric space is closed, you only need to take any convergent sequence in the subspace and prove that the limit belongs to the subspace. In this case take a sequence $\left( x^{(n)}\right)_{n=1}^{\infty} \subset c_0$, such that $x^{(n)} \underset{n \to \infty}{\longrightarrow} x \in \ell^{\infty}$. Since each $x^{(n)} \in c_0$, we have \begin{equation} \lim_{j \to \infty } x_j^{(n)} = 0 \ \ \text{ for each $n \in \mathbb{N}$ } \ \ \ \ \ (1) \end{equation} Also, for every $\varepsilon >0$, there exist $n=N(\varepsilon) \in \mathbb{N}$, such if $n > N$, then $$ \|x^{(n)}- x \|< \varepsilon $$ So, for each $j \in \mathbb{N}$, if $n>N$ $$ |x^{(n)}_j - x_j| \leq \sup_{j \in \mathbb{N}} |x^{(n)}_j - x_j| = \|x^{(n)}- x \|< \varepsilon $$ that is, for all $j \in \mathbb{N}$, $x_j^{(n)} \underset{n \to \infty}{\longrightarrow} x_j \in \ell^{\infty} \ \ (2)$, an it converges uniformly ${(3)}$. Then $$ \lim_{j\to \infty} x_j \overset{(2)}{=} \lim_{j\to \infty} \left( \lim_{n \to \infty} x^{(n)}_j\right) \overset{(3)}{=}\lim_{n\to \infty}\left(\lim_{j \to \infty} x^{(n)}_j\right) \overset{(1)}{=} \lim_{n \to \infty} 0 = 0 $$ that is , $x=\left( x_j\right)_{j=1}^{\infty} \in c_0$, so we conclude that $c_0$ is a closed subspace of $\ell^{\infty}$.

Alonso Delfín
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HINT: Use a "cantor diagonal argument" to show that $(a_i)_i$ admits subsequence that converges to zero, then apply this result to every subsequence of $(a_i)_i$.

Daniel
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