Upward Löwenheim-Skolem theorem without equality I am confused about the following argument for a simplified proof of the upward Löwenheim-Skolem theorem for first-order logic without equality:
Let $\mathcal M$ be an infinite first-order structure. Given some element $a\in M$, we add a family of new elements $a_i$, $i\in I$ and stipulate that they behave as $a$ for every function and predicate symbol, creating a new structure $\mathcal M'$. Now, $\mathcal M$ and $\mathcal M'$ satisfy the same equality-free formulas if we allows variables assignments in $M$.
I don't quite understand how it is necessary that $\mathcal M$ is infinite here. But if the restriction would be dropped, then the result would be that any satisfiable set has models of arbitrary large cardinalities.
 A: Indeed, the requirement that $\mathcal{M}$ be infinite is superfluous, and the more general result you suggest holds: if $T$ is a satisfiable set of first-order sentences without equality, then $T$ has arbitrarily large models. The proof goes through unchanged.
In more detail, given $\mathcal{M},M,a,I$ as in the question, we define $\mathcal{M}'$ as follows:

*

*The underlying set $M'$ of $\mathcal{M}'$ is $M\sqcup\{\langle a,i\rangle: i\in I\}$.


*Let $i:M'\rightarrow M$ be the identity on $M$ and send $\langle a,i\rangle$ to $a$; basically, $i$ is the "collapse" map corresponding to the "blow-up" we're performing. If we were to "blow-up" more than one element of $\mathcal{M}$, which we could but don't need to, this $i$ would correspondingly be (slightly) more interesting.


*For each $n$-ary relation symbol $R$ in the language of $\mathcal{M}$ we set $$R^\mathcal{M'}=\{(c_1,...,c_n)\in ((M')^n: (i(c_1),...,i(c_n))\in R^\mathcal{M}\}.$$


*For each $n$-ary function symbol $f$ in the language of $\mathcal{M}$ we set $$f^\mathcal{M'}(c_1,...,c_n)=f^\mathcal{M}(i(c_1),...,i(c_n)).$$ (Note that $M\subseteq M'$: we've added new things corresponding to $a$, but we haven't gotten rid of $a$ itself. So the output of $f^\mathcal{M'}$ is in fact guaranteed to be an element of $\mathcal{M}'$.)
It's then easy to check that $\mathcal{M'}\equiv_{\mathsf{w/o=}}\mathcal{M}$.
