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Let $\Gamma$, $\Sigma$ be two curves with ranges in $(\{0\}\cup\mathbb{R}_{+})^2$. $\Gamma$ starts on the $y$ and ends on the $x$ axis: $\Gamma(0)=(0,\gamma_2),\Gamma(1)=(\gamma_1,0)$. $\Sigma$ is a "positive unit step" curve, in the sense that it is a concatenation of horizontal and vertical segments of unit length in the positive $x$ or $y$ directions (it "moves" by a unit step either to the right or upwards). Moreover, $\Sigma$ starts on one of the axes, and, if $\Sigma(0)=(\sigma_1,\sigma_2)$, then $\sigma_j\le \gamma_j$.

The following statement is clear, but I haven't proved it rigorously yet: $\Gamma$ and $\Sigma$ intersect. Consider the closed curve formed by $\Gamma$, the origin, and the two segments on the axes in between. If a point moving along $\Sigma$ is in the (bounded) region enclosed by this closed curve, it will eventually exit it, since $\Sigma$ is unbounded.

Could someone please assist me in constructing a rigorous proof?

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  • $\begingroup$ If $\Sigma$ has initially just horizontal segments that go beyond $\gamma_2$, would that cause any problem? $\endgroup$ Apr 14 '15 at 22:48
  • $\begingroup$ Thanks for the comment. I just corrected a typo: switched the indices of $\gamma$ the first time they appear. $\endgroup$
    – pitaya
    Apr 14 '15 at 22:53
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You are pretty close. As you said, consider the closed curve $\Lambda$ which starts at the origin, goes straight to $(0, \gamma_2)$, follows $\Gamma$, and then goes straight from $(\gamma_1, 0)$ to the origin. By the Jordan curve theorem, this divides $\mathbb{R}_{\geq 0}^2$ into two regions. Let us call the bounded region $A$ and the unbounded region $B$, and note $\mathbb{R}_{\geq 0}^2 = A \cup \operatorname{im}(\Lambda) \cup B$, which are disjoint. Also note the Jordan curve theorem says that $\partial A = \operatorname{im}(\Lambda)$.

We now construct the regions $X$ and $Y$, where $X = A \cup \operatorname{im}(\Lambda) - \operatorname{im}(\Gamma)$, and $Y = B$. It is easy to see that we have a partition $\mathbb{R}_{\geq 0}^2 = X \cup \operatorname{im}(\Gamma) \cup Y$ and that $\partial X = \operatorname{im}(\Gamma)$. Now we know that $\Sigma$ starts in $X \cup \operatorname{im}(\Gamma)$ and ends up in $Y$ from the givens.

If $\Sigma$ starts within $\operatorname{im}(\Gamma)$ we are done. So let us assume $\Sigma(0) \in X$. Since $\Sigma$ leaves $X$, we know $\exists s \in \mathbb{R}$ s.t. $\Sigma(s) \in \partial X$. But $\partial X = \operatorname{im}(\Gamma)$, so $\Sigma(s) \in \operatorname{im}(\Gamma)$, providing us with our intersection.

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  • $\begingroup$ Thank you. The curve $\Lambda$ you define is closed, but do we have to assume that it is simple? My question does not assume this. $\endgroup$
    – pitaya
    Apr 14 '15 at 23:45
  • $\begingroup$ @graviola Ah my mistake. If it is not simple, then you can reduce the curve by eliminating all the loops created by any self-intersections. The resulting curve's image is a subset of the original curve's image, so proving the result for the modified curve implies it for the original curve. $\endgroup$ Apr 14 '15 at 23:48
  • $\begingroup$ Thank you. Is there a standard, rigorous procedure for finding a simple curve that is a subset of a non-simple curve, with the same endpoints (the loops may be infinitely many?)?. It is not very clear to me how to handle this rigorously. More importantly, could you please explain how you conclude, in your argument above, that there is an $s\in \mathbb{R}$ such that $\Sigma(s)$ is on the boundary of $X$? Thanks again! $\endgroup$
    – pitaya
    Apr 14 '15 at 23:55
  • $\begingroup$ Oh, I see, it's because otherwise the curve wouldn't be connected, but it has to be connected by continuity? $\endgroup$
    – pitaya
    Apr 15 '15 at 0:05
  • $\begingroup$ @graviola Exactly. $\endgroup$ Apr 15 '15 at 0:12

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