Partial order given by a stratification on a topological space.

Question

A (good) stratification of a topological $X$ is a partition $\{X_i\}_{i\in I}$ ($I$ finite) of $X$, with $X_i$ locally closed, such that $X_i \cap \overline{X_j}\neq \emptyset \Rightarrow X_i \subset \overline{X_j}$. It follows that each $\overline{X_i}$ is the union of some $X_j$. We define the relation $"i \leq j$ if and only if $X_i \subset \overline{X_j}"$ on $I$. Is this a partial order relation? In other words, does $i \leq j$ and $j \leq i$ imply $i=j$? I don't know why, i cannot see it.

Answer 1

The key is that if $X$ and $Y$ are (non-empty and) locally closed, and if $X\cap Y=\emptyset$, then it is impossible to have $X\subset \overline Y$ and $Y\subset \overline X$.

Proof:

Write $X=O\cap C$ with $O$ open and $C$ closed.

Since $X\cap Y=\emptyset$ and $Y\subset \overline{X}$, it follows that $Y\subset \left(\overline{X}\setminus X\right)$. But $\left(\overline{X}\setminus X\right)\cap O=\emptyset$ (easy exercise), and so $$Y\cap O=\emptyset$$ Therefore it is impossible for any point in $O$ (and therefore in $X$) to lie in the boundary of $Y$.