Topological Dimension

Abstract

In this chapter, we introduce the topological dimension $\dim (X)$, also called the covering dimension, of a topological space X. The definition of $\dim (X)$ involves the combinatorics of the finite open covers of X. We establish some basic properties of the topological dimension and give first examples of topological spaces for which it can be explicitly computed.

Appendices

Notes

Covering dimension is one among many other invariants that were introduced by mathematicians all along the twentieth century in order to give a precise definition for the intuitive notion of dimension in the category of topological spaces. The branch of general topology that studies these invariants is known as “dimension theory”. This is also the title of the most famous monograph devoted to the subject, namely the book by Hurewicz and Wallman [50], which was first published in 1941. There are several other excellent books entirely devoted to dimension theory, e.g., [9, 33, 79, 80, 86]. The reader interested in the history of the developments of dimension theory is referred to [7, 8, 33, 56, 57, 92].

The covering dimension $\dim (X)$ was introduced by Čech [111]. Its definition was directly inspired by a topological characterization of the dimension of the n-cube $[0,1]^n$ formulated by Lebesgue [66, 67] (see Lemma 3.5.2).

The idea of using induction for defining the dimension of a topological space was popularized by Poincaré (see for example [89, p. 73]). This approach led in particular to the definition of the small inductive dimension ${{\mathrm{ind}}}(X)$ and of the large inductive dimension ${{\mathrm{Ind}}}(X)$ . The small inductive dimension ${{\mathrm{ind}}}(X) \in \{-1\}\cup \mathbb {N}\cup \{\infty \}$ of a topological space X, also called the Menger-Urysohn dimension , is inductively defined by the following conditions: (1) ${{\mathrm{ind}}}(X) = -1$ if and only if $X = \varnothing$, (2) ${{\mathrm{ind}}}(X) \le n$ if and only if X admits a base of open subsets $\mathcal {B}$ such that ${{\mathrm{ind}}}(\overline{U}{\setminus }U) \le n -1$ for all $U \in \mathcal {B}$. The large inductive dimension ${{\mathrm{Ind}}}(X) \in \{-1\} \cup \mathbb {N}\cup \{\infty \}$, also called the Brouwer-Čech dimension , is defined by: (1) ${{\mathrm{Ind}}}(X) = -1$ if and only if $X = \varnothing$, (2) ${{\mathrm{Ind}}}(X) \le n$ if and only if, for every pair of disjoint closed subsets F and G of X, there exist disjoint open subsets U and V of X such that $F \subset U$, $G \subset V$ and ${{\mathrm{Ind}}}(X{\setminus }(U \cup V)) \le n - 1$. An easy induction shows that every accessible space X satisfies ${{\mathrm{ind}}}(X) \le {{\mathrm{Ind}}}(X)$. Urysohn’s theorem asserts that $\dim (X) = {{\mathrm{ind}}}(X) = {{\mathrm{Ind}}}(X)$ for every separable metrizable space X (see for example [50]). Katětov [54, 55] and independently Morita [78] proved that one has ${{\mathrm{ind}}}(X) \le \dim (X) = {{\mathrm{Ind}}}(X)$ for every metrizable space X. The question whether every metrizable space X satisfies $\dim (X) = {{\mathrm{ind}}}(X)$ remained open for a long time (cf. [7, p. 3]). Finally, it was answered in the negative in [96, 98] by Roy who provided an example of a metrizable space X with ${{\mathrm{ind}}}(X) = 0$ and $\dim (X) = 1$. It was shown by Pasynkov [84] that the equalities $\dim (X) = {{\mathrm{ind}}}(X) = {{\mathrm{Ind}}}(X)$ remain true when X is the underlying space of a locally compact Hausdorff topological group. A theorem of Alexandroff [6] asserts that every compact Hausdorff space X satisfies $\dim (X) \le {{\mathrm{ind}}}(X) \le {{\mathrm{Ind}}}(X)$. Filippov [36] gave an example of a compact Hausdorff space X with $\dim (X) = {{\mathrm{ind}}}(X) = 2$ and ${{\mathrm{Ind}}}(X) = 3$. When X is a normal space, one always has $\dim (X) \le {{\mathrm{Ind}}}(X)$ (see for example [33, 79, 86]) but the inequality may be strict. In [79, p. 114], Nagami gives an example of a normal Hausdorff space X such that ${{\mathrm{ind}}}(X) = 0$, $\dim (X) = 1$, and ${{\mathrm{Ind}}}(X) = 2$.

The notion of a normal space goes back to the work of Vietoris [112] (see [94, p. 1233]) and Tietze [105]. However, the main results about general properties of normal spaces are due to Urysohn [109].

