# Finding the closure of some metric spaces

• Mr Davis 97
In summary, the homework statement is to identify ##\bar{c}##, ##\bar{c_0}## and ##\bar{c_{00}}## in the metric spaces ##(\ell^\infty,d_\infty)##. The homework equations state that the ##\ell^\infty## sequence space is$$\ell^\infty:=\left\{\{x_n\}_{n=1}^\infty\in\mathbb{R}^\mathbb{N}\,:\, \sup(\{|x_n|:n\in\mathbb{N}\})<\infty Mr Davis 97 ## Homework Statement Identify ##\bar{c}##, ##\bar{c_0}## and ##\bar{c_{00}}## in the metric spaces ##(\ell^\infty,d_\infty)##. ## Homework Equations The ##\ell^\infty## sequence space is$$
\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}\},
\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.

Last edited:
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
andrewkirk said:
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."

Mr Davis 97 said:
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.

fresh_42 said:
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}##?

Mr Davis 97 said:
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.

Mr Davis 97
Mr Davis 97 said:
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##?

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

## 1. What is the closure of a metric space?

The closure of a metric space is the smallest closed set that contains all the points in the metric space. It includes all the limit points of the metric space and is often denoted by the symbol "cl".

## 2. How is the closure of a metric space calculated?

To find the closure of a metric space, you can use the definition of closure, which states that it is the intersection of all closed sets that contain the metric space. Alternatively, you can also use the closure operator, which is a function that maps a set to its closure.

## 3. Why is finding the closure of a metric space important?

Finding the closure of a metric space is important because it helps in understanding the topological properties of the metric space. It also plays a crucial role in various mathematical proofs and has applications in fields such as analysis, topology, and geometry.

## 4. Can the closure of a metric space be equal to the metric space itself?

Yes, the closure of a metric space can be equal to the metric space itself, but this is not always the case. If the metric space is already closed, then its closure will be equal to the metric space. However, if the metric space is not closed, its closure will be a larger set that includes all the limit points.

## 5. How does the closure of a metric space relate to continuity?

The closure of a metric space is closely related to continuity. A function is continuous at a point if and only if the closure of the preimage of that point is equal to the preimage of the closure of that point. In other words, a function is continuous if and only if the closure of its domain is equal to its range.

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