Local dimension at a point and dimension of irreducible components passing through that point

Question

I'm using this definition for the (Krull) dimension of a topological space $X$ and (Krull) dimension at a point $x\in X$. In general, given a topological space $X$, one always has $$ \dim X=\max\{\dim T\mid T\subset X\text{ is an irreducible component}\}. $$ If $X$ is a scheme locally of finite type over a field, one has $$ \dim_x X=\max\{\dim T\mid T\subset X\text{ is an irreducible component passing through }x\}. $$ (See 0A21(5).) Now, if $X$ is just some arbitrary topological space, is the last formula true? My guess is that it is not, but I don't know right now what a counterexample might look like. Besides, can we show that one quantity is always bounded by the other one? Or are there examples of both types of behaviors?

It seems one cannot obtain a bound: let $T\subset X$ be an irreducible component passing through $x$, and let $U\subset X$ be an open neighborhood of $x$. Then $\dim T$ and $\dim U$ are both greater or equal that $\dim T\cap U$, but I don't see a way of comparing the values $\dim T$ with $\dim U$.

Answer 1

The formula does not, indeed, hold in the general case. Consider the space $X=\{a,b,c\}$ with the open subsets $\emptyset, X, \{a\}, \{a,b\}$.

Then $\dim_b{X}=1$. But $X$ is irreducible of dimension $2$, so the RHS of the formula is $2$.

To find a situation where $LHS > RHS$, consider a topological space $X=L \cup \bigcup_{n \in \mathbb{Z}}{P_n}$, where

1. $L$ is closed, and equal to $\mathbb{C}$ with the Zariski topology,
2. every $P_n$ is closed and isomorphic to $\mathbb{C}^2$ with the Zariski topology,
3. a subspace $F \subset X$ is closed iff its intersection with $L$ (resp. $P_n$ for every $n$) ia closed in $L$ (resp. $P_n$),
4. $P_n \cap L$ is the unique point $n \in L$,
5. if $n \neq m$, $P_n$ and $P_m$ are disjoint.

Take some $x \in L \backslash \mathbb{Z}$.

Every open subset $U$ containing $x$ contains a cofinite subset of $L$, and in particular contains some $n \in L \cap \mathbb{Z}$. In particular, $U \cap P_n$ is a nonempty Zariski-open subset of $P_n$, so $\dim{U} \geq \dim{U \cap P_n} \geq 2$.

On the other hand, the only irreducible component going through $x$ is $L$, which has dimension $1$ (this isn’t entirely obvious, but I’d be very surprised if it was false). QED.