The Proof of Birkhoff's Theorem

Question

Hi i'm a maths undergrad currently writing an essay on Universal algebra for my second year project.

I'm primarily using "A Course in Universal Algebra" by Burris and Sankappanavar. I can't seem to understand the proof of lemma 11.8 from chapter 2. The lemma essentially covers the $\Leftarrow$ part of the proof of Birkhoff's theorem (which is stated right below the lemma).

I'm struggling to understand why a second infinite set of variables Y is introduced, as the set X of variables is already infinite.

If it is because we need an infinite set of variables where we can choose any cardinality so that it is always larger or equal to the algebra A, as per the requirements of 10.11. Then why do we need an infinite set X?

Thank you.

Lemma 11.8 If V is a variety and X is an infinite set of variables, then $V=M(Id_V(X) )$

Proof: Let $$V′ = M(IdV (X))$$. Clearly $V′$ is a variety by 11.3, $V′ ⊇ V$, and
$$Id_V ′(X) = Id_V (X)$$ So by 11.4,
$$\mathbf{F}_V ′(\overline{X}) = \mathbf{F}_V(\overline{X})$$ Now given any infinite set of variables Y, we have by 11.6
$$Id_{V′}(Y) = Id_{\mathbf{F}_{V′(X)}}(\overline{Y}) = Id_{\mathbf{F}_{V(X)}} (\overline{Y}) = Id_V(Y).$$ Thus again by 11.4, $$θ_{V′}(Y) = θ_{V}(Y)$$ hence $$\mathbf{F}_{V′}(\overline{Y}) = \mathbf{F}_{V}(\overline{Y})$$ Now for $\mathbf{A}∈ V'$ we have (by 10.11), for suitable infinite Y, $$\mathbf{A} ∈ H(\mathbf{F}_{V′}(\overline{Y}));$$ hence $$\mathbf{A} ∈ H(\mathbf{F}_V(\overline{Y}))$$ so $A ∈ V$ ; hence $V′ ⊆ V$, and thus $V′ = V$.

Answer 1

The reason is just to apply Corollary 10.11.
Notice that $X$ could perhaps not be large enough.

Corollary 10.11 tell us that

If $K$ is a class of algebras of type $\mathscr F$ and $\mathbf A \in K$, then for sufficiently large $X$, $\mathbf A \in H(\mathbf F_K(\overline{X}))$.

The proof starts with "let $|X|\geq |A|$..." and proceeds to prove that that is enough.
So in this case (in the proof of Lemma 11.8), you need $|Y|\geq |A|$, and it could be that $|X| < |A|$ (there's always such an algebra in any non-trivial variety—if $\mathbf B$ is a non-trivial algebra, then $\mathbf A=\mathbf B^X$ will be such an algebra), so that 10.11 wouldn't apply.