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Added explanation after comment by original poster.
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Karl Fabian
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Amplitwist's solution just with Jordan: Consider the curves on the Riemann sphere $S$ and connect $\pm 1$ along the real interval $R$ inside the disk such that $R \cup B$ is a closed curve on $S$. Then $A$ is a curve connecting inside and outside of $R \cup B$ and thus intersects $R \cup B$ by Jordan curve theorem, but not inside the disk on $R$. Therefore $A$ and $B$ must intersect.

Update: $i$ and $-i$ are on different sides of $R\cup B$ because the straight line $[-i,i] \subset D$ intersects $R$ exactly once and does not intersect $B$.

Amplitwist's solution just with Jordan: Consider the curves on the Riemann sphere $S$ and connect $\pm 1$ along the real interval $R$ inside the disk such that $R \cup B$ is a closed curve on $S$. Then $A$ is a curve connecting inside and outside of $R \cup B$ and thus intersects $R \cup B$ by Jordan curve theorem, but not inside the disk on $R$. Therefore $A$ and $B$ must intersect.

Amplitwist's solution just with Jordan: Consider the curves on the Riemann sphere $S$ and connect $\pm 1$ along the real interval $R$ inside the disk such that $R \cup B$ is a closed curve on $S$. Then $A$ is a curve connecting inside and outside of $R \cup B$ and thus intersects $R \cup B$ by Jordan curve theorem, but not inside the disk on $R$. Therefore $A$ and $B$ must intersect.

Update: $i$ and $-i$ are on different sides of $R\cup B$ because the straight line $[-i,i] \subset D$ intersects $R$ exactly once and does not intersect $B$.

Source Link
Karl Fabian
  • 1.7k
  • 10
  • 15

Amplitwist's solution just with Jordan: Consider the curves on the Riemann sphere $S$ and connect $\pm 1$ along the real interval $R$ inside the disk such that $R \cup B$ is a closed curve on $S$. Then $A$ is a curve connecting inside and outside of $R \cup B$ and thus intersects $R \cup B$ by Jordan curve theorem, but not inside the disk on $R$. Therefore $A$ and $B$ must intersect.