Krull Dimension of Topological Subspace is Smaller

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 $N$ be a chain of closed irreducible sets of $Y$.

Let:
 * $N' := \set { \map \cl A : A \in N }$

where:
 * $\map \cl A$ denotes the closure of $A$ in $X$

By Closure of Irreducible Subspace is Irreducible, all elements of $N'$ are irreducible.

Thus $N'$ is a chain of closed irreducible sets of $X$.

Moreover, $N$ and $N'$ have the same length.

Indeed:
 * $\forall A, B \in N : A \ne B \implies \map \cl A \ne \map \cl B$

since by Closure of Subset in Subspace:

As $N$ was arbitrary, the claim follows.