Theorems 1.7.1 and 1.8.3 were obtained by Čech [111].

In [110], Čech introduced the following definition. A topological space X is called perfectly normal if X is normal and every open subset of X is an $F_\sigma$-set. For example, every metrizable space is perfectly normal by Proposition 1.5.3 and Lemma 1.8.2. It turns out that every perfectly normal space X is completely normal , that is, every subset $Y \subset X$ is normal (see [18 exerc. 7, 9 and 11 p. IX. 102–103], [64, 111]). Thus, the proof of Theorem 1.8.3 can be extended to perfectly normal spaces. Consequently, every subset Y of a perfectly normal space X satisfies $\dim (Y) \le \dim (X)$ [111, par. 28]. Alexandroff (see [8, p. 28]) conjectured that $\dim (Y) \le \dim (X)$ whenever Y is a normal subspace of a normal space X. This conjecture was disproved by Dowker [30] who gave an example of a normal Hausdorff space X with $\dim (X) = 0$ containing a normal open subset Y such that $\dim (Y) = 1$.

The idea of finding a homological interpretation of the dimension of a topological space was developed in the work of Alexandroff [4, 5] in the late 1920s. It subsequently led to the investigation of various notions of homological and cohomological dimension (see the books [9, 50, 79, Appendix by Kodama] and the survey papers [31, 32, 65]). Given a topological space X and an abelian group G, the cohomological dimension ${{\mathrm{cdim}}}_G(X)$ is defined as being the smallest integer $n \ge - 1$ such that $\check{\mathrm {H}}^{n + 1}(X,A;G) = 0$ for all closed subsets $A \subset X$, or $\infty$ if there is no such integers. Here $\check{\mathrm {H}}^*$ denotes relative Čech cohomology. It was shown by Alexandroff that every compact metrizable space X with $\dim (X) < \infty$ satisfies $\dim (X) = {{\mathrm{cdim}}}_\mathbb {Z}(X)$. In the first International Topological Conference held in Moscow in September 1935, Alexandroff asked if this equality remains true in the case when $\dim (X) = \infty$. This question was answered in the negative in the late 1980s by Dranishnikov [31] who proved, by using methods from K-theory, the existence of a compact metrizable space X with topological dimension $\dim (X) = \infty$ and integral cohomological dimension ${{\mathrm{cdim}}}_\mathbb {Z}(X) = 3$.

Fig. 1.3

Construction of a Sierpinski triangle

Exercises

 

• 1.1 Show that every finite topological space X satisfies $\dim (X) < \infty$.

• 1.2 Show that the topological space X described in Example 1.1.11 is connected but not accessible.

• 1.3 Let X be an infinite set and $x_0 \in X$. The set X is equipped with the topology for which the open subsets are $\varnothing$ and all the subsets of X containing $x_0$. Show that $\dim (X) = \infty$.

• 1.4 Let X be the topological space whose underlying set is $\mathbb {R}$ and whose open subsets are $\varnothing$, $\mathbb {R}$, and all the intervals of the form $(a,+\infty )$, where $a \in \mathbb {R}$. Show that X is connected and that $\dim (X) = 0$.

• 1.5 Let X be a non-empty set and $\pi$ a partition of X. The set X is equipped with the topology for which the open subsets are $\varnothing$ and all the subsets of X that can be written as a union of elements of $\pi$. Show that $\dim (X) = 0$.

• 1.6 Show that every finite accessible topological space is discrete.

• 1.7 Let X and Y be topological spaces with Y accessible and non-empty. Show that $\dim (X \times Y) \ge \dim (X)$.

• 1.8 Let $\alpha = \{U,V\}$ be the open cover of $\mathbb {R}$ defined by

  $$ U := \bigcup _{n \in \mathbb {Z}} (n,n+1) \text { and } V := \bigcup _{n \in \mathbb {Z}} (n - \frac{1}{\vert n \vert + 1}, n + \frac{1}{\vert n \vert + 1}). $$

  Show that $\alpha$ admits no Lebesgue numbers, i.e., for every $\lambda > 0$, there exists a subset $Y \subset \mathbb {R}$ such that ${{\mathrm{diam}}}(Y) \le \lambda$ that is contained in no element of $\alpha$.

• 1.9 Let $\mathbb {S}^1 := \{(x,y) \in \mathbb {R}^2 \mid \; x^2 + y^2 = 1\}$ denote the unit circle in $\mathbb {R}^2$. Show that $\dim (\mathbb {S}^1) = 1$.

