Countable sequence groups and tail filtrations

The abelian group identities for coordinate addition on $P$ hold at each index by the identities in $k$. The zero sequence has empty support; negatives preserve support and the support of a sum is contained in the union of the two supports, so $S$ is a subgroup. For maps $f_j:G\to k$, the unique map $G\to P$ is $g\mapsto (f_j(g){)}_j$. For maps $g_j:k\to G$, define $S\to G$ by $x\mapsto {\sum }_{j\in \mathrm{supp}(x)}{g}_j(x_j)$. This sum is finite; extending the summation set by zero terms proves additivity, and the identity $x={\sum }_j{\iota }_j(x_j)$ proves uniqueness. These are precisely the product and coproduct properties.

The inclusion is injective. The constant-one sequence belongs to $P$ but has infinite support, so does not belong to $S$. Each unit sequence belongs to $S$, and different indices give different unit sequences.

Encode $x\in S$ by $b(x)={\sum }_j{2}^j{x}_j\in \mathbb{N}$. If $x\mathrm{\ne }y$, their finite union of supports has a largest differing index $r$. The magnitude of the contribution there is $2^r$, while the sum of the magnitudes at lower indices is at most ${\sum }_{j<r}{2}^j=2^r-1$; the latter identity follows by starting with $0=1-1$ and adding $2^r$ at the next index. Hence $b(x)\mathrm{\ne }b(y)$. Conversely, successive unique divisions of any nonnegative integer by $2$ give its binary digits; the nonzero quotients strictly decrease, so after finitely many divisions the quotient is zero. Substituting the equations $a_j=2a_{j+1}+r_j$ back gives $a_0={\sum }_j{2}^j{r}_j$. Thus $b$ is a bijection $S\to \mathbb{N}$.

The first-$m$-coordinates map $A\to k^m$ is onto by extension by zero, for either $A=S$ or $P$, and has kernel $T_{m}A$. It therefore induces a bijective homomorphism $A\mathrm{/}{T}_{m}A\to k^m$: equality of images means the difference lies in $T_{m}A$, and every tuple is represented by its zero extension. For $m=0$ the quotient and empty tuple group are zero. If $x\in {\bigcap }_m{T}_{m}A$, take $m=j+1$ to conclude $x_j=0$ at each index, so $x=0$.

For any map $e:\mathbb{N}\to P$, the sequence $y_j=1-e(j{)}_j$ lies in $P$ and differs from $e(j)$ at coordinate $j$. Thus $e$ is not onto. If an injection $P\to \mathbb{N}$ existed, inversion on its image and the zero sequence as value off that image would define a surjection $\mathbb{N}\to P$, which has just been excluded. Hence $P$ is uncountable and cannot be bijective with $S$.

Let $L$ be the subgroup of ${\prod }_{m\ge 0}{k}^m$ consisting of tuples $(z^{(m)}{)}_m$ for which truncating $z^{(m+1)}$ gives $z^{(m)}$. For any compatible cone of homomorphisms into $k^m$, the map sending an element to its tuple of cone images is the unique homomorphism into $L$ inducing that cone. Thus $L$ is the inverse limit. The homomorphism $P\to L$ sends a sequence to its initial segments. Its inverse sends a compatible tuple to $x_j=z_j^{(j+1)}$; compatibility proves that all its first-$m$ coordinates equal $z^{(m)}$. These formulas are mutually inverse and select no representatives.

The quotient identifications in step 2.2 commute with truncation, so they identify both $\widehat{S}$ and $\widehat{P}$ with $L\cong P$. The completion map sends $x$ to its initial segments, hence becomes the original inclusion for $S$ and the identity for $P$. Step 2.2 proves separatedness for both, and step 1.2 proves the first inclusion is proper. All constructions use explicit coordinates, finite sums, or uniquely specified digits; AC has not been used.