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 $Y = X$.

$\blacksquare$

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