Let \(\{U_1,U_2,\ldots\}\) be a cover of \(\mathbb R^n\) by open sets. Prove that there is a cover \(\{V_1,V_2,\ldots\}\) such that
\(V_j\subset U_j\) for each \(j\);
each compact subset of \(\mathbb R^n\) is disjoint from all but finitely many of the \(V_j\).
Solution
Define \[ V_1=U_1 \] and, for \(j\geq2\), \[ V_j = U_j\setminus\bigcup_{i<j}U_i. \]
1. For every \(j\geq1\),\[ V_j\subseteq U_j. \]
This follows immediately from the definition of \(V_j\).
2. The family \(\{V_1,V_2,\ldots\}\) covers \(\RR^n\).
Let \(x\in\RR^n\). Since \(\{U_1,U_2,\ldots\}\) covers \(\RR^n\), the set \[ \{j\geq1:x\in U_j\} \] is nonempty. Let \(j\) be its least element. Then \(x\in U_j\) and \(x\notin U_i\) for every \(i<j\). Hence, by definition, \[ x\in V_j. \] Thus every point of \(\RR^n\) belongs to some \(V_j\).
3. If \(K\subseteq\RR^n\) is compact, then there is an integer \(M\) such that\[ K\cap V_j=\varnothing \]for every \(j>M\).
The sets \(U_j\cap K\) form an open cover of the compact space \(K\). Hence there are indices \[ j_1,\ldots,j_r \] such that \[ K\subseteq U_{j_1}\cup\cdots\cup U_{j_r}. \] Let \[ M=\max\{j_1,\ldots,j_r\}. \]
Suppose \(j>M\) and \(x\in K\). Then \(x\in U_{j_s}\) for some \(s\), with \[ j_s\leq M<j. \] Therefore \(x\in\bigcup_{i<j}U_i\), so the definition of \(V_j\) implies \(x\notin V_j\). Thus \(K\cap V_j=\varnothing\) for every \(j>M\).
4. Each compact subset of \(\RR^n\) is disjoint from all but finitely many of the \(V_j\).
For a compact set \(K\), step 3 shows that only \[ V_1,\ldots,V_M \] can meet \(K\).
5. Q.E.D.
Step 1 gives property (1), step 2 shows that the \(V_j\) form a cover, and step 4 gives property (2).