Is $\varnothing$ an affine variety with these definitions?

Question

A topological space $X$ is called reducible if $X=X_1 \cup X_2$ where $X_1, X_2$ are non-empty, proper subsets of $X$ and closed.

An irreducible algebraic set in $\mathbb{A}^n$ is called an affine variety.

$\varnothing = Z(1)$, so $\varnothing$ is an algebraic set, and obviously $\varnothing \subseteq \mathbb{A}^n$.

If we suppose that $\varnothing$ is reducible, then the only options for $X_1$ and $X_2$ are $\varnothing$, but this can't happen, because $X_1$ and $X_2$ must be non-empty.

So, am I right saying that

$\varnothing$ is an affine variety?

possible duplicate of is the empty set an (irreducible) variety? — Alex Kruckman, Mar 09 '15 at 21:04
I saw that question but the definition of ''reducible'' is different, I don't suppose that $X \neq \varnothing$ — Leafar, Mar 09 '15 at 21:08
@Leafar. You are not entitled to change the meaning of definitions. This is like asking "if I call $3$ the number $4$, is it true that $2+2=3$" ? So, yes, $\emptyset=Z(1)$ is algebraic, not irreducible and that's the end of the (uninteresting) story. — Georges Elencwajg, Mar 09 '15 at 21:23
Why uninteresting? There are many things associated, like the dimension, we have to consider all pathological cases... — Leafar, Mar 09 '15 at 21:27

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

With your definition of "irreducible" and "affine variety", the answer is yes.

However, there are good reasons for considering this to be the wrong definition of "irreducible", as explained in this question. Here's another: the defining property of sober spaces (examples include algebraic varieties with the Zariski topology and Hausdorff spaces) is "every irreducible closed set has a unique generic point". This is nonsense if we allow $\emptyset$ to be irreducible.

Is $\varnothing$ an affine variety with these definitions?

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