Asymptotic dimension

Let ${\displaystyle X}$ be a metric space and ${\displaystyle n\geq 0}$ be an integer. We say that ${\displaystyle \operatorname {asdim} (X)\leq n}$ if for every ${\displaystyle R\geq 1}$ there exists a uniformly bounded cover ${\displaystyle {\mathcal {U}}}$ of ${\displaystyle X}$ such that every closed ${\displaystyle R}$-ball in ${\displaystyle X}$ intersects at most ${\displaystyle n+1}$ subsets from ${\displaystyle {\mathcal {U}}}$. Here 'uniformly bounded' means that ${\displaystyle \sup _{U\in {\mathcal {U}}}\operatorname {diam} (U)<\infty }$.

We then define the asymptotic dimension ${\displaystyle \operatorname {asdim} (X)}$ as the smallest integer ${\displaystyle n\geq 0}$ such that ${\displaystyle \operatorname {asdim} (X)\leq n}$, if at least one such ${\displaystyle n}$ exists, and define ${\displaystyle \operatorname {asdim} (X):=\infty }$ otherwise.

Also, one says that a family ${\displaystyle (X_{i})_{i\in I}}$ of metric spaces satisfies ${\displaystyle \operatorname {asdim} (X)\leq n}$ uniformly if for every ${\displaystyle R\geq 1}$ and every ${\displaystyle i\in I}$ there exists a cover ${\displaystyle {\mathcal {U}}_{i}}$ of ${\displaystyle X_{i}}$ by sets of diameter at most ${\displaystyle D(R)<\infty }$ (independent of ${\displaystyle i}$) such that every closed ${\displaystyle R}$-ball in ${\displaystyle X_{i}}$ intersects at most ${\displaystyle n+1}$ subsets from ${\displaystyle {\mathcal {U}}_{i}}$.

• If ${\displaystyle X}$ is a metric space of bounded diameter then ${\displaystyle \operatorname {asdim} (X)=0}$.
• ${\displaystyle \operatorname {asdim} (\mathbb {R} )=\operatorname {asdim} (\mathbb {Z} )=1}$.
• ${\displaystyle \operatorname {asdim} (\mathbb {R} ^{n})=n}$.
• ${\displaystyle \operatorname {asdim} (\mathbb {H} ^{n})=n}$.

• If ${\displaystyle Y\subseteq X}$ is a subspace of a metric space ${\displaystyle X}$, then ${\displaystyle \operatorname {asdim} (Y)\leq \operatorname {asdim} (X)}$.
• For any metric spaces ${\displaystyle X}$ and ${\displaystyle Y}$ one has ${\displaystyle \operatorname {asdim} (X\times Y)\leq \operatorname {asdim} (X)+\operatorname {asdim} (Y)}$.
• If ${\displaystyle A,B\subseteq X}$ then ${\displaystyle \operatorname {asdim} (A\cup B)\leq \max\{\operatorname {asdim} (A),\operatorname {asdim} (B)\}}$.
• If ${\displaystyle f:Y\to X}$ is a coarse embedding (e.g. a quasi-isometric embedding), then ${\displaystyle \operatorname {asdim} (Y)\leq \operatorname {asdim} (X)}$.
• If ${\displaystyle X}$ and ${\displaystyle Y}$ are coarsely equivalent metric spaces (e.g. quasi-isometric metric spaces), then ${\displaystyle \operatorname {asdim} (X)=\operatorname {asdim} (Y)}$.
• If ${\displaystyle X}$ is a real tree then ${\displaystyle \operatorname {asdim} (X)\leq 1}$.
• Let ${\displaystyle f:X\to Y}$ be a Lipschitz map from a geodesic metric space ${\displaystyle X}$ to a metric space ${\displaystyle Y}$ . Suppose that for every ${\displaystyle r>0}$ the set family ${\displaystyle \{f^{-1}(B_{r}(y))\}_{y\in Y}}$ satisfies the inequality ${\displaystyle \operatorname {asdim} \leq n}$ uniformly. Then ${\displaystyle \operatorname {asdim} (X)\leq \operatorname {asdim} (Y)+n.}$ See^[3]
• If ${\displaystyle X}$ is a metric space with ${\displaystyle \operatorname {asdim} (X)<\infty }$ then ${\displaystyle X}$ admits a coarse (uniform) embedding into a Hilbert space.^[4]
• If ${\displaystyle X}$ is a metric space of bounded geometry with ${\displaystyle \operatorname {asdim} (X)\leq n}$ then ${\displaystyle X}$ admits a coarse embedding into a product of ${\displaystyle n+1}$ locally finite simplicial trees.^[5]