Axiomatizing convergent sequences and limits

Question

The usual exposition of limits of sequences starts with a definition and then derives properties like linearity. I'm curious if, conversely, a reasonable set of these properties characterize the subspace $C_{\text{std}} \subset \mathbf{R}^\mathbf{N}$ of convergent sequences and the map $L_{\text{std}}: C_{\text{std}} \to \mathbf{R}, (a_n) \mapsto \lim_{n \to \infty} a_n$. Here is an example list of properties (when I say equivalently, I mean equivalent assuming the previous conditions):

1. $C$ is a vector subspace and $L: C \to \mathbf{R}$ is $\mathbf{R}$-linear.
2. $C$ contains all monotone bounded sequences.
3. If $a$ is eventually the constant $a_\infty$, ie $a_n = a_\infty$ for large $n$, then $La = a_\infty$ (equivalently, $L$ vanishes on the subspace of eventually $0$ sequences and takes the constant $1$ sequence to $1$).
4. If $a \in (\mathbf{R}_{\ge 0})^\mathbf{N} \cap C$ then $La \ge 0$ (equivalently $L$ is monotonic).
5. If $a, b \in C$ then $ab := (a_n b_n) \in C$ and $L(ab) = (L a)(L b)$.
6. For $a \in C$ every subsequence $a'$ of $a$ is also in $C$ and $La = La'$.

But all of these are also satisfied by $$C = \operatorname{span}\{a \text{ monotone bounded}\} = \{ a - b \mid a, b \text{ increasing and bounded} \}$$ (see https://math.stackexchange.com/a/4377050/32766 for an alternate characterization) and $L = L_{\text{std}}|C$. So, is there a similar list of "well-known" properties of $(C_{\text{std}}, L_{\text{std}})$ that implies $C \supseteq C_{\text{std}}$ and $L$ is an extension of $L_{\text{std}}$? The implication should of course be "non-trivial", ie the list should not just include a definition of $\lim$. Given such a list, we could define $(L_{\text{std}}, C_{\text{std}})$ as the minimal pair satisfying that list.

Note that (1)-(4) are enough to show that $L$ restricted to $C \cap C_{\text{std}}$ agrees with $L_{\text{std}}$: if $a_\infty = \lim a_n$ then $a_n$ is eventually bounded between $a_\infty - \epsilon$ and $a_\infty + \epsilon$ and hence so is $La$. We don't even need all of (2), just that eventually constant sequences are in $C$.

(1), (3), (4) and (6) also imply that $C \subseteq C_{\text{std}}$. First note that if $a \in C$ then it has to be bounded, otherwise for each $N > 0$ either $a$ or $-a$ has a subsequence above $N$, so $|La| \ge N$ by monotonicity. Now if $a \in C$ is not convergent then it must have convergent subsequences with distinct limits, which forces a contradiction between (6) and that $L$ agrees with $L_{\text{std}}$ on $C \cap C_{\text{std}}$, as shown above. However I'd be generally interested if an answer provides a list that allows for $C \supsetneq C_{\text{std}}$.

Remarks:

• (1), (3) and (5) can be combined to: $C$ is a unital subalgebra of $\mathbf{R}^\mathbf{N}$ containing the ideal $C_0$ of eventually $0$ sequences and $L: C \to C/C_0 \to \mathbf{R}$ is a unital algebra homomorphism.
• It is plausible to me that (4) is implied by the other conditions. It could also be replaced by the (not obviously equivalent) condition $a \in C\cap (\mathbf{R}_{\ge 0})^{\mathbf{N}} \implies (\sqrt{a_n}) \in C$.

Answer 1

[Upgarding my comment to an answer.]

[Edit: Elaborating a little more on the final axiom list.]

We can start with your axioms (1)-(4) and add a squeezing axiom (S) that says if $a=\{a_n\},b=\{b_n\}\in C$, and $L(a)=L(b)$, then for all $c=\{c_n\}$ such that $a_n\leq c_n\leq b_n$ for all $n$, we have $c\in C$ with $L(c)=L(a)=L(b)$.

Since $C$ is already guaranteed to contain bounded monotone sequences, this guarantees we get all convergent sequences, since an arbitrary convergent $\{a_n\}$ is sandwiched between $\{\sup_{k\geq n}a_k\}$ and $\{\inf_{k\geq n}a_k\}$.

Minimality

You remarked in the original post that axioms (1)-(4) guarantee $L$ agrees with $L_{\text{std}}$ on the intersection $C\cap C_{\text{std}}$. We conclude therefore that any $(C,L)$ satisfying the axioms (1)-(4) and (S) must extend $(C_{\text{std}}, L_{\text{std}})$, and so $(C_{\text{std}}, L_{\text{std}})$ is the minimal pair satisfying (1)-(4) and (S).

Nontrivial Extensions

Cesàro Convergence also satisfies the axioms (1)-(4) and (S), so the axioms do allow for nontrivial extension. Note that Cesàro convergence fails your axioms (5) (via $a= \{(-1)^n\}, b=\{(-1)^{n+1}\}$) and (6) (via $a=\{(-1)^n\}, a'=\{1\}$).

