Why is uncountable union of $\mathbb{R}$ the same as this space Can anyone give an intuitive reasoning as to why the uncountable disjoint union of copies of $\mathbb{R}$ is the same as $\mathbb{R}$ with discrete topology product with $\mathbb{R}$ with the usual topology?
 A: It’s not just any uncountable union: it’s the union of $|\Bbb R|$ copies of $\Bbb R$.
For each $r\in\Bbb R$ let $X_r$ be a copy of $\Bbb R$ with the usual topology, and let $X$ be the disjoint union $\bigsqcup_{r\in\Bbb R}X_r$. Let $Y$ denote $\Bbb R$ with the discrete topology. For each $r\in\Bbb R$ define
$$h_r:X_r\to Y\times\Bbb R:x\mapsto\langle r,x\rangle\;;$$
check that $h_r$ is a homeomorphism from $Y_r$ to $\{r\}\times\Bbb R$. (In other words, it’s the natural homeomorphism between two copies of $\Bbb R$.
Now define $h:X\to Y\times\Bbb R$ by setting $h(x)=h_r(x)$ if $x\in X_r$; since $X$ is the disjoint union, for each $x\in X$ there is a unique $r\in\Bbb R$ such that $x\in X_r$, so $h$ is well-defined. (As a set of ordered pairs $h=\bigcup_{r\in\Bbb R}h_r$.) Now check that $h$ is a homeomorphism.
In more intuitive terms, for each $y\in Y$ the set $\{y\}\times\Bbb R$ is a clopen subset of $Y\times\Bbb R$, and it’s homeomorphic to $\Bbb R$. Similarly, for each $r\in\Bbb R$, the set $X_r$ is a clopen subset of $X$ that’s homeomorphic to $\Bbb R$. Thus, each of $X$ and $Y\times\Bbb R$ is in effect the disjoint union of $|\Bbb R|=|Y|$ copies of $\Bbb R$.
A: Your uncountable disjoint union must have exactly $c = |\mathbb{R}|$ summands. I would visualise it as the plane $\mathbb{R} \times \mathbb{R}$ thought of as the disjoint union of $c$ vertical lines. Give each vertical line the usual topology and accept any union of vertical open sets as open. You can view this either as the product of $\mathbb{R}$ with the discrete topology and $\mathbb{R}$ with the usual topology or as the disjoint union of the $c$ vertical lines each of which is homeomorphic to $\mathbb{R}$.
