1

I need to show that the set $A=\left\{ x \in l_{2} : |x_{n}| \leq \frac{1}{n}, n=1, 2, ...\right\}$ is a closed subset of $l_{2}$

I'm assuming the best way to show this is to have a sequence in A that converges to x, and show that this implies x is in A. But the manipulation is evading me.

Tryss
  • 14,490
  • 20
  • 35
KyleW
  • 592
  • If a sequence converges in $l^2,$ then it converges in each component. – zhw. Jul 04 '15 at 20:00
  • 1
    Here's a related question with many others linked: http://math.stackexchange.com/q/175251/ (it gives a more general result, not a duplicate) – Jonas Meyer Jul 04 '15 at 20:05

4 Answers4

5

Consider continuous functionals $$ f_n:\ell_2\to\mathbb{R}:x\mapsto x_n $$ for each $n\in\mathbb{N}$. Note that $A_n:=(f_n)^{-1}([-1/n,1/n])$ is closed as preimage of closed set under continuous map $f_n$. Since $A=\bigcap_{n=1}^\infty A_n$, then $A$ is closed as intersection of closed sets.

Norbert
  • 58,398
2

Let $\{ x^{(m)} = (x_1^{(m)}, x_2^{(m)},\ldots ) \}_{m=1}^{\infty}$ be a sequence in $A$ and assume that $x^{(m)} \to x$ in $\ell^2$ as $m \to \infty$. To show that $A$ is closed, we need to show that $x \in A$. Write $x = (x_1,x_2,\ldots)$, then $x^{(m)} \to X$ in $\ell^2$ means that $$ \| x^{(m)}-x \|_{\ell^2}^2 = \sum_{n=1}^{\infty} |x^{(m)}_n - x_n |^2 \to 0 \textrm{ as $m \to \infty$.} $$ In particular, for each fixed $n$, $|x_n^{(m)} - x_n | \to 0$ as $m \to \infty$. Therefore, $$ |x_n| \leq |x_n^{(m)} - x_n | + |x_n^{(m)}| \leq |x_n^{(m)} - x_n| + \frac{1}{n}, $$ and taking the limit as $m \to \infty$ gives that $|x_n| \leq \frac{1}{n}$. Since this holds for all $n$, we may conclude that $x \in A$, and hence that $A$ is closed.

msteve
  • 6,238
  • This is basically what I had, but in taking the limit we have, for a fixed $\epsilon$, $\exists m$ such that $|x^{(m)}{n} - x{n}| < \epsilon$. Which implies that for all $\epsilon$, $|x_{n}| < \epsilon + \frac{1}{n}$. How does this imply that $|x_{n}| \leq \frac{1}{n}? – KyleW Jul 04 '15 at 20:16
  • Indeed, so what you would effectively have shown is that $|x_n| \leq \epsilon + \frac{1}{n}$ for all $\epsilon > 0$, hence $|x_n| \leq \frac{1}{n}$. – msteve Jul 04 '15 at 20:18
  • I'm not following that last implication – KyleW Jul 04 '15 at 20:20
  • If $|x_n| > \frac{1}{n}$, then there has to be some $\epsilon > 0$ (take, e.g., $\epsilon = \frac{|x_n| - \frac{1}{n}}{2}$) so that $|x_n | > \epsilon + \frac{1}{n} $... – msteve Jul 04 '15 at 20:27
2

Obviously $$ A=\left[-1,1\right]\times\left[-\frac12,\frac12\right]\times\ldots\times\left[-\frac1n,\frac1n\right]\times\ldots=\prod_{n=1}^\infty\left[-\frac1n,\frac1n\right]. $$ Given $a\in \bar{A}$, there is a sequence $(a^k)$ of elements of $A$ whose limit is $a$, i.e. $\lim_{k\to\infty}\|a-a^k\|_2=0$. Therefore, for every $k,n\ge 1$ we have: $$ |a_n|\le |a_n^k|+|a_n-a_n^k|\le \frac1n+\|a-a^k\|_2. $$ It follows that $$ |a_n|\le \frac1n+\lim_{k\to\infty}\|a-a^k\|_2=\frac1n \quad \forall n\ge 1, $$ i.e. $a=(a_n)\in A$. Hence $\bar{A}\subset A\subset \bar{A}$, i.e. $A=\bar{A}$ and $A$ is closed.

HorizonsMaths
  • 16,846
  • 1
  • 29
  • 42
0

Hint: The complement of $A$ is $$ A^c=\left\{y\in l_2: |y_i|>\frac{1}{i}\text{ for some }i\right\}. $$ If you are having trouble with your method, you could show that $A^c$ is open.

TomGrubb
  • 13,177