Let $K$ be a field and $K\{X\}$ be the free non-associative algebra, freely generated by the countably infinite set $X$. We consider elements of $K\{X\}$ as (non-associative) polynomials in the variables of $X$.


Given $\mathfrak{F}\subseteq K\{X\}$, one can consider the variety of algebras defined by $\mathfrak{F}$. This is the class of all algebras which satisfy all identities of $\mathfrak{F}$, i.e., an algebra $A$ is an algebra in such variety if given $p\in \mathfrak{F}$, for any $a\_i\in A$, we have $p(a\_1,\dots,a\_n)=0$.


For example, if one consider the polynomial $p=x(yz)-(xy)z$, the variety of algebras defined by $\{p\}$ is the variety of associative algebras. 


My question is the following:



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> Is the variety of algebras defined by some $\mathfrak{F}\subseteq K\{X\}$, a set?
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A simpler question:
Is a variety of associative algebras a set?



Is a variety of algebras a set?
Summarizing the above, the answer is no. 


