I am studying topology on my own, and I am having trouble proving the following.

For a Hausdorff, connected, locally euclidean paracompact space $X$, there exists a countable basis for $X$.

I think if I possibly get any countable open cover of $X$ which consists of coordinate balls (that is, homeomorphic to open ball in euclidean space) since each coordinate balls are second countable. However, I cannot get a clue of how to find them. Am I doing right?


Start with an open cover of $X$ by coordinate balls, and let $\mathscr{U}$ be a locally finite open refinement. Define a relation $\sim$ on $\mathscr{U}$ as follows: for $U,V\in\mathscr{U}$, $U\sim V$ iff there is a finite subset $\{U_1,\dots,U_n\}$ of $\mathscr{U}$ such that $U=U_1$, $V=U_n$, and $U_k\cap U_{k+1}\ne\varnothing$ for $k=1,\dots,n-1$. Clearly $\sim$ is an equivalence relation on $\mathscr{U}$. For $U\in\mathscr{U}$ let $[U]$ denote the $\sim$-equivalence class of $U$.

  1. Show that for each $U\in\mathscr{U}$, $\bigcup[U]=\bigcup\{V\in\mathscr{U}:V\sim U\}$ is open in $X$. Conclude that for all $U,V\in\mathscr{U}$, $U\sim V$.

  2. Show that $[U]$ is countable.

Conclude by using the idea that you described in the question.

  • $\begingroup$ By 'for all $U,V\in\mathscr{U}$', you mean it has only one equivalence class? $\endgroup$ – Nemesis Aug 11 '12 at 9:38
  • $\begingroup$ @Nemesis: Yes, that’s exactly right. (Here’s where you use connectedness.) $\endgroup$ – Brian M. Scott Aug 11 '12 at 9:40
  • $\begingroup$ I'm sorry but I am confused again. How do we know that? I think it is true if we admit infinite subset of $\mathscr{U}$ but is it also true for finite subset?? $\endgroup$ – Nemesis Aug 11 '12 at 9:55
  • $\begingroup$ @Nemesis: How do we know what? $\endgroup$ – Brian M. Scott Aug 11 '12 at 12:33

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