Whitney stratification

Remarks

Condition (b) implies condition (a). The same definition applies to subsets of $\C^N = \R^{2N}$, and for complex analytic sets one asks the strata to be complex submanifolds.

Example: the cone $X = \set{x^2 + y^2 = z^2} \subset \R^3$ with three strata: $\set{0}$ and the two connected components of $X \setminus \set{0}$, where $z > 0$ and $z < 0$. The frontier condition holds, and condition (b) holds at $0$. Indeed, the cone is a union of lines through $0$, so the line from $0$ to $x_k$ lies in the tangent plane $T_{x_k} X$, hence also in the limit plane.

Non-example: the complex Whitney umbrella $X = \set{x^2 = z y^2} \subset \C^3$ with the two strata $S'$, the $z$-axis, and $S = X \setminus S'$. The surface $S$ is smooth, since the gradient $(2x, -2zy, -y^2)$ of $x^2 - z y^2$ is nonzero off the $z$-axis. Every point $(0, 0, c)$ is a limit of points $(\sqrt{c}\, t, t, c)$ of $S$ as $t \to 0$, so the frontier condition holds. For small real $s > 0$, the point $x_s = (s^{3/2}, s^{1/2}, s^2)$ lies on $S$ and tends to $0$. There the normal vector is $(2 s^{3/2}, -2 s^{5/2}, -s)$, which, divided by $s$, tends to $(0, 0, -1)$. So $T_{x_s} S$ tends to the complex plane $z = 0$, which does not contain $T_0 S'$, the $z$-axis. Condition (a) fails at $0$, so condition (b) fails too. The origin must be made a stratum of its own: the strata $S$, the $z$-axis minus $0$, and $\set{0}$ form a Whitney stratification.