Intermediate Value for Connected Subset of $\mathbb{C}$

Question

Show that a path going from one vertex of the unit square (in $\mathbb{C}$) to its opposite and a second path going between the other pair of opposite vertices must intersect. By path I mean continuous function $f:[0,1] \to \mathbb{C}$ with $f(0)$ one vertex and $f(1)$ the other. [EDIT] Both are paths in the unit square (no going outside)

I looked through other questions and didn't quite find an answer to this. I know that the intermediate value theorem says that $\textrm{Im}((f-g)(t))$ and $\textrm{Re}((f-g)(t))$ must be $0$ at some point, but I'm not sure how to show that this must occur at the same point.

Duplicate of Why do “path diagonals” of a rectangle need to intersect? ? — Martin R, Mar 04 '18 at 20:53
Wait this is not a duplicate really because that post appealed to more advanced machinery than I am looking for in its solutions. — Jon Hillery, Mar 05 '18 at 04:31

score 1 · Answer 1 · answered Mar 04 '18 at 20:21

1

Assume that such paths exist, for a contradiction. Let $S$ be the unit square, and $I(f),I(g)$ be the images of the paths. Let $v_{1},v_{2}$ be the vertices of $S$ which are end-points of $I(f)$. Note that $v_{1},v_{2} \notin I(g)$, since $I(g) \cap I(f)$ is empty.

One can proceed by showing that $v_{1},v_{2}$ lie in different connected components of $S \setminus I(g)$ by applying the intermediate value theorem "along the diagonal". Then one gets a contradiction since the existence of the path $f$ shows that these connected components are the same.

answered Mar 04 '18 at 20:21

Nick L

4,290

That's surely the right idea but there's no intermediate value theorem in $\Bbb C $ as shown by $f (t)=e^{it} $ – krirkrirk Mar 04 '18 at 20:24
1

I mean apply intermediate value theorem to the real function D(x+iy) = x+y. – Nick L Mar 04 '18 at 20:30
Can you elaborate? I don't see how this shows that $v_{1}$ and $v_{3}$ are in different connected components. – Jon Hillery Mar 05 '18 at 04:38

Intermediate Value for Connected Subset of $\mathbb{C}$

1 Answers1