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.



Covering dimension of uncountable union of compact spaces
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.


