Closed Subset of Irreducible Space with Same Krull Dimension is Itself
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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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