Solution for Exercise 22-26 from Chapter 02 in Baby Rudin

Books/BabyRudin/Chapter02/ex22.cecilia.tex

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+\subsection*{Exercise 22 (Cecilia)}
+Consider the set $ E \subset \mathbb{R}^n $ of points having rational coordinates.
+
+$ E $ is countable because it is a finite Cartesian product of countable sets.
+
+For any point $ p = (p_1, p_2, \cdots, p_n )$ is $ \mathbb{R}^n $, for any neighborhood $ N_p(r) $ of $ p $, we claim that there exists a point $ q \in E $ such that $ q \in N_p(r) $.
+
+Indeed, by the Archimedean property, there exists a rational number $ q_i $ such that $ p_i - r < q_i < p_i $ all $ i = 1, 2, \cdots, n $, and so $ q = (q_1, q_2, \cdots, q_n) \in N_p(r) $.
+
+Therefore $ p $ is a limit point of $ E $, and so $ E $ is dense in $ \mathbb{R}^n $.
+
+$ E $ is both countable and dense, and so $ \mathbb{R}^n $ is separable.

Books/BabyRudin/Chapter02/ex23.cecilia.tex

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+\subsection*{Exercise 23 (Cecilia)}
+For a separable metric space $ X $, let $ E $ be a countable dense subset of $ X $. We claim that the set of all neighborhood $ B $ with centers in $ E $ having rational coordinates is a base.
+
+The set $ B $ is countable is obvious, we can map 1:1 from $ \mathbb