### Arithmetic progressions in space?

Here is a neat question that I stumbled upon during my academic random walk.

A sequence of $k+1$ vectors $x_1, \ldots, x_{k+1} \in \mathbb{Z} ^ n$ is called an arithmetic progression of length k if there exists a vector $d \in \mathbb{Z}^n$ such that $x_{i+1} = x_i + d$. In other words, an arithmetic progression in $\mathbb{Z}^n$ is exactly the generalization you would expect of your good ol’ high-school arithmetic progression in $\mathbb{Z}$. For example, in $\mathbb{Z}^2$, the sequence $(11,12), (15,17), (19,22), (23,27)$ is an arithmetic progression of length 3, with difference vector $(4,5)$.

An infinite path in $\mathbb{Z}^n$ is a sequence of distinct points $x_1, x_2, \ldots \in \mathbb{Z}^n$ such that $|| x_i - x_{i+1}|| = 1$. In other words, you start at some $x_0$, and then progress by going one step up or down or left or right or forward or backward or whatever it is they call it in general $n$ dimensions, while making sure not to intersect yourself. For example, here is the beginning of an infinite path in $\mathbb{Z}^2$, which also just happens to contain as a subset the arithmetic progression given above (shown in red):

Question: Does every infinite path in $\mathbb{Z}^n$ contain arbitrarily long arithmetic progressions? That is, given a path $x_1, x_2, \ldots \in \mathbb{Z}^n$, is it true that it contains a subsequence $x_{i_1}, \ldots, x_{i_{k+1}}$ which is an arithmetic progression, for every $k$?

I really liked this question, since at first I had no clue as to what the answer was; while working on it I kept changing my mind, at one time certain that arithmetic progressions are unavoidable, and at the next imagining that an example of a progressionless path is nearly within my grasp.

You too, O’ faithful invisible readers, can tell me what you think, in the form of a non-binding referendum! Post your thoughts and ideas in the comments; I will post the answer in the not-too-distant future.