Borromean rings

Example

Take three ellipses in the three coordinate planes of $\R^3$: $$C_1 = \set{(x, y, 0) : \tfrac{x^2}{4} + y^2 = 1}, \quad C_2 = \set{(0, y, z) : \tfrac{y^2}{4} + z^2 = 1}, \quad C_3 = \set{(x, 0, z) : x^2 + \tfrac{z^2}{4} = 1}.$$ They are pairwise disjoint. For instance, a common point of $C_1$ and $C_2$ has $x = 0$ and $z = 0$. Then $y = \pm 1$ from $C_1$ and $y = \pm 2$ from $C_2$, which is impossible. The other pairs are similar. So $C_1 \cup C_2 \cup C_3$ is a link with $3$ components, the Borromean rings. Any two of the components form the unlink of two components. For example, $C_2$ meets the plane $z = 0$ only at $(0, \pm 2, 0)$, outside the region bounded by $C_1$. So $C_1$ bounds a flat disk disjoint from $C_2$, and it can be shrunk in that disk to a tiny circle far from $C_2$. But the three components together cannot be pulled apart: the link is not equivalent to the unlink of three components.