Counterexample to sum theorem for small inductive dimension ind?

Question

Whereas the small inductive dimension $\operatorname{ind}(X)$ agrees with the covering dimension $\operatorname{cov}(X)$ when $X$ is a separable metrizable space, that is no longer the case for more general topological spaces. And that raises the issue as to why the covering dimension is preferred to the small inductive dimension for such more general spaces.

My question: What is some standard property of $\operatorname{ind}$ — for example, the sum (or addition) theorem — for separable metrizable spaces that no longer holds more generally, and what is an example of such a space?

I've tried to dig this out of Engelking's "Dimension Theory" book but so far have not been successful.

Answer 1

Theorem (Roy): There is a metric space $X$ with $ind\;X=0$ and $Ind\;X=1$. $\square$

For this space there are disjoint closed sets $A,B\subseteq X$ such that $X$ cannot be written $X=U\cup V$ with $U,V$ open, $A\subset U$, $B\subset V$, and $U\cap V=\emptyset$ (i.e. the empty set is not a partition between $A,B$). Choose a continuous function $f:X\rightarrow I$ with $f^{-1}(0)=A$ and $f^{-1}(1)=B$. Now take a point $a\not\in X\setminus A$ and equip $$\widehat X=(X\setminus A)\cup\{p\}$$ with the topology generated by the collection of all open sets in $X\setminus A$ together with the sets $\widehat X\setminus f^{-1}[1/n,1]$ for $n\geq1$. Note that $X\setminus A\cong\widehat X\setminus\{a\}$.

Then $B$ is a a closed subset of $\widehat X$ and $a\not\in B$. Since any partition in $\widehat X$ between $a$ and $B$ gives a partition in $X$ between $A$ and $B$, we see that $ind\;\widehat X>0$. In fact a brief argument yields $ind\;\widehat X=1$.

Now write $\widehat X=E\cup F$, where $$E=\{a\}\cup\bigcup_\mathbb{N}f^{-1}[1/(2n+1),1/2n],\qquad F=\{a\}\cup\bigcup f^{-1}[1/2n,1/(2n+1)].$$ Then $E,F$ are closed in $\widehat X$ and $$ind\;E=0=ind\;F.$$ Thus we have a space with small inductive dimension $1$ which is the union of two of its zero-dimensional closed subspaces. Thus we witness the failure of the sum theorem for $ind$. However, this would be none too impressive without the following.

It's easy to see that $\widehat X$ is regular $T_1$, and a subsequent argument with the Nagata-Smirnov metrisation theorem yields

$\widehat X$ is metrisable.

Consequently:

The sum theorem for small inductive dimension fails in metrisable spaces. $\square$

The construction above is from van Douwen's paper. While P. Roy's example of a metric space with noncoinciding $ind$ and $Ind$ is historically the first, a simpler example was provided by Kulesza, with further simplification coming from Levin.