Closed Subset of Irreducible Space with Same Krull Dimension is Itself

Definition
Let $X$ be an irreducible topological space.

Let $Y \subseteq X$ be a closed subset.

Suppose:
 * $\map \dim Y = \map \dim X < + \infty$

where $\dim$ denotes the Krull dimension.

Then:
 * $Y = X$

Proof
Let $n = \map \dim Y$.

Then there exists a chain of closed irreducible sets of $Y$:
 * $A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_n$

If $Y \subsetneq X$, then:
 * $A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_n \subsetneq X$

would be a chain of closed irreducible sets of $X$, which implies that:
 * $n + 1 \le \map \dim X$

which is a contradiction.

Thus $Y = X$.