# Every point of Ext(C) lies on some tangent line to C

In our differential geometry class our professor gave the following problem as a homework problem:

Let $C$ be a smooth non-singular simple closed curve in plane. Prove that every point of the unbounded component of the complement $\mathbb{R}^2 \setminus C$ (Ext($C$)) lies on some tangent line to $C$.

The statement is simple but I have no idea how to proceed. Any hint is really appreciated.

## 1 Answer

Hint: For a point in the unbounded component, pick some random line through that point not intersecting your curve. Now move that line towards your curve $C$ until it touches it. At the moment it touches $C$ it will be tangent.

Further hint: This is a transversality statement if you like, the condition of being a transverse intersection is stable under small pertubation. So since a small pertubation from the first moment of touching the curve takes you off of it, then you must not have been transversally intersecting. But now dimension considerations force the intersection to be tangential.

PS As far as I know, Ext(C) is not standard notation. I am guessing that you mean the unbounded component of $\mathbb{R}^2 \setminus C$.

• Yes, that's what I mean. Thank you for your hint, although I don't understand your further hint. We just started differential geometry for a few weeks. Maybe when we get further I'll get what you mean. – user228960 Feb 21 '16 at 16:58
• @idioteque For two plane curves to intersect transversally at a point p just means that their tangent spaces at p together span the tangent space of the plane at p. This property is preserved if one of the curves is moved a little (you can express this stability in terms of the continuity of the determinant, like so many things). In particular, small changes don't disconnect transverse intersections. Guilleman and Pollack explain this well in their introductory text on differential topology. It is just a hint for how to show tangency of the first intersection. – Lorenzo Najt Feb 21 '16 at 18:15