Understanding $c_0$ is closed subset in $\ell^{\infty}$

Question

Prove the set of sequences $c_0$ which converge to zero in $l_{\infty}$ is closed. I came across this proof and have a question: Since all $x_n(k)$ is in $c_0$, it's limit point $x(k)$ should be 0, now we only need to show that the sequence of all zeros is in $c_0$ to show that $c_0$ is closed. Is my understanding correct? Any help would be helpful.

Answer 1

To show a subset $Y$ of a normed linear space $X$ (or metric space, more generally) is closed, we need to show the following:

If $y_n$ is a sequence of points in $Y$, such that $y_n$ converges to $x$ in $X$, then $x \in Y$.

What's confusing here is that, in this case $X$ and $Y$ consist of sequences themselves. In particular, they consist of sequences of real numbers: functions from $\Bbb{N}$ to $\Bbb{R}$ (or $\Bbb{C}$, but I'm just going to consistently use $\Bbb{R}$). Each point is a real sequence, and we will be considering sequences of sequences.

In our case we have \begin{align*} X &= \ell^\infty = \{\text{Real sequences that are bounded}\} \\ Y &= c_0 = \{\text{Real sequences that converge to } 0\}. \end{align*}

So, in order to prove $c_0$ is closed in $\ell^\infty$, we assume that we have a sequence of points $(x_n)_{n=1}^\infty$ in $c_0$ that converge to $x$. To clarify, $x$ and $x_1$ and $x_2$ and $x_3$, etc, are each real sequences. We are assuming that $x_1$ and $x_2$ and $x_3$, etc, all converge to the real number $0$. We are not assuming this of $x$; that $x$ limits to $0 \in \Bbb{R}$ is precisely what we want to prove.

The notation used in the linked question is important to clarify too. The $k$th entry of a point (read: real sequence) $y$ (in $\ell^\infty$ and/or $c_0$) is denoted $y(k)$. As we are considering a sequence of points $(x_n)_{n=1}^\infty$, be prepared to see notation like $x_n(k)$, referring to the $k$th entry (a real number) of the $n$th point in the sequence $(x_n)$.

Now, let's quote the linked question:

Let $\varepsilon>0$. Then there exists an $N>0$ such that $$\parallel x_N-x\parallel_{\infty}:=\sup_{k\in\mathbb{N}}|x_N(k)-x(k)|\leq\varepsilon.$$

Then we have, \begin{align} |x(k)| &= |x(k) - x_N(k) + x_N(k)| \\\\ &\leq |x_N(k) - x(k)| + |x_N(k)| \\\\ &\leq \varepsilon + |x_N(k)|\rightarrow \varepsilon\;\; \text{as}\;\; k\rightarrow\infty. \end{align}

Therefore since $\varepsilon$ was chosen arbitrarily we can conclude that $x(k)\rightarrow0$ and thus that $x(k)\in c_0$.

In the above argument, the point was to show that $x$, the limit (in $\ell^\infty$) of the sequence of points $x_n$ in $c_0$, was also in $c_0$. In particular, the membership requirement of $c_0$ is that $x$, a sequence of real numbers, must converge to $0$. That is, $x(k) \to 0$ as $k \to \infty$. Hopefully you can see that the above argument proves this is the case.

To answer your question specifically, no, the limit $x$ does not need to be the $0$ sequence. It just has to be a sequence (of real numbers) that converges to the real number $0$.

To illustrate this, consider an example. Let $$x_n(k) = \frac{n}{nk+1}.$$ So, for example, the first point $x_1$ is $$x_1 = \left(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots \right).$$ Note this converges to $0$. The second point $x_2$ is $$x_2 = \left(\frac{2}{3}, \frac{2}{5}, \frac{2}{7}, \frac{2}{9}, \ldots \right),$$ which also converges to $0$. Similarly \begin{align*} x_3 &= \left(\frac{3}{4}, \frac{3}{7}, \frac{3}{10}, \frac{3}{13}, \ldots \right) \\ x_4 &= \left(\frac{4}{5}, \frac{4}{9}, \frac{4}{13}, \frac{4}{17}, \ldots \right), \end{align*} etc. Note that each of these points is a sequence of real numbers, approaching the real number $0$.

Now, let $x(k) = \frac{1}{k}$, that is $$x = \left(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots \right).$$ I claim that $x$ is the limit (in $\ell^\infty$) of $(x_n)_{n=1}^\infty$. That is, $$\|x_n - x\|_\infty \to 0$$ as $n \to \infty$. Note that, \begin{align*} \|x_n - x\|_\infty &= \sup_{k \in \Bbb{N}} |x_n(k) - x(k)| \\ &= \sup_{k \in \Bbb{N}} \left|\frac{n}{nk + 1} - \frac{1}{k}\right| \\ &= \sup_{k \in \Bbb{N}} \left|\frac{nk - (nk - 1)}{k(nk + 1)}\right| \\ &= \sup_{k \in \Bbb{N}} \frac{1}{k(nk + 1)}. \end{align*} The supremum is attained at $k = 1$, since this expression is decreasing with respect to $k$, hence $$\|x_n - x\|_\infty = \frac{1}{n+1} \to 0$$ as $n \to \infty$. Thus, $x_n \to x$ in $\ell^\infty$.

