Problem 1.2 (Disjoint union of uncountably many copies of real line)

Let $I$ be an uncountable index set and let ${\coprod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}ℝ$ be the disjoint union of uncountably many copies of $ℝ$. Recall that an open set of the disjoint union is declared to be open iff its intersection with each $ℝ$ is open in $ℝ$ (cf. page 604).

Hausdorff. By Proposition A.25, every disjoint union of Hausdorff spaces is Hausdorff. The assertion then follows since $ℝ$ is Hausdorff.

Locally Euclidean. Let $\bar{x}=(x,i)\in {\coprod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}ℝ$ be arbitrary. We must find an open set $\bar{U}\subseteq {\coprod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}ℝ$ and an open set $U\subseteq ℝ$ such that there is a homeomorphism $\phi :\bar{U}\to U$ between them. Take $U$ to be any open neighbourhood of $x$ in $ℝ$ (for instance $ℝ$ itself) and define the open set $\bar{U}:=\{(u,i)\mid u\in U\}$, $\bar{x}\in \bar{U}$. Then there is an obvious homeomorphism between the two sets.

Not second countable. Let $B$ be a basis for ${\coprod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}ℝ$. Any open subset $U\subseteq {\coprod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}ℝ$ is representable in terms of the basis sets

where we always have at least one $B\in B$ with property $B\subseteq U$. In particular, for each fiber $U={ℝ}_i:=\{(x,i)\mid x\in ℝ\}$, $B$ contains at least one open set $B_i\in B$ such that $B_i\subseteq {ℝ}_i$. But $B_i\cap B_j=\varnothing$ since ${ℝ}_i\cap {ℝ}_j=\varnothing .$ Thus, $B$ contains at least one open set for each fiber ${ℝ}_i$, i.e., $\#B\ge \mathit{\#I}$.