Covering dimension of uncountable union of compact spaces

Question

It is well-known that the (covering) dimension of countable union of compact spaces is the superium dimension of these spaces. I would like to understand certain uncountable union of compact spaces as follows. Let $b>a$. Let $\{X_r\}_{r\in [a,b]}$ be a family of compact spaces with index $r\in [a,b]$. Suppose that for each $r\in (a,b)$, for any $r_k\in [a,b]$ with $r_k\to r$ and $x_{r_k}\in X_{r_k}$, the limit points of $\{x_{r_k}\}_{k\in \mathbb{N}}$ belong to $X_r$. The question is what the covering dimension of $\cup_{r\in [a,b]}X_r$ is.

The trivial example is the product space $Y\times [a,b]$ and the dimension is then dim$(Y)+1$. I guess the dimension of $\cup_{r\in [a,b]}X_r$ is bounded above by $\sup_{r\in [a,b]} \dim X_r+1$. But I have no idea how to show it.

Answer 1

There is no upper bound for dimension in this case.

Just consider any surjective continuous map $f:[0,1]\to [0,1]^\omega$ onto the Hilbert cube. Such a map exists since the Hilbert cube is a Peano continuum. Then for every $r\in[0,1]$ put $X_r=\{f(r)\}$. It is clear that $\dim(X_r)=0$ for all $r\in[0,1]$ but $\dim([0,1]^\omega)=\infty$. The continuity of $f$ implies that the family $(X_r)_{r\in[0,1]}$ has the continuity property you require.

Answer 2

If you don't mind the correspondence $r\mapsto X_r$ being non-measurable, then you can put $[a,b]$ in correspondence with any other uncountable set, including $[a,b]^n$, and thereby get a set of arbitrary dimension.