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Let $M$ be a smooth manifold and $X\subset M$ be a Whitney object, i.e. a subset with a Whitney stratification $\mathcal{S}$. If $W\subset X$ is a closed subset of $X$ such that for each stratum $S\in\mathcal{S}$ the intersection $W\cap S$ is a smooth submanifold of $M$, then is it true that $W$ is also a Whitney object with the decomposition $\mathcal{S}_{W}=\{S\cap W\mid \forall S\in\mathcal{S}\}$?

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    $\begingroup$ Whose definition of "Whitney object" are you using? $\endgroup$ Jan 4, 2017 at 19:57

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I think the answer is 'No'.

A very simple counterexample is as follows:

Let $M = \mathbb R^2$. Let $X$ be the stratification of $M$ given by two strata, the origin $(0,0)$ and $\mathbb R^2 \setminus (0,0)$.

Now consider the spiral $\mathcal A$ in $\mathbb R^2$ defined by the condition that the tangent to the spiral makes a constant angle with the radial vector. In polar coordinates the equation of such a spiral can be given by $r - b\theta =$ constant. Take $W$ to be this spiral $\mathcal A$ union the origin and notice that $S_W = \{\mathcal A, (0,0)\}$ is not a Whitney stratification.

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