Is $A = \{ \{x_n\}_{n=1}^{\infty} = x \in l \ ^ 2: \sum_{n=1}^{\infty} |x_n| < \infty \} $ open? closed?

Question

Let $l \ ^ 2 = \{ \{x_n\}_{n=1}^{\infty} = x \in R \ ^ N : \sum_{n=1}^{\infty} |x_n| \ ^ 2 < \infty \}$

Define $||x||_2 = ( \sum_{n=1}^{\infty} |x_n| \ ^2 ) \ ^ {1/2} $

I have proved that $(l \ ^ 2 , \|\ \|_2) $ perform a norm space.

The question is:

Let $$A = \bigg\{ \{x_n\}_{n=1}^{\infty} = x \in l \ ^ 2: \sum_{n=1}^{\infty} |x_n| < \infty \bigg\} $$

Is $A$ open ? closed? both? none?

My attempt to show A is not closed is to take

$a_n = \bigg\{\dfrac{1}{k \ ^{1 + 1/n}} \bigg\}_{k=1}^{\infty}$ and to show that $a_n \to 1/k $ which does not belong to A.

But, I need to show that $d(a_n , \{1/k\}_{k=1}^{\infty}) \to 0 $

and im not sure how to do it.

$d(a_n,a) = ||a_n-a||_2 = \bigg\|\{ \dfrac{1}{k} (\dfrac{1}{k ^{1/n} }-1 ) \}_{k=1}^{\infty}\bigg\|_2$

If im right so far, and A is indeed not closed, im not sure how to continue.

Thanks for helping

Answer 1

First you wrote: $$d(a_n,a) = ||a_n-a||_2 = \bigg\|\dfrac{1}{k} (\dfrac{1}{k ^{1/n} -1} )\bigg\|_2$$

But the last equation is not true: $||a_n-a||_2$ is the difference of two sequences but $\dfrac{1}{k} (\dfrac{1}{k ^{1/n} -1} )$ is (btw your conversion here is also wrong) is the difference of two elements of the sequences so something completely different.

But ok, we have to show: $||a_n-a||_2 \to 0$, let's start with a simple calculation:

$$\begin{align*}||a_n-a||_2 &= \sum_{k=1}^{\infty} |a_n^k - a^k|^2 \\ &= \sum_{k=1}^{\infty} \left|\dfrac{1}{k \ ^{1 + 1/n}} - \frac{1}{k}\right|^2 \\ &= \sum_{k=1}^{\infty} \left|\frac{1}{k}\left(\dfrac{1}{k^{1/n}} - 1\right)\right|^2 \\&= \sum_{k=1}^{\infty} \left|\frac{1}{k}\right|^2 \left|\left(\dfrac{1}{k^{1/n}} - 1\right)\right|^2 \end{align*}$$

But we have $$\left|\left(\dfrac{1}{k^{1/n}} - 1\right)\right|^2 \le 1$$ so $$||a_n-a||_2 \le \sum_{k=1}^{\infty} \frac{1}{k^2} < \infty$$

So we can use dominated convergence theorem and interchange limit and summation to get:

$$\begin{align*}\lim_{n\to\infty} ||a_n-a||_2 &= \lim_{n\to\infty} ||a_n-a||_2 \\ &= \lim_{n\to\infty} \sum_{k=1}^{\infty} \left|\dfrac{1}{k \ ^{1 + 1/n}} - \frac{1}{k}\right|^2 \\ &= \sum_{k=1}^{\infty} \lim_{n\to\infty} \left|\dfrac{1}{k \ ^{1 + 1/n}} - \frac{1}{k}\right|^2 \\ &= \sum_{k=1}^{\infty} 0 = 0 \end{align*}$$