• 1.10 Construct, for every $\varepsilon > 0$, an open cover $\alpha$ of $\mathbb {R}^2$ with order ${{\mathrm{ord}}}(\alpha ) = 2$ and Euclidean mesh ${{\mathrm{mesh}}}(\alpha ) \le \varepsilon$. Deduce that every compact subset $X \subset \mathbb {R}^2$ satisfies $\dim (X) \le 2$.

• 1.11 Deduce from the previous exercise and the countable union theorem (Theorem 1.7.1) that $\dim (\mathbb {R}^2) \le 2$. More generally, show that $\dim (\mathbb {R}^n) \le n$ for every $n \in \mathbb {N}$. (The fact that $\dim (\mathbb {R}^n) = n$ will be proved in Corollary 3.5.7 below).

• 1.12 Deduce from the previous exercise that one has $\dim (Y) \le n$ for every subset $Y \subset \mathbb {R}^n$.

• 1.13 Show that the topological space X of Example 1.1.11 is normal if and only if $n = 0$.

• 1.14 Let X be a normal Hausdorff space. Show that any two distinct points of X admit disjoint closed neighborhoods.

• 1.15 Let $\mathcal {T}$ denote the set consisting of all subsets of $\mathbb {R}$ of the form $U{\setminus }C$, where U is an open subset of $\mathbb {R}$ for the usual topology and $C \subset U$ is a countable subset.

  • (a) Show that $\mathcal {T}$ is the set of open sets of a topology on $\mathbb {R}$. Let X denote the topological space whose underlying set is $\mathbb {R}$ and whose set of open subsets is $\mathcal {T}$.

  • (b) Show that any two distinct points of X admit disjoint closed neighborhoods.

  • (c) Show that X is not normal. Hint: consider the sets $A := \{0\}$ and $B := \{1/n \mid \; n \ge 1\}$.

• 1.16 Let $X \subset \mathbb {R}$. Show that the following conditions are equivalent: (1) $\dim (X) = 1$, (2) X contains a subset homeomorphic to the unit segment [0, 1], (3) the interior of X in $\mathbb {R}$ is not empty.

• 1.17 Let $Y := [0,1]$ denote the unit segment in $\mathbb {R}$ and let $X := Y \cup \{x_0\}$ be the set obtained from Y by adjoining an element $x_0 \notin Y$. Equip X with the topology for which the open subsets are X and all the subsets $\Omega \subset Y$ such that $\Omega$ is an open subset for the usual topology on Y.

  • (a) Show that X is not accessible.

  • (b) Show that X is compact and connected.

  • (c) Show that every subspace of X is normal.

  • (d) Show that $\dim (X) = 0$ but $\dim (Y) = 1$.

• 1.18 Let X be a compact metric space. Let $(\alpha _k)_{k \in \mathbb {N}}$ be a sequence of finite open covers of X such that $\lim _{k \rightarrow \infty } {{\mathrm{mesh}}}(\alpha _k) = 0$. Show that $\dim (X) = \lim _{k \rightarrow \infty } D(\alpha _k)$.

• 1.19 Let T be a triangle in the Euclidean plane (i.e., the convex hull of three non-collinear points in $\mathbb {R}^2$). The middle-triangle of T is the interior in $\mathbb {R}^2$ of the triangle whose vertices are the midpoints of the sides of T. We inductively construct a decreasing sequence $(K_n)_{n \in \mathbb {N}}$ of subsets of $\mathbb {R}^2$ in the following way. We start by setting $K_0 := T$. Then we define $K_1$ as being the set obtained from $K_0 = T$ by removing its middle-triangle. Thus, $K_1$ is the union of three triangles that are the images of T by the homotheties of ratio 1 / 2 centered at each of the vertices of T. More generally, assuming that $K_n$ has already been constructed and is the union of $3^n$ triangles that are all pairwise disjoint except at some of their vertices, we define $K_{n + 1}$ as being the set obtained from $K_n$ by removing all the middle-triangles of these $3^n$ triangles. The Sierpinski triangle associated with T is the set $S := \bigcap _{n \in \mathbb {N}} K_n$ (see Fig. 1.3).

  • (a) Show that the homeomorphism type of S does not depend on the initial choice of T.

  • (b) Show that S is a connected compact subset of $\mathbb {R}^2$.

  • (c) Show that $\dim (S) = 1$. Hint: observe that the construction yields a sequence $(\beta _n)_{n \in \mathbb {N}}$ of finite closed covers of S with ${{\mathrm{ord}}}(\beta _n) = 1$ and Euclidean mesh

    $$ {{\mathrm{mesh}}}(\beta _n) = \frac{{{\mathrm{diam}}}(T)}{2^n} $$

    and then apply Corollary 1.6.7 to get $\dim (S) \le 1$.