# Show that three pairwise non-intersecting lines in $\mathbb{R}\mathbb{P}^{3}$ have a transversal.

Let $\mathbb{P}(U1)$ and $\mathbb{P}(U2)$ be two non-intersecting lines in the 3-dimensional projective space $\mathbb{R}\mathbb{P}^{3}$ = $\mathbb{P}(\mathbb{R}^{4})$. Show that $\mathbb{R}^{4}$ equals the direct sum U1 $\oplus$ U2. Deduce that three pairwise non-intersecting lines in $\mathbb{R}\mathbb{P}^{3}$ have a transversal, that is, a projective line meeting all three.

I'm fine with the first part but I really don't know how to deduce that there must exist a transversal.

The first part states that every point in $\mathbb{RP}^3$ can be expressed as a linear combination of points on these two lines, i.e. is collinear with two such points. So choose an arbitrary point on the third line, apply that result and you have a transversal. So in fact the three lines will not only have one transversal, but a whole family of them, one for every point of the third line.