Krull Dimension of Topological Subspace is Smaller
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Definition
Let $X$ be a topological space.
Let $Y \subseteq X$ be a subspace.
Then:
- $\map \dim Y \le \map \dim X$
where $\dim$ denotes the Krull dimension.
Proof
Let:
- $A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_n$
be a chain of closed irreducible sets of $Y$.
Let $\map \cl {A_i}$ be the closure of $A_i$ in $X$ for each $i = 0, \ldots, n$.
By Closure of Irreducible Subspace is Irreducible, each $\map \cl {A_i}$ is irreducible.
Furthermore:
- $\map \cl {A_0} \subsetneq \map \cl {A_1} \subsetneq \cdots \subsetneq \map \cl {A_n}$
since, by Closure of Subset in Subspace:
- $A_i = Y \cap \map \cl {A_i}$
Thus:
- $n \le \map \dim X$
As $\map \dim Y$ is the spremum of such $n$, we have:
- $\map \dim Y \le \map \dim X$
$\blacksquare$
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