Another nontrivial extension arises with the set of sequences which converge on some subset (depending on the sequence) of asymptotic density $1$, with $L$ defined as the limit of the sequence restricted to any such subset on which it converges. This satisfies (5) as well. We could also replace asymptotic density $1$ subsets with any other nontrivial family of infinite subsets of $\mathbb N$ closed under pairwise intersection.

Remark 1

Assuming we have linearity (i.e., axiom (1)), the squeezing axiom (S) is equivalent to the axiom that if $a=\{a_n\}\in C$ is a non-negative sequence with $L(a)=0$, and $0\leq b_n \leq a_n$ for all $n$, then $b=\{b_n\}\in C$ with $L(b)=0$.

Remark 2

As you suggested in the comments, axiom (1) can be relaxed to say that $L$ is $\mathbb Z$-linear. If this is done, we still have $(C_{\text{std}},L_{\text{std}})$ as the minimal pair satisfying the axioms (1)-(4) and (S). In this case, I do not know if $\mathbb R$-linearity is implied by the axioms.

We could further relax axiom (1) to say that $C$ is an additive subgroup of $\mathbb R^{\mathbb N}$, and $L$ is $\mathbb Z$-linear. With this further weakening, we yet again have $(C_{\text{std}},L_{\text{std}})$ minimal among pairs satisfying the axioms, but now, axioms (1)-(4) and (S) are insufficient to recover the vector space assumption.

In fact, let $c=\{(-1)^n\}$, let $C=C_{\text{std}} + \mathbb Z c$, define $L(a+\lambda c)=\lim_{n\to \infty}a_n$ for $a\in C_{\text{std}}$ and $\lambda\in \mathbb Z$, and observe that $(C,L)$ satisfies the axioms, yet is not a vector subspace.

(1)-(4) are easily satisfied, and to prove (S), we argue as follows, using the equivalent formulation of (S) from the first remark: Suppose $0\leq b \leq a+\lambda c $, for some $\lambda\in \mathbb Z$ and $a\in C_{\text{std}}$, with $L(a+\lambda c)=0$, i.e., for all $n$ we have $$0\leq b_n \leq a_n +\lambda(-1)^n\text{,}$$ and $a_n\to 0$.

Then looking only at the even terms and passing to the limit, we see that $0\leq \lambda$, and likewise the odd terms show us that $0\leq -\lambda$, whereby $\lambda=0$ and $b\in C_{\text{std}}$, and $L(b)=0$.

Answer 2

Proposal 1, which reuses some of your ideas:

• (1) $C$ contains all constant sequences. If $a$ is the constant sequence with value $a_\infty$, $La = a_\infty$.
• (2) If $a \in C$, every subsequence $a'$ of $a$ is also in $C$, and $La'=La$.
• (3) A partial order on sequences is defined by $a \le b$ iff $\forall n, a_n \le b_n$. Then $\forall a, b \in C, a \le b \Rightarrow La \le Lb$.
• (4) Let $M_<(a) = \{x \in \mathbb R \mid \exists b \in C, b \le a, Lb=x\}$ and $M_>(a) = \{x \in \mathbb R \mid \exists b \in C, b \ge a, Lb=x\}$.
  We can prove (below) that all elements of $M_<(a)$ are $\le$ to all elements of $M_>(a)$.
  Then if $M_<(a)$ and $M_>(a)$ are both non-empty, and $M_<(a) \cup M_>(a) = \mathbb R$, then $a \in C$ and $La$ is defined to be the element that separates $M_<(a)$ from $M_>(a)$ (same as Dedekind cuts, this is were $\mathbb R$ being complete is used).

$C$ is defined to be the smallest set (for inclusion) that verifies these 4 properties.

Proof that all elements of $M_<(a)$ are $\le$ to all elements of $M_>(a)$:
$\forall x \in M_<(a), \exists b \in C, b \le a, Lb=x$
$\forall x' \in M_>(a), \exists b' \in C, b' \ge a, Lb'=x'$
By transitivity, $b \le b'$, so $Lb = x \le Lb' = x'$

These 4 properties define convergent subsequences and their limits, because:

• Eventually constant sequences are in $C$ due to (1) and (2).
• For any convergent sequence $a$ we can be bound it below by an eventually constant sequence whose limit is as close to the limit of $a$ as we want; and similarly, we can bound above as close as we want; so $M_<(a) \cup M_>(a) = \mathbb R$.
• For any unbounded sequence $a$, at least one of $M_<(a)$ or $M_>(a)$ will be empty, so $a \notin C$.
• For any bounded, non-convergent sequence $a$, there exists at least 2 different accumulation points $x \lt x'$ (there is at least one because the sequence is bounded, and if there is only one, the sequence is convergent). So open interval $(x, x')$ is not in $M_<(a) \cup M_>(a)$, so $a \notin C$.

Proposal 2
Define $T(a)$ as the set of open intervals that contain $a$.
Define $T^∗(a)$ as the union of $T(b)$ for all subsequences $b$ of $a$.
Then by definition $a$ is convergent iff $T^∗(a)$ is stable by finite intersection.
Note that we get also sequences that "converge" to $+\infty$ or $-\infty$; to remove them, use only bounded sequences.

That defines $C$. Then for $a \in C$, define $L(a)$ as the (infinite) intersection of all elements of $T^*(a)$.