Example of covering dimension of a finite space

Question

Consider $X=\lbrace 1,2,3\rbrace\newcommand{\dim}{\operatorname{dim}}$ endowed with the topology $T$ generated by $\lbrace \lbrace 1,2\rbrace,\lbrace 2,3\rbrace\rbrace$. It is claimed (in Exercise 2.20, Dimension Theory, M. Charalambous) that the covering dimension $\dim(X)$ is $1$. According to Proposition 2.10 of the same book that implies that any open cover of $X$ consisting of three sets has an open shrinking $\lbrace B_1, B_2, B_3\rbrace$ with $B_1\cap B_2\cap B_3 = \emptyset$. I am sure I am overlooking something very obvious here, but is it not the case that the intersection of any collection of non-empty elements of $T$ is non-empty (and wouldn't this imply that the statement $\dim(X)=1$ is false)?

Answer 1

The basic idea of the (Lebesgue) covering dimension is that if you a space is covered by open sets, each point in the space will be contained in one or more of those open sets. The covering dimension, in some sense, looks at the most efficient covers (i.e. the covers which have the least redundancy) and looks for the points which are most redundantly covered. More formally:

Definition: Let $X$ be a topological space, and let $\mathscr{U}=\{U_{\alpha}\}$ be an open cover of $X$. For each point $x \in X$, let $n_x$ be the number of sets $U_\alpha\in \mathscr{U}$ which contain $x$. That is, $$ n_x = \bigl| \{ \alpha : x \in U_{\alpha} \} \bigr|. $$ The order of the cover $\mathscr{U}$ is $$\operatorname{ord}(\mathscr{U}) = \max_{x\in X} n_x.$$

The covering dimension $\newcommand{\dim}{\operatorname{dim}}\dim(X)$ is the smallest number $n$ such that every open cover of $X$ has a refinement of order $n+1$. This is equivalent to the statement that every open cover has a refinement $\mathscr{V} = \{ V_{\beta}\}$ such that $$ V_{\beta_1} \cap V_{\beta_2} \cap \dotsb \cap V_{\beta_{n+1}} \cap V_{\beta_{n+2}} = \varnothing $$ for any distinct $V_{\beta_j}$ with $j=1,2,\dotsc, n+2$ (this is a restatement of Proposition 2.10).

The space in the question is $X = \{a,b,c\}$ (I am using letters here to avoid confusion with numbers below), the topology is $$ \mathscr{T} = \{\varnothing, \{b\}, \{a,b\}, \{b,c\}, X \}. $$ Any open cover of $X$ must contain either $\{a,b\}$ or $X$ (in order to cover $a$), and must contain either $\{b,c\}$ or $X$ (in order to cover $c$).

• Any cover which contains $X$ can be refined to the cover $\{X\}$, which has order $1$.

• Any cover which does not contain $X$ must contain both $\{a,b\}$ and $\{b,c\}$. Any such cover can be refined to $\{ \{a,b\}, \{b,c\} \}$, which has order $2$ (since $b$ is contained in both sets).

Therefore the smallest $n$ such that every open cover has a refinement of order $n+1$ is $1$ (since every open cover has a refinement of order at most $2 = 1+1$). Therefore $\dim(X) = 1$ (as claimed).

With respect to Proposition 2.10, notice that every open cover of $X$ has a refinement of at most two sets. It is therefore vacuously true that if $V_1$, $V_2$, and $V_3$ are distinct sets in any of these refinements, it must be the case that $$ V_1 \cap V_2 \cap V_3 = \varnothing.$$ Since every open cover can be refined to either a one- or two-element cover, any statement which assumes the exitence of three-element covers is vacuously true.

On the other hand, the (refined) open cover $\mathscr{V} = \{ \{a,b\}, \{b,c\} \}$ has the property that if $V_1, V_2 \in \mathscr{V}$, then $$ V_1 \cap V_2 = \{b\}, $$ which is nonempty. Per Proposition 2.10, this implies that $\dim(X) = 1$.