This is from Hartshorne's Algebraic Geometry:
I don't understand how he concludes that $\text{height } m=n$. I agree that $\overline{Z_0}\subset...\subset\overline{Z_n}$ is maximal and gives rise to a maximal chain of primes in $A(\overline{Y})$ with $m$ as the greatest element. But that is just one maximal chain; if we wanted to conclude $\text{height }m=n$, we would have to check all possible maximal chains involving $m$, right?
I don't see the need to use heights. Isn't it possible to speak only in topological terms? Here is my attempt to prove Proposition 1.10:
Let $n=\dim \overline{Y}$ and $S_0\subset...\subset S_n$ a maximal chain of irreducible, closed subsets of $\overline{Y}$. In particular, by the subset topology, $S_i=\overline{Y}\cap C_i$ for some closed $C_i\subset\mathbb{A}^n$ (Furthermore, since $S_i$ is irreducible, so must be $C_i$). Defining $D_i:=Y\cap C_i$, we also have that $D_0\subset...\subset D_n$ is a maximal chain of irreducible, closed subsets of $Y$. This means $\dim \overline{Y}=n\leq \dim Y$. The reverse inequality is trivial.
Is this correct?