Qing Liu Proposition 2.5.5, $\dim X=\sup\{\dim_x X \mid x \in X\}$

Question

I've been trying to understand Liu's proof of the statement contained in Proposition 2.5.5 (d): "for X topological space, we have $\dim X=\sup\{\dim_x X \mid x \in X\}$." He basically intersects a closed, irreducible chain $X_0 \subsetneq X_1 \subsetneq \dots \subsetneq X_n$ in $X$ with any open neighborhood $U$ of $x$, saying that the $X_i \cap U$ form a closed, irreducible closed chain of $U$. I understand that all $X_i \cap U$ are closed and irreducible, but I don't see why the inclusion $X_i \cap U \subsetneq X_{i+1} \cap U$ is strict.

Answer 1

It’s clear that for any $x \in X$, $\dim{X} \geq \dim_x{X}$.

Now let $X_0 \subset X_1 \ldots \subset X_n$ be an increasing sequence of closed irreducible subsets, and let $x \in X_0$. I claim that $\dim_x{X} \geq n$.

To do that, for every $i$, $X_i \cap U$ is a nonempty open subset of the irreducible $X_i$, so $X_i \cap U$ is irreducible; it’s a closed subset of $U$.

Now, the goal is to show that the sequence of the $X_i \cap U$ is increasing. Well, let $Z_i\subset X_i$ be the closure in $X$ of $X_i \cap U$. Then $X_i=Z_i \cup [X_i \cap (X \backslash U)]$ is irreducible and the reunion of two closed subsets.

Now $x \notin X \backslash U$, so the second closed subset is proper, so that $X_i=Z_i$ is the closure of $X_i \cap U$ in $X$.

It follows that the sequence of $(X_i\cap U)_i$ is increasing.