Finding the closure of some metric spaces

[Mr Davis 97]

Homework Statement

Identify $\bar{c}$, $\bar{c_0}$ and $\bar{c_{00}}$ in the metric spaces $(\ell^\infty,d_\infty)$.

Homework Equations

The $\ell^\infty$ sequence space is
$$
\ell^\infty:=\left\{\{x_n\}_{n=1}^\infty\in\mathbb{R}^\mathbb{N}\,:\, \sup(\{|x_n|:n\in\mathbb{N}\})<\infty\right\},
$$
with metric $d_\infty:\ell^\infty\times\ell^\infty \to [0,\infty)$ given by
$$
d_\infty(x,y)=\sup(\{|x_n-y_n|:n\in\mathbb{N}\}),
\qquad\text{ for each $x=\{x_n\}_{n=1}^\infty\in\ell^\infty$ and $y=\{y_n\}_{n=1}^\infty\in\ell^\infty$.}
$$
Set
\begin{align*}
c
&:=
\left\{x=\{x_n\}_{n=1}^\infty\in\ell^\infty:x=\{x_n\}_{n=1}^\infty\text{ converges in $\mathbb{R}$}\right\},\\
c_0
&:=
\left\{x=\{x_n\}_{n=1}^\infty\in c:\,\lim_{n\to\infty}x_n=0\right\},\\
c_{00}
&:=
\left\{x=\{x_n\}_{n=1}^\infty\in c_0\,:\,\text{ there is an $N\in\mathbb{N}$ such that $x_n=0$ for all $n\geq N$}\right\},\\
\ell^\infty_a
&:=
\{x=\{x_n\}_{n=1}^\infty\in\ell^\infty
:|x_n|\le|a_n|,\text{ for each }n\in\mathbb{N}\},
\quad\text{ given }a=\{a_n\}_{n=1}^\infty\in\ell^\infty.
\end{align*}

The Attempt at a Solution

It's simpler to find the closure of subsets of the real numbers: you just try to find all of elements of a set to which a sequence has a limit. I am having more trouble doing this for spaces which are not subsets of the real numbers. How should I go about doing this? Would the fact that $\{\text{closure points}\} = \{\text{accumulation points}\} \cup \{\text{isolation points}\}$ help? Note that I don't need rigorous proofs; just some intuition that would allow me to suppose what the closures are.

 

[andrewkirk]

There is a total order via the order relation of containment, over the four spaces $c,c_0,c_{00},\ell^\infty$. What is that order?

Having established the order, the natural guess for a problem like this is that the closure of each space is either the one immediately above it in the order, or itself.

Play around with a few examples and then make guesses as to which would apply to each of $c,c_0,c_{00}$.

For the ones that are already closed, try to prove that for any Cauchy sequence of sequences in the space, the limit must be in the space itself.

For the ones that are not closed, start by finding an example of a convergent sequence of sequences in the space, whose limit is not in the space but is in the next one up.

 

[Mr Davis 97]

There is a total order via the order relation of containment, over the four spaces $c,c_0,c_{00},\ell^\infty$. What is that order?

Having established the order, the natural guess for a problem like this is that the closure of each space is either the one immediately above it in the order, or itself.

Play around with a few examples and then make guesses as to which would apply to each of $c,c_0,c_{00}$.

For the ones that are already closed, try to prove that for any Cauchy sequence of sequences in the space, the limit must be in the space itself.

For the ones that are not closed, start by finding an example of a convergent sequence of sequences in the space, whose limit is not in the space but is in the next one up.

What is the best way to construct examples? I am having a hard time wrapping my head around the idea of "sequence of sequences."

 

[fresh_42]

What is the best way to construct examples? I am having a hard time wrapping my head around the idea of "sequence of sequences."

You don't need this imagination. Simply think of a usual sequence $(x_n)_{n\in \mathbb{N}}$ of certain elements which have certain properties. That $x_n \in \ell^\infty$ is only needed to have these certain properties. To think about them as $((x_{n,m})_{m\in \mathbb{N}})_{n \in \mathbb{N}}$ is indeed confusing. Therefore it's better to treat them as sequences in some abstract space. You can probably imagine a sequence of vectors $(v_n)_{n\in \mathbb{N}}$ without thinking about their components. This is the same, only infinitely many components. They are even a vector space.

 

[Mr Davis 97]

You don't need this imagination. Simply think of a usual sequence $(x_n)_{n\in \mathbb{N}}$ of certain elements which have certain properties. That $x_n \in \ell^\infty$ is only needed to have these certain properties. To think about them as $((x_{n,m})_{m\in \mathbb{N}})_{n \in \mathbb{N}}$ is indeed confusing. Therefore it's better to treat them as sequences in some abstract space. You can probably imagine a sequence of vectors $(v_n)_{n\in \mathbb{N}}$ without thinking about their components. This is the same, only infinitely many components. They are even a vector space.

So I think that $c$ and $c_0$ are closed, but that $c_{00}$ is not. If this is the case then $\bar{c}=c$ and $\bar{c_0}=c_0$. So then how do I find $\bar{c_{00}}$? Will it be the case that $\bar{c_{00}} = c_{0}$?

 

[fresh_42]

So I think that $c$ and $c_0$ are closed, but that $c_{00}$ is not. If this is the case then $\bar{c}=c$ and $\bar{c_0}=c_0$. So then how do I find $\bar{c_{00}}$? Will it be the case that $\bar{c_{00}} = c_{0}$?

"So I think" is not an argument. But I don't know the answers by mind. One way to investigate a closure is to consider whether the set is dense in the assumed closure.

 

[andrewkirk]

What is the best way to construct examples? I am having a hard time wrapping my head around the idea of "sequence of sequences."

The first question that occurs to me is whether there can be a sequence of sequences in $c_{00}$ whose limit is not in $c_{00}$.

Try to think of a sequence $L_1, L_2, ...$ of sequences in $c_{00}$ such that $L_{ij}$ is the $j$th element of sequence $L_i$.
For its limit to not be in $c_{00}$ it must have nonzero elements infinitely far into the tail. So we could look at sequences $L_i$ for which the nonzero part of the sequence gets longer with $i$. Consider the case where for each $L_i$ only the first $i$ elements are nonzero.

Now since all sequences in $c_{00}$ converge to zero, we would expect a limit sequence to converge to zero as well. So pick any sequence in $L\in c_0\smallsetminus c_{00}$, that converges to zero. Can you think a way that uses the idea of the previous paragraph to construct a sequence of sequences in $c_{00}$ that converges to $L$?

 

[WWGD]

It may be helpful too, just to include in your toolbox, to consider what $\epsilon$ -balls look like in $l^{\infty}$, tho, of course, when possible use the approach that is most familiar for you.:

Sup $| x_n -y_n| < r$. This way you can consider the complement to see if it is closed , e.g., the difference for each term