数学分析原理 练习2.6

Let \(E'\) be the set of all limit points of a set \(E\). Prove that \(E'\) is closed. Prove that \(E\) and \(\overline{E}\) have the same limit points. (Recall that \(\overline{E}\) = \(E \cup E'\).) Do \(E\) and \(E'\) always have the same limit points?

Proof: (1) Let \(P\) be a limit point of \(E'\). For every positive real number \(r\), there exists a point \(Q\) such that \(Q \neq P\) and \(Q \in E' \cap N_{\frac{r}{2}(P)}\). Since \(Q \in E'\), there exists a point \(M\) such that \(M \neq Q\) and \(M \in E \cap N_{d(P, Q)}(Q)\), hence \(d(P, M) < \frac{r}{2} \cdot 2 = r\). Then we have \(M \in E \cap N_{r}(P)\), hence \(P \in E'\), and \(E'\) is closed.

(2) We've got \(E'\) as the set of limit points of \(E\). And the points in \(E'\) are also limit points of \(\overline{E}\). Suppose that there exists a point \(P\) such that \(P\) is a limit point of \(\overline{E}\) but not a limit point of \(E\), then \(P \notin E'\). Hence for \(P\), there exists a positive real number \(r\) such that \(N_r(P) \cap (E \backslash \{P\}) = \emptyset\). Because \(P\) is a limit point of \(\overline{E}\), we have that \(P\) is a limit point of \(E'\). Then there exists a point \(Q\) such that \(Q \neq P\) and \(Q \in E' \cap N_{\frac{r}{2}}(P)\). But since \(Q \in E'\), there exists a point \(M\) such that \(M \neq Q\) and \(M \in E \cap N_{d(P, Q)}(Q)\), hence \(d(P, M) < \frac{r}{2} \cdot 2 = r\), namely \(M \in N_r(P) \cap (E \backslash \{P\})\), which contradicts the assumption that \(N_r(P) \cap (E \backslash \{P\}) = \emptyset\). Hence \(E\) and \(\overline{E}\) have the same limit points.

(3) Let \(E = \{ \frac{1}{n} \mid n \in N_+ \}\) , then \(E' = {0}\), \((E')' = \emptyset\), hence \(E\) and \(E'\) don't always have the same limit points.