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It seems there is a distinction between the eventually-zero sequences $$c_{00} := \{(x_n)_{n\in \mathbb{N}} \mid x_n \in \mathbb{R};\ x_m = 0 \text{ for all $m$ greater than some $k$}\} $$ and the finite sequences $$\mathbb{R}^\infty := \{(x_n)_{n\in \mathbb{N}} \mid x_n \in \mathbb{R};\ x_m \neq 0 \text{ for finitely many } m\}.$$ For the sake of completeness, the square-summable sequences are: $$\ell^2 := \{(x_n)_{n\in \mathbb{N}} \mid x_n \in \mathbb{R};\, \sum_n |x_n|^2 < \infty\}.$$

Originally, I don't see why there is any distinction, based on the following argument: Any finite sequence will necessarily be zero after its final non-zero element, hence it is also eventually-zero. Conversely, any eventually-zero sequence will have a final non-zero element $x_k$, hence the number of non-zero elements is at most $k$, which has to be finite. Every finite sequence is obviously square-summable, so this would then imply that $$c_{00} = \mathbb{R}^\infty \subsetneq \ell^2.$$

However, it then occurred to me that the assumption of a finite $k$ may not be true, based on the following construction: Given any integer $N$, define a tuple $x^{(N)} := (x_n)_{0 \leq n \leq 2N}$ with the following entries: $$x_n = \begin{cases}1 & \text{if } n \leq N\\ 0 & \text{if } n > N.\end{cases}$$ Then $\lim_{N\to\infty} x^{(N)} \in c_{00}$, but not in $\mathbb{R}^\infty$. This sequence is also not square-summable, which would suggest $$\mathbb{R}^\infty \subsetneq \ell^2, \quad \mathbb{R}^\infty \subsetneq c_{00}, \quad c_{00} \not\subseteq \ell^2 \text{ and } c_{00} \not\supseteq \ell^2.$$ So which is correct, or neither?

MrArsGravis
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  • Your claim that $\lim_{n \to \infty} x^{(N)} \in c_{00}$ is false. Why? It is the constant sequence: $x_n = 1$ for all $n \geq 0$. Why is that? For any index $n$, once $N \geq n$ we have $x^{(N)}n = 1$, and hence in the limit as well (in any reasonable topology on $c{00}$). As the answer below suggests, $c_{00}$ and $\mathbb{R}^\infty$ are two different descriptions of the same space of sequences, which is a proper subspace of $\ell^2$. – Sammy Black Sep 07 '22 at 19:01
  • Every term in your sequence is square-summable and resides in $c_{00} = \mathbb{R}^{\infty}$. Whether the limit exists or not depends on what topology you're taking the limit with respect to. The pointwise limit is not in any of those three spaces, and most other limits (e.g. the $\ell^2$ limit) won't exist. – Qiaochu Yuan Sep 07 '22 at 19:01

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The way you've written the definitions of $c_{00}$ and $\mathbb{R}^{\infty}$, they are exactly the same set.

It's not super common - and so in most contexts should probably be explicitly stated - but in my experience usually the notation $\mathbb{R}^{\infty}$ is a specific topological space, i.e. this set is being equipped with a certain topology that comes from viewing it as the limit of the finite dimensional Eucldiean spaces (See Unit sphere in $\mathbb{R}^\infty$ is contractible? and https://mathoverflow.net/questions/119362/unit-sphere-in-r-infty-is-contractible )

Whereas $c_{00}$ might just be notation for a set and if you're doing analysis/topology then talking about the subspace $c_{00} \subset l^2$ is different from talking say about the subspace $c_{00} \subset l^{\infty}$.

SBK
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  • Yes, I just realized that in the construction that I wrote, the limit does not exist, in the sense that there is no well-defined mapping $\mathbb{N} \to \mathbb{R}$ if the limit $N \to \infty$ is taken. – MrArsGravis Sep 07 '22 at 18:57