# Closure of subset of $\ell^\infty$

I am doing some assignment and I am trying to understand the nature of the closure of $A\subset \ell^\infty$ as I am having trouble exactly getting an image of how the elements in it is.

Our set A is defiend as $$A=\{(x_1,x_2-x_1,x_3-x_2,\ldots):x=(x_1,x_2,x_3,\ldots)\in\ell^\infty\}$$ I know it's a normed space and that the elements in the closure of $A$ are those elements we can get arbitrarily close to within $A$, I know also that $||x||=\sup |x_i|$. As such I thought that means that the difference $x-y$ must have all it's elements approching $0$, as for a sequence $\{x_i\}$ to converge to $y$ the $\sup |x_i-y_i|\to 0$, which gives that each element element one by one converges to that of $y$ and as such I'd get $x_1\to y_1$, $x_2-x_1\to y_2$, but from the first one we can also see that $x_2\to y_1+y_2$ and so on such we get $x_n\to\sum_{i=0}^n y_i$. I do know however that this implies of course that our $(x_1,x_2,\ldots)\in\ell^\infty$ which means for some $m$ we have $|x_m|$ being the greatest and such that the sum is also bounded by what they can take.

This is about as far as I get, is there anything I have done wrong or is there anything else I can use to get a better grasp over these?

## 1 Answer

Consider $T(x_1,x_2,\dots) = (x_1, x_2-x_1,x_3-x_2,\dots)$. Then $A = T(l^{\infty})$. Note that $T$ may be inverted by the following operation $(y_1, y_2, \dots) \to (y_1, y_1 + y_2, y_1+y_2 +y_3,\dots)$.

From your argument, if $y \in A$, you must have $|\sum_{i=0}^n y_i| \leq K$ for all $n$ for some fixed $K$. Let $B = \{y \in l^{\infty} : |\sum_{i=0}^n y_i| \leq K \text{ for all }n\text{ for some } K\}$. $B$ is a closed set and the above shows $A \subseteq B$. But $A$ is equal to this set because if $y \in B$, we can find some $x \in l^{\infty}$ with $T(x) = y$ simply by using the above inverting operation. So $A = B$.