Sawtooth-like polygonal chains in $\mathbb R^2$ must self-intersect. We consider closed polygonal chains in the 2-dimensional plane with an even number of sides, say $2n$, numbered as $A_1B_1A_2B_2\dots A_nB_nE$, where $E = A_1$.
We require additionally, that each $B_i$ lies right below $A_i$, like with sawtooth.
Edit: Since right below seems ambiguous, let's state this line explicitly as that the $y$-coordinate of each $B_i$ be strictly less than that of $A_i$.
Is it true that any such polygonal chain must self-intersect?
In the 3-dimensional case, there is the obvious counterexample of traversing the vertices of any prism in the obvious order. I have however yet to find a proof or counterexample in the 2-dimensional case.

For convenience, extend the indices cyclically to all of $\Bbb Z$, so that e.g. $A_0=A_n$, $A_{n+1}=A_1$.
Proof by induction on $n$.
The case $n=1$ is trivial and the case $n=2$ is clear: $B_1A_2$ and $B_2A_1$ are the diagonals of the trapezoid $A_1B_1B_2A_2$.
With $n>2$ consider a sawtooth polygonal chain $A_1B_1A_2\ldots B_nA_1$ and assume it is a simple polygon. Introduce coordinates $A_k=(x_k,y_k)$ and $B_k=(x_k,z_k)$ (where $z_k<y_k$).  If for some $k$, we have $x_{k+1}=x_k$ (i.e., $A_{k+1}$ is on the line $A_kB_k$), then either $y_{k+1}>z_k$ and we have an immediate self-intersection, or $y_{k+1}\le z_k$ and we can drop $B_k$, $A_{k+1}$ from the vertices and describe the same sawtooth polygonal chain with only $2\cdot(n-1)$ vertices - which is self-intersecting by induction hypothesis, contradiction! Hence we may assume that $x_{k+1}\ne x_k$ for all $k$.
As $x_{n+1}=x_1$, the sequence $x_1,x_2,\ldots$ cannot be monotonic. Hence there must exist $k$ with $x_{k-1}<x_k>x_{k+1}$; call such $k$ locally rightmost. Among all locally rightmost $k$, pick one with minimal $x_k$. Depending on the situation, we consider the following trapezoid $A_kB_kPQ$:

*

*If $x_{k-1}=x_{k+1}$, let $P=A_{k+1}$, $Q=B_{k-1}$

*If $x_{k-1}>x_{k+1}$, let $Q=B_{k-1}$ and $P$ the point on $B_{k}A_{k+1}$ on the same vertical line as $Q$

*If $x_{k-1}<x_{k+1}$, let $P=A_{k+1}$ and $Q$ the point on $B_{k-1}A_k$ on the same vertical line as $P$
As $B_kA_{k+1}$ does not intersect $B_{k-1}A_k$, we have $P$ below $Q$ in each case, i.e., $A_kB_kPQ$ is a convex trapezoid. Let $U$ be the interior of $A_kB_kPQ$ together with the interior of line segment $PQ$ (in other words, the closed filled trapezoid $A_kB_kPQ$ but with the three edges $QA_kB_kP$ removed). We make two observations:

*

*Not all $A_i$ are $\in U$. In fact, at least $A_{k-1},A_k,A_{k+1}\notin U$.

*$A_i\in U\iff B_i\in U$. This follows because the line segment $A_iB_i$ can enter/leave $U$ only across the also vertical line segment $PQ$.

Assume some $A_i$ is in $U$. Extend this to a maximal set $I=\{r,r+1,\ldots, s\}$ of (cyclically) consecutive indices with $A_i\in U$ for all $i\in I$, but $A_{r-1}, A_{s+1}\notin U$. As observed, then also $B_{r-1}\notin U$ and $B_s\in U$. As the only way "allowed" to enter/leave $U$ is across $PQ$,  we conclude $x_{r-1}<x_r$ and $x_s>x_{s+1}$. It follows that some $j\in I$ is locally rightmost. By minimality of $x_k$, we have $x_j\ge x_k$, but that contradicts $A_j\in U$! We conclude that no $A_i$ is $\in U$.

By our observation, also no $B_i$ is $\in U$.
Finally, if some arbitrary point $R$ on the polygonal chain is $\in U$, then at least one of the endpoints of the line segment containing $R$ must be $\in U$, but we just saw that no $A_i,B_i$ is $\in U$.
We conclude that the interior of line segment $PQ$ has no point in common with our polygonal chain. This allows us to take $QP$ as a shortcut, i.e.,
the sawtooth polygonal chain $A_1B_1\ldots A_{k-1}B_{k-1}QPA_{k+1}B_{k+1}\ldots B_nA_1$, or rather


*

*$A_1B_1\ldots A_{k-1}B_{k+1}\ldots B_nA_1$ if $x_{k-1}=x_{k+1}$

*$A_1B_1\ldots A_{k-1}PA_{k+1}B_{k+1}\ldots B_nA_1$ if $x_{k-1}=x_{k+1}$ if $x_{k-1}>x_{k+1}$

*$A_1B_1\ldots A_{k-1}B_{k-1}QB_{k+1}\ldots B_nA_1$ if $x_{k-1}=x_{k+1}$ if $x_{k-1}<x_{k+1}$
is not self-intersecting - whereas by induction hypothesis it is!
