Dimension of union of two varieties Suppose $X$ and $Y$ are two varieties. By varieties, I mean affine varieties or quasi-affine varieties or projective varieties or quasi-projective varieties. Suppose Krull dimension of $X$ is $n$ and Krull dimension of $Y$ is $m$ and, without lost of generality, assume $n\leq m$. Is it true that Krull dimension of $X\cup Y$ is $m$? Intuitively, if I add a point or a curve to a curve, it should still look like a curve.
A complete proof or counter example would be greatly appreciated.
 A: Despite appearances these general questions on dimension have nothing to do with algebraic geometry, but are pure general topology.
Here are two results which answer vastly more  than what you ask.
A) If the irreducible components of the topological space $X$ are the closed subsets $X_i, i\in I $ (with $I$ not supposed finite) then we have the equality of Krull dimensions $$\dim X=\sup_{i\in I} (\dim X_i)$$   B) If the topological space $X$ is a finite union of its closed subspaces  (not supposed irreducible) $F_1,F_2,\cdots , F_n$, then $$\dim X=\max (\dim F_1,\cdots, \dim F_n)$$
Although these results are fairly easy to prove, I'll give a reference (alas not in English):
Bourbaki, Algèbre commutative, Chapitre 8 , pages VIII.2 and VIII.3.
A: For any chain of irreducible closed subset of $Y$, their closure in $X\cup Y$ is a chain of irreducible and closed subset of $X\cup Y$, hence the dimension of $X\cup Y$ is at least $m$.
Conversely, suppose $V$ is an irreducible closed subset of $X\cup Y$, then $V=\overline{(V\cap X)}\cup \overline{(V\cap Y)}$, which means either $V=\overline{(V\cap X)}\subset V\cap \overline{X}$ or $V=\overline{(V\cap Y)}\subset V\cap \overline{Y}$. Therefore either $V\subset \overline{X}$ or  $V\subset \overline{Y}$. Hence any chain of irreducible closed subsets of $X\cup Y$ is a chain of irreducible closed subsets of either $\overline{X}$ or $\overline{Y}$. Notice that $X$ and $Y$ are varieties, we have dimension of $Y$ equal dimension of $\overline{Y}$. Hence the length of chain is at most $m$.
