Each component of $X$ is second-countable This problem relates paracompactness to connected components in a space $X$, which I find it fairly hard to do. Hope some one can help me solve this or give me some useful hints. Thanks.


If $X$ is a Hausdorff space that is locally compact and paracompact, then each component of $X$ is second-countable


 A: This is false. According to Spacebook, the following are compact (hence paracompact and locally compact), connected Hausdorff spaces that are not second-countable:


*

*Alexandroff Square

*Helly Space

*Lexicographic Ordering on the Unit Square

*The Extended Long Line

*Uncountable Cartesian Product of Unit Interval


The last seems easiest to deal with. Let $I=[0,1]$ and let $X=I^S$, where $S$ is uncountable. Then $X$ is compact because it is a product of compact spaces, connected because it is a product of connected spaces, and Hausdorff because it is a product of Hausdorff spaces. Suppose $\mathcal B$ is a countable basis of $X$. Then because a countable union of finite sets is countable, there is some $i\in S$ such that every element of $\mathcal B$ has a basic (in the sense of the usual basis for the product topology) subset whose $i$th projection is $I$, contradicting the claim that $\mathcal B$ is a basis.
A: Concerning Munkres, problem 41-10.  Take any 
uncountable Cartesian product of [0,1]; it is compact + Hausdorff 
as noted (by Kelley pp. 92,143).  Hence, paracompact.
It is connected by Kelley, problem 3.O; and, non-second countable 
by Kelley problem 3.M.  Since second-countable implies 
first-countable, non-second countability also follows from 
Kelley p.92 (theorem 6).     