The quoted argument proves now that $x(k) = \frac{1}{k} \to 0$ as $k \to \infty$. Note that $x$ is not the constant $0$ sequence, just a real sequence that (by necessity) limits to the real number $0$.

Answer 2

As I am just a student, be carefull with my solution.

$C_0 = \{a_n \in l_{_\infty}( \mathbb{N}) : lim_{n \to \infty}a_n = 0 \}$

1- In order to prove that $C_0$ is closed we need to show that any "sequence" of $C_0$ that converges, converge to a limit in $C_0$.
The difficulty here is to understand what do we call a "sequence" in $C_0$. A sequence in $C_0$ is not a sequence of $ \mathbb{C}$ numbers but rather a sequence of sequence $ \mathbb{C}$ numbers. More simply you visualise (that how I do it at least) a sequence $(a_{m;n}) \in C_0$ as a sequence of column vectors with an infinite number of rows.
$ \{\underline{a}_{m;n} \} = \{ \underline{a}_{1;n}; \underline{a}_{2;n}; ...; \underline{a}_{m;n}; ... \} = \{ \begin{pmatrix} a_{1;1}\\ a_{1;2}\\ ...\\ ... \end{pmatrix} ; \begin{pmatrix} a_{2;1}\\ a_{2;2}\\ ...\\ ... \end{pmatrix} ;...; \begin{pmatrix} a_{m;1}\\ a_{m;2}\\ ...\\ ... \end{pmatrix} ; ... \} $
So basically $ lim_{m \to \infty} a_{mn} = l_n $ can be seen as $lim_{m \to \infty} \begin{pmatrix} a_{m;1}\\ a_{m;2}\\ ...\\ ... \end{pmatrix} = \begin{pmatrix} l_1\\ l_2\\ ...\\ ... \end{pmatrix} $
And in order for $C_0$ to be close we must have $ lim_{n \to \infty} l_n = 0 $ and this is what we want to prove.
Rem: Do not forget that for each of the sequence (in the "sequence of sequence" )$a_{mn}$ you have that $lim _{n \to \infty } a_{mn} = 0$

2- If $lim_{m \to \infty} a_{mn} = l_n$ it means that $\forall \epsilon > 0, \exists M > 0 : \forall m > M \Rightarrow || a_{mn} - l_n||_{\infty} = Sup_{n \in \mathbb{N}} \{ |a_{mn} - l_n| \} < \epsilon $.
If it is true $ \forall \epsilon > 0 $ it is in particular true for $\epsilon$ of the form $ \frac{\epsilon}{2}$ and we take the corresponding $M_{\epsilon/2}$.

3- We know that $\forall m , lim_{n \to \infty} a_{m;n} = 0 $ so by definition $\forall \epsilon > 0 , \exists N > 0 : \forall n >N \Rightarrow |a_{m;n}|< \epsilon $. If it is true $ \forall \epsilon > 0 $ it is in particular true for$\epsilon$ of the form $ \frac{\epsilon}{2}$ and we take the corresponding $N_{\epsilon/2}$

4- Combining "2-" and "3-" we have for $\forall \epsilon >0 , \exists N' = Max \{ M_{\epsilon/2}, N_{\epsilon/2} \} : \forall n,m > N' \Rightarrow |l_n| = |l_n -a_{mn}+a_{mn}| \leq |l_n -a_{mn}| + | a_{mn} | < |l_n -a_{mn}| + \epsilon/2 \leq Sup_{n>N'} \{ |l_n -a_{mn}| \} + \epsilon/2 < \epsilon/2 + \epsilon/2 = \epsilon $
(i) Because according to "3-" $N'\geq N_{\epsilon/2} \Rightarrow | a_{mn} | < \epsilon/2$
(ii) Because according to "2-" as $N' \geq M_{\epsilon/2} \Rightarrow Sup_{n>N'} \{ |l_n -a_{mn}| \} < \epsilon/2$
Rem: When we write something like $ |l_n - a_{mn}| $ it means any element of the sequence $(l_n)$ with a index greater than $N'$ minus any $n$-th element of a $m$-th sequence (with $m > N'$) of the sequence of sequence $(a_{mn})$. Or if you prefer in refer to "1-" any specific $n$-th element of a $m$-th column vector but $m > N'$.

Q.E.D.