If $X$ is an irreducible affine variety and $U \subseteq X$ a non-empty and open (in $X$), then $\dim U = \dim X$.
The part $\dim U \le \dim X$ is clear, but I do not see how to prove the other part. All I can think of that is somehow similar is the common lemma:
Lemma: For a general topological space $X$ holds that if $\{U_i\}_{i \in I}$ is an open cover of $X$, then $\dim X = \sup \{\dim U_i \mid i \in I \}$.
But I am not sure if we can apply that here. Could you please give me a hint?
REMARK: We defined the dimension via chains.