For each $t \in R$, let $E_t$ be a subset of $R$. Suppose that if $s<t$ then $E_s$ is a proper subset of $E_t$. $$\bigcup_{t\in R} E_t$$ Is countable. How?
I see that the union runs over R which is an uncountable set. Each of the set $E_t$ contains infinitely many subsets of R. It seems that this union equals R itself which is uncountable. I Don't get how this union become countable??
Here's a very simple example. For each $t\in \mathbb{R}$, let $E_t = [0, t]\cap \mathbb{Q}$. From the density of rationals, we know that $E_s \subset E_t$ when $s < t$ and their union is countable since it's a subset of $\mathbb{Q}$.