Equivalence of metrics

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In the study of metric spaces in mathematics, there are various notions of two metrics on the same underlying space being "the same", or equivalent.

In the following, ${\displaystyle X}$ will denote a non-empty set and ${\displaystyle d_{1}}$ and ${\displaystyle d_{2}}$ will denote two metrics on ${\displaystyle X}$.

Topological equivalence[]

The two metrics ${\displaystyle d_{1}}$ and ${\displaystyle d_{2}}$ are said to be topologically equivalent if they generate the same topology on ${\displaystyle X}$. The adjective "topological" is often dropped.^[1] There are multiple ways of expressing this condition:

• a subset ${\displaystyle A\subseteq X}$ is ${\displaystyle d_{1}}$-open if and only if it is ${\displaystyle d_{2}}$-open;
• the open balls "nest": for any point ${\displaystyle x\in X}$ and any radius ${\displaystyle r>0}$, there exist radii ${\displaystyle r',r''>0}$ such that

${\displaystyle B_{r'}(x;d_{1})\subseteq B_{r}(x;d_{2}){\text{ and }}B_{r''}(x;d_{2})\subseteq B_{r}(x;d_{1}).}$

• the identity function ${\displaystyle I:X\to X}$ is both ${\displaystyle (d_{1},d_{2})}$-continuous and ${\displaystyle (d_{2},d_{1})}$-continuous.

The following are sufficient but not necessary conditions for topological equivalence:

• there exists a strictly increasing, continuous, and subadditive ${\displaystyle f:\mathbb {R} \to \mathbb {R} _{+}}$ such that ${\displaystyle d_{2}=f\circ d_{1}}$.^[2]
• for each ${\displaystyle x\in X}$, there exist positive constants ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ such that, for every point ${\displaystyle y\in X}$,

${\displaystyle \alpha d_{1}(x,y)\leq d_{2}(x,y)\leq \beta d_{1}(x,y).}$

Strong equivalence[]

Two metrics ${\displaystyle d_{1}}$ and ${\displaystyle d_{2}}$ are strongly equivalent if and only if there exist positive constants ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ such that, for every ${\displaystyle x,y\in X}$,

${\displaystyle \alpha d_{1}(x,y)\leq d_{2}(x,y)\leq \beta d_{1}(x,y).}$

In contrast to the sufficient condition for topological equivalence listed above, strong equivalence requires that there is a single set of constants that holds for every pair of points in ${\displaystyle X}$, rather than potentially different constants associated with each point of ${\displaystyle X}$.

Strong equivalence of two metrics implies topological equivalence, but not vice versa. An intuitive reason why topological equivalence does not imply strong equivalence is that bounded sets under one metric are also bounded under a strongly equivalent metric, but not necessarily under a topologically equivalent metric.

When the two metrics ${\displaystyle d_{1},d_{2}}$ are those induced by norms ${\displaystyle \|\cdot \|_{A},\|\cdot \|_{B}}$ respectively, then strong equivalence is equivalent to the condition that, for all ${\displaystyle x\in X}$,

${\displaystyle \alpha \|x\|_{A}\leq \|x\|_{B}\leq \beta \|x\|_{A}}$

In finite dimensional spaces, all metrics induced by the p-norm, including the euclidean metric, the taxicab metric, and the Chebyshev distance, are strongly equivalent.^[3]

Properties preserved by equivalence[]

• The continuity of a function is preserved if either the domain or range is remetrized by an equivalent metric, but uniform continuity is preserved only by strongly equivalent metrics.^[4]
• The differentiability of a function ${\displaystyle f:U\to V}$, for ${\displaystyle V}$ a normed space and ${\displaystyle U}$ a subset of a normed space, is preserved if either the domain or range is renormed by a strongly equivalent norm.^[5]
• A metric that is strongly equivalent to a complete metric is also complete; the same is not true of equivalent metrics because homeomorphisms do not preserve completeness. For example, since ${\displaystyle (0,1)}$ and ${\displaystyle \mathbb {R} }$ are homeomorphic, the homeomorphism induces a metric on ${\displaystyle (0,1)}$ which is complete because ${\displaystyle \mathbb {R} }$ is, and generates the same topology as the usual one, yet ${\displaystyle (0,1)}$ with the usual metric is not complete, because the sequence ${\displaystyle (2^{-n})_{n\in \mathbb {N} }}$ is Cauchy but not convergent. (It is not Cauchy in the induced metric.)

Notes[]

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