# Show that a closed curve formed by two disjoint paths contains the corner of a unit square?

I am looking for a reference, or a topologically/analytically rigorous way of showing the following:

Consider two injective paths in $$\mathbb{R}^2$$ parametrized as $$p_1(t)$$, $$p_2(t)$$, $$0 \leq t \leq 1$$ respectively, such that $$p_1$$ and $$p_2$$ are disjoint everywhere except at the endpoints (i.e. $$p_1(x) = p_2(x)$$ iff $$x \in \{0,1\}$$). Let $$S$$ be a unit square somewhere in the plane. Suppose that $$p_1$$ enters $$S$$ through one of its edges and exits through a different one, and that $$p_1$$ enters $$S$$ exactly once. Further suppose that $$p_2$$ never touches any point of $$S$$. Show that the interior or boundary of the closed curve formed by $$p_1$$ and $$p_2$$ contains at least one corner of $$S$$.

This seems completely obvious to me, but I am not sure how one would show this rigorously.

If it helps, one may assume $$p_1$$ and $$p_2$$ are composed of polygonal lines.

• The composed path $p = p_2^{-1} * p_1$, where $p_2^{-1}(t) = p_2(1-t)$, is a Jordan curve. Now we may apply the well-known Jordan curve theorem en.wikipedia.org/wiki/Jordan_curve_theorem. This is a non-trivial result although it seems to be so obvious. – Paul Frost Oct 28 '18 at 10:40