if f is continuous in a sequence of sets is it continuous in the union of them?

Question

Let f be a continuous function on each sequence of compact sets {$K_i$}. Does it have to be continuous in the finite union of these sets $K= \bigcup\limits_{i=1}^n K_i $ ?

Answer 1

This is the notion of a space $X$ being coherent with respect to some cover $S_\alpha$ of $X$. We say that $X$ is coherent with respect to $S_\alpha$ if $f$ is continuous on $X$ as soon as $f$ is continuous when restricted to each $S_\alpha$.

One particular case when $X$ is always coherent with respect to $S_\alpha$ is when $S_\alpha$ is a locally finite closed cover. So in particular, if $X$ is Hausdorff, then any compact subset of $X$ is closed, and a finite compact cover $K_i$ of $X$ will in particular be a locally finite closed cover, and so $X$ will be coherent with respect to $K_i$.

Answer 2

Let $S$ be the Sierpiński space, the set $\{0,1\}$ whose open sets are $\emptyset$, $\{1\}$, and $S$. You can think of it this way: $0$ is infinitely close to $1$, but not vice versa.

Consider the function $f:S\to S$ defined by $f(x)=1-x$. It's continuous on the compact sets $K_1=\{0\}$ and $K_2=\{1\}$, but it's not continuous on $\bigcup\limits_{i=1}^2K_i=S$. (You can check that it's not continuous using the standard definition. Alternatively, we see that $0$ is infinitely close to $1$, but $f(0)=1$ is not infinitely close to $f(1)=0$.)