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I know $\mathbb{Q}$ is metric, and thus paracompact. But is there a way to show that every covering has a locally finite refinement?

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Can you do it for $\mathbb R$? The same refinement should work for the rationals. – Olivier Bégassat Jul 19 '12 at 20:06
@OlivierBégassat I can divide the real line in closed intervals, as big as I wish, extract a finite sub-covering. Refine it so that the covering of an interval don't cover "too many other intervals" and this should suffice. Is it correct? The refinement is possible thanks to the characteristics of the real line, I think (i.e. choping every open set at epsilon-distance from the interval he is covering). And paracompactness hold for every type of subspace... – Temitope.A Jul 19 '12 at 20:19
up vote 2 down vote accepted

Here is a direct proof that $\Bbb Q$ is paracompact.

Let $\mathscr{U}$ be an open cover of $\Bbb Q$; without loss of generality assume that the elements of $\mathscr{U}$ are open intervals. For $A\subseteq\Bbb Q$ let $\operatorname{st}(A,\mathscr{U})=\bigcup\{U\in\mathscr{U}:U\cap A\ne\varnothing\}$; write $\operatorname{st}(x,\mathscr{U})$ for $\operatorname{st}(\{x\},\mathscr{U})$. For $x\in\Bbb Q$ let $\operatorname{st}^1(x,\mathscr{U})=\operatorname{st}(x,\mathscr{U})$, and for $n\in\Bbb Z^+$ let $\operatorname{st}^{n+1}(x,\mathscr{U})=\operatorname{st}\big(\operatorname{st}^n(x,\mathscr{U}),\mathscr{U}\big)$. Finally, for $x\in\Bbb Q$ let $I(x)=\bigcup\{\operatorname{st}^n(x,\mathscr{U}):n\in\Bbb Z^+\}$. Clearly $I(x)$ is open and order-convex, and $y\in I(x)$ iff $x\in I(y)$, so $\mathscr{I}=\{I(x):x\in\Bbb Q\}$ is a partition of $\Bbb Q$ into clopen, order-convex sets.

Fix $x_0\in\Bbb Q$, let $L=(\leftarrow,x_0]\cap I(x_0)$, and let $R=[x_0,\to)\cap I(x_0)$. If $\operatorname{st}(x_0,\mathscr{U})\nsupseteq R$, there must be an $x_1\in R\cap\operatorname{st}^2(x_0,\mathscr{U})\setminus\operatorname{st}(x_0,\mathscr{U})$. Given $x_n\in R\cap\operatorname{st}^{n+1}(x_0,\mathscr{U})\setminus\operatorname{st}^n(x_0,\mathscr{U})$, if $\operatorname{st}^{n+1}(x_0,\mathscr{U})\nsupseteq R$, there is an $x_{n+1}\in R\cap\operatorname{st}^{n+2}(x_0,\mathscr{U})\setminus\operatorname{st}^{n+1}(x_0,\mathscr{U})$. There are now two cases.

  1. There is some $m_R\in\Bbb N$ such that $\operatorname{st}^{m_R+1}(x_0,\mathscr{U})\supseteq R$. Then for $n=0,\dots,m_R$ there are $U_n\in\mathscr{U}$ such that $x_n\in U_n$, and moreover $U_n\cap U_{n+1}\ne\varnothing$ if $n\ne m_R$; clearly $\{U_0,\dots,U_{m_R}\}$ covers $[x_0,x_{m_R}]$. If there is some $U_{m_R+1}\in\mathscr{U}$ such that $x_{m_R}\in U_{m_R+1}\supseteq R\cap[x_{m_R},\to)$, then $\{U_0,\dots,U_{m_R+1}\}$ is a finite and hence locally finite open cover of $R$ refining $\mathscr{U}$.

    Otherwise there must be a strictly increasing sequence $\langle z_n:n\in\Bbb N\rangle$ in $R\cap[x_{m_R},\to)$ and sets $V_n\in\mathscr{U}$ for $n\in\Bbb N$ such that $x_{m_R},z_n\in V_n$, $z_{n+1}\notin V_n$, and $R\cap[x_{m_R},\to)\subseteq\bigcup_{n\in\Bbb N}V_n$. Moreover, we may assume that $z_0>x_{m_R}$. Let $W_0=V_0\cap(x_{m_R},z_1)$, and for $n\in\Bbb Z^+$ let $W_n=V_n\cap(z_{n-1},z_{n+1})$. $\mathscr{W}=\{W_n:n\in\Bbb N\}$ refines $\mathscr{U}$ and covers $R\cap[x_{m_R},\to)$, and each member of $\mathscr{W}$ meets at most two other members of $\mathscr{W}$, so $\mathscr{W}\cup\{U_0,\dots,U_{m_R}\}$ is a locally finite open cover of $R$ refining $\mathscr{U}$.

  2. We end up with a sequence $\langle x_n:n\in\Bbb N\rangle$. Then for $n\in\Bbb N$ there are $U_n\in\mathscr{U}$ such that $x_n\in U_n$ and $U_n\cap U_{n+1}\ne\varnothing$. The choice of the points $x_n$ ensures that $U_m\cap U_n\ne\varnothing$ iff $|n-m|\le 1$, so $\{U_n:n\in\Bbb N\}$ is a locally finite open cover of $R$ refining $\mathscr{U}$.

Thus, in all cases we can find a locally finite open refinement $\mathscr{R}_R$ of $\mathscr{U}$ covering $R$. In similar fashion we can find a locally finite open refinement $\mathscr{R}_L$ of $\mathscr{U}$ covering $L$, and $\mathscr{R}=\mathscr{R}_L\cup\mathscr{R}_R$ is then a locally finite open refinement of $\mathscr{U}$ whose union is precisely $I(x_0)$.

Since we can do this for each $I\in\mathscr{I}$, $\mathscr{U}$ has a locally finite open refinement covering $\Bbb Q$, and $\Bbb Q$ is paracompact.

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