Remark 5.28.5. Given a locally finite stratification $X = \coprod X_ i$ of a topological space $X$, we obtain a family of closed subsets $Z_ i = \bigcup _{j \leq i} X_ j$ of $X$ indexed by $I$ such that

$Z_ i \cap Z_ j = \bigcup \nolimits _{k \leq i, j} Z_ k$

Conversely, given closed subsets $Z_ i \subset X$ indexed by a partially ordered set $I$ such that $X = \bigcup Z_ i$, such that every point has a neighbourhood meeting only finitely many $Z_ i$, and such that the displayed formula holds, then we obtain a locally finite stratification of $X$ by setting $X_ i = Z_ i \setminus \bigcup _{j < i} Z_ j$.

Comment #2122 by UT on

Why is $Z_i = \bigcup_{j \leq i} X_j$ closed? If the stratification is good, this holds by definition. But in general? Conversely, if we have a family of closed subsets Z_i satisfying the above equation then is it true that the induced stratification as described is good?

Comment #2141 by on

Note that $Z_i$ is the union of the closures of the $X_j$ for $j \leq i$ (this uses that we start with a stratifictation). Since the stratification is locally finite, then we see $Z_i$ is locally a finite union of closed sets. The answer to your second question is no. I suggest you make an example for yourself.

There are also:

• 2 comment(s) on Section 5.28: Partitions and stratifications

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09Y2. Beware of the difference between the letter 'O' and the digit '0'.