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Let $f, g$ be injective paths from the interval $I$ to the closed unit disc $\overline{D}^2$ such that $f(0)=(-1,0), f(1)=(1,0), g(0)=(0,-1)$ and $g(1)=(1,0).$ Show that $f(I) \cap g(I) \ne \emptyset$.

Define an injective path $\gamma : I \to \mathbb{R}^2 \setminus D^2$ such that $\gamma(0)=f(0)$ and $\gamma(1)=f(1)$. If we now consider the closed curve $C$ formed by "gluing" the paths $\gamma$ and $f$ we can use the Jordan curve theorem and have that $\mathbb{R}^2\setminus C$ has two components $U_1$ and $U_2$ which another one is bounded and another unbounded. Wlog let $U_2$ be the unbounded component. It's now enough to show that $C$ meets $g(I)$, but I don't know how to prove this. What could be done here?

Michele
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  • You wrote that $f(1)=(1,0)=g(1)$ so the intersection trivially holds; I think you meant $g(1)=(0,1)$. The reality is Jordan Curve Theorem is a lot of machinery that doesn't get you much here... in particular the same result holds for not necessarily injective $f,g$. This is the Lovers and Haters problem, typically solved with a winding number or Brouwer Fixed Point argument. See e.g. https://math.stackexchange.com/questions/3717288/lovers-and-haters-via-the-jordan-curve-theorem – user8675309 Mar 15 '23 at 16:27

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Let us suppose to the contrary that $f(I) \cap g(I) = \emptyset.$ Since $f$ and $g$ are continuous, their images $f(I)$ and $g(I)$ are compact subsets of $\overline{D}^2,$ and hence they are disjoint compact subsets of $\overline{D}^2.$

Consider the function $h: I \to \mathbb{R}$ given by $$h(t) = |f(t)-g(t)|,$$ Since $h(t) \geq 0$ for all $t \in I,$ the function $h$ attains its minimum value at some point $t_0 \in I,$ say $h(t_0) = d > 0.$

Now, consider the circle $C$ centered at $f(t_0)$ with radius $d/2.$ Since $f$ is injective, $f(t_0) \not \in f(I \setminus {t_0}),$ so $C$ intersects $f(I)$ only at $f(t_0).$ Similarly, the circle $C$ centered at $g(t_0)$ with radius $d/2$ intersects $g(I)$ only at $g(t_0).$ But since $d > 0,$ $f(t_0)$ and $g(t_0)$ are distinct points, and so we have shown that $f(I)$ and $g(I)$ have nonempty intersection with $C.$

Since $C$ is a connected set, it follows that $f(I)$ and $g(I)$ have nonempty intersection, a contradiction. Therefore, our assumption that $f(I) \cap g(I) = \emptyset$ must be false, and so $f(I) \cap g(I) \neq \emptyset.$

Wrloord
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