An exercise in the first chapter of Hartshorne asks you to give an example of a topological space $X$ and a dense open subset $U$ in $X$ such that $\dim U < \dim X$. I have an example, but it feels very arbitrary, and I'm wondering if there is a more insightful example.
My example: Take $\mathbb{A}^1$ over an infinite field, and adjoin a single point $\infty$ (let $X := \mathbb{A}^1 \cup \{\infty\}$). Let the open sets be: $\emptyset, X, \mathbb{A}^1$, and subsets of $\mathbb{A}^1$ with finite complement. Then $\mathbb{A}^1$ is dense and open, and the topology on $\mathbb{A}^1$ is the usual Zariski topology, so $\dim \mathbb{A}^1 = 1$, but if $q$ is any point of $\mathbb{A}^1$, then $\{q\} \subset \{q, \infty\} \subset X$ is a strictly increasing sequence of closed irreducible spaces, so $\dim X = 2$.
This example feels forced. Does anyone know a better example that appears naturally (in any area of math)?
Best Answer
Consider the Sierpinski space $S$: this is the set $\{0,1\}$ with the topology $\{\emptyset,\{1\},\{0,1\}\}$. Then $\{1\}$ is open, dense, and of dimension zero, while $S$ is of dimension one.
The way this comes up in algebraic geometry is as the spectrum of a discrete valuation ring: a DVR $R$ has two ideals, $(0)$ and $\mathfrak{m}$, with $(0)\subset\mathfrak{m}$ giving that the underlying topological space of $\operatorname{Spec} R$ is homeomorphic to $S$. (You'll meet this in chapter II of Hartshorne.)