How should I think of $dim_x(X)=\inf\{dim(U)|$ U open neighborhood of $x\}$ where $dim(U)$ is defined by longest chain of irreducible closed sets.

Question

This is related to Iitaka's Algebraic Geometry a.Definition on pg 22 of Sec 1.9.

$X$ is a non-empty topological space. Dimension of $X$ is supremum of all non-negative integers $l$ for which $F_i$ are closed irreducibles s.t. $\emptyset\neq F_0\subset F_1\subset\dots\subset F_l\subset X$ where we require inclusion is proper here.

Set $x\in X$. Define dimension of $X$ at $x$ by $dim_x(X)=\inf\{dim (U)\vert U$ is an open neighborhood of $x\}$

$\textbf{Q:}$ How should I think of $dim_x(X)$? What is a good example? It looks like that $dim_x(X)$ only sees the smallest irreducible component of $X$ containing $x$. Note that $dim_x(X)$ is defined by taking infinimum rather than supremum. What do I expect so?