I usually think of a locally finite abstract simplicial complex as having that any simplex has only finitely many higher dimensional simplices containing it. Namely if $\sigma$ is a $d$-simplex, then for any $d'\geq d$, the number of $d'$-simplices containing $\sigma$ is finite. I noticed that my definition does not explicitly prohibit infinite simplices, but I wanted to verify whether the following argument is correct in showing that there are no infinite simplices in such a complex:
Let $X$ be an abstract simplicial complex, and define for any $d$-simplex $\sigma\in X$ its $d'$-degree, for $d'> d$, by $$ \deg_{d'}(\sigma):=\vert \{ \tau\in X(d'): \; \tau \supset \sigma \} \vert, $$ where $X(d')$ is the collection of $d'$-simplices in $X$. Then if $X$ is locally finite, i.e., if $\deg_{d'}(\sigma)<\infty$ for any $d$-simplex $\sigma\in X(d)$ and any $d'> d$, then $X$ has no infinite simplices.
I assume by contradiction that there is an infinite simplex $\tau_\infty$, then any of its finite subsets is a simplex. Take some $d\in \mathbb{N}\cup \{0\}$, and we have a $d$-simplex $\sigma\subset \tau_\infty$. Thus,
$$ \deg_{d+1}(\sigma)=\vert \{ \tau\in X(d+1): \sigma \subset \tau \} \vert= \vert \{ v\in \tau_\infty: \; v\notin \sigma \} \vert=\infty, $$ which is a contradiction. Therefore we can't have any infinite simplices.
My question is whether this reasoning is sound, and whether the following derivative argument is also true:
Let $X$ be an abstract simplicial complex. If there exist $m,d\in\mathbb{N}$ such that $\deg_m(\sigma)<\infty$ for all $\sigma \in X(d)$, then $X$ has no infinite simplices.