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Let $S^1$ be a unit circle whose center is the origin and radius is $1$.

And let $B^1$ denote $\{(x,y) : x^2 + y^2 < 1 \}$, which is an open ball.

Also let $D^1$ denote $\{(x,y) : x^2 + y^2 \leq 1 \}$, which equals $B^1 \cup S^1$.

Let $f : [0,1] \rightarrow D^1$ be a continuous map such that $f(0) = (-1,0)$, $f(1) = (1,0)$ and $f(t) \in B^1$ for each $0 < t < 1$.

Let $g : [0,1] \rightarrow D^1$ be a continuous map such that $g(0) = (0,-1)$, $g(1) = (0,1)$ and $g(t) \in B^1$ for each $0 < t < 1$.

In my opinion, there should be an intersection point between $f$ and $g$.

In other words, (in my opinion), there exists $0 < t_1, t_2 < 1$ such that $f(t_1) = g(t_2)$.

It seems obvious, however I'm in stuck.

For example, if $f$ is defined as $f(t) = (-1 + 2t, 0)$, then by intermediate value theorem, the problem is easily solved.

Joshua Tilley
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with-forest
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1 Answers1

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As state in the comments, while it looks obvious, this requires non-trivial theorems to prove rigorously. Can you come up with a proof for your statement if you take the Jordan curve theorem as given? This theorem states that any closed continuous curve separates the plane into an inside and an outside.

If you don't want to use this theorem or a variant thereof you essentially have to reprove the theorem. I think one can show that your statement implies the Jordan curve theorem so you would need the same kind of machinery and techniques to prove it from scratch as are used to prove the Jordan curve theorem.

quarague
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