## Walks that cannot hit the boundary

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I've casually proved, as application of some ideas that I am developing, a result that might be of interest in itself. I am completely new in this field and then I would like to ask your help to understand: 1) might it be of interest? 2) Is it trivial, in the sense that it can be proved directly? 3) is it well-known?

Let me fix a planar and regular setting, even if I could state the result in more general settings: let $n\geq1$ be a fixed integer and $P=[-n,n]^2\subseteq\mathbb Z^2$. Let me fix the following notation: given $(x,y)\in P$, I will denote by $A(x,y)$ the set formed by the following at most five points: $(x-1,y),(x,y),(x+1,y),(x,y-1),(x,y+1)$, where at most means that if one of those points does not belong to $P$, then I will not consider it.

Now, the situation is the following: for any point $p\in P$, let $\gamma_p$ a walk starting on $p$ and ending on $p^+$. I suppose that: if a walk hits the boundary, then it ends. In particular, if $p\in\partial P$, then $p^+=p$.

Definition: A flow is a family of walks $\gamma_p$, one for each $p\in P$, such that: for all $p\in P$, whenever $q\in A(p)$, then $q^+\in A(p^+)$.

My result would be: Given a flow of walks, there is at least one walk which does not hit the boundary.

I was thinking that it might be useful to prove that some walks are bounded, but I repeat that I am really new in this field.

Every comment is welcome and also references are appreciated.

Thanks in advance,

Valerio

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## 1 Answer

In fact, you can prove that $p^+=p$ for all $p$.

Actually, consider the shortest paths from $(-n,-n)$ to $(n,n)$ and from $(-n,n)$ to $(n,-n)$ passing through $p$. Taking `pluses' of them, you should also obtain the paths of the same lengths connecting the same points. Now from the first path you obtain that the sum of coordinates of $p^+$ is the same as for $p$, and from the second path you see that the difference of coordinates is the same for $p$ and $p^+$. Hence $p=p^+$.

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Jesus! I'm doing a mess. It was enough to draw a stupid picture to realize it! – Valerio Capraro Oct 14 2011 at 17:23
I was thinking that actually it would be enough for me that the points on the boundary are mapped bijectively in points on the boundary (veryfying the property in the definition). But probably also in this case the result remains trivial: it seems to me that in this case every square of the shape $[-k,k]^2$ is mapped in itself – Valerio Capraro Oct 14 2011 at 17:32
Well, if your map is a bijection on the boundary, then some two points should come to opposite vertices of the square. Then they should be the vertices themselves, otherwise there would exist a path between them which is shorter than $4n$. Hence it is also a bijection on the vertices, and by the same reasons of the shotrest path the whole map is some symmetry of the square. – Ilya Bogdanov Oct 14 2011 at 17:40
Yes, on the squares everything get very intuitive. Actually my proof would work for more general subsets of $\mathbb Z^2$; intuitively those subsets without holes whose boundary has a hole. Well, maybe I will put it just as a remark. Many thanks for the clarifications. – Valerio Capraro Oct 14 2011 at 21:43
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