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We say that a function $f: \mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}$ is uniformly continuous if there is an integer $K\geq 1$ such that whenever $(x,y),(x',y')\in \mathbb{Z}\times \mathbb{Z}$ with $|(x,y)-(x',y')| = 1$ in the Euclidean distance, then $|f(x,y)-f(x',y')| < K$.

Is there an injective uniformly continuous function $f:\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}$?

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    $\begingroup$ This is not what's usually called "uniformly continuous". In this precise setting (functions from $\mathbf{Z}^k$ to any metric space), this is just "Lipschitz". In the usual definition, every function from a uniformly discrete metric space to any metric space is uniformly continuous. $\endgroup$
    – YCor
    Commented Jul 8, 2019 at 12:01

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No. A uniformly continuous function takes $O(N)$ distinct values on an $N\times N$ grid.

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