# Metric dimension

A numerical characteristic of a compact set, defined in terms of coverings of "standard measure" , the number of which defines the metric dimension. Let $ F $
be a compact set, and let $ N _ {F} ( \epsilon ) $
be the minimal number of sets with diameter not exceeding $ \epsilon $
that are needed in order to cover $ F $.
This function, depending on the metric in $ F $,
takes integer values for all $ \epsilon > 0 $,
and increases without bound as $ \epsilon \rightarrow 0 $;
it is called the volume function of $ F $.
The metric order of the compact set $ F $
is the number

$$ k = fnnme \underline{lim} \ \left ( - \frac{ \mathop{\rm ln} N _ {F} ( \epsilon ) }{ \mathop{\rm ln} \epsilon } \right ) . $$

This quantity is not yet a topological invariant. Thus, the metric order of a curve in the sense of Jordan (cf. Line (curve)) with the Euclidean metric is equal to 1, but for a curve in the sense of Jordan passing through a perfect totally-disconnected set in $ \mathbf R ^ {n} $ of positive measure, this value is equal to $ n $. However, the greatest lower bound of the metric orders for all metrics on $ F $( called the metric dimension) is equal to the Lebesgue dimension (the Pontryagin–Shnirel'man theorem, 1931, see [1]).

#### References

[1] | W. Hurevicz, G. Wallman, "Dimension theory" , Princeton Univ. Press (1948) |

#### Comments

Metric dimension makes sense for non-compact separable metrizable spaces (using totally bounded metrics), and the Pontryagin–Shnirel'man theorem extends to them. This was shown by E. Szpilrajn-Marczewski. See [a2].

There are also other types of metric-dependent dimension functions.

One example is the Hausdorff dimension.

Another example is obtained by modifying the definition of the covering dimension $ \mathop{\rm dim} $( see Dimension): If $ ( X , d ) $ is a metric space, one defines $ \mu \mathop{\rm dim} ( X , d ) $ by $ \mu \mathop{\rm dim} ( X , d ) \leq n $ if and only if for every $ \epsilon > 0 $ there is an open covering $ \mathfrak U $ of $ X $ with $ \textrm{ mesh } \mathfrak U \leq n + 1 $ and $ \mathop{\rm ord} \mathfrak U < \epsilon $. Here $ \textrm{ mesh } \mathfrak U = \sup \{ { \mathop{\rm diam} ( U) } : {U \in \mathfrak U } \} $ and $ \mathop{\rm ord} \mathfrak U \leq n + 1 $ means that no point of $ X $ is an element of more than $ n + 1 $ elements of $ \mathfrak U $. One can show that $ \mu \mathop{\rm dim} ( X , d ) \leq \mathop{\rm dim} X \leq 2 \mu \mathop{\rm dim} ( X , d ) $ and that these inequalities are best possible, see [a1].

#### References

[a1] | R. Engelking, "Dimension theory" , North-Holland & PWN (1978) pp. 19; 50 |

[a2] | J.-I. Nagata, "Modern dimension theory" , Interscience (1965) |

**How to Cite This Entry:**

Metric dimension.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Metric_dimension&oldid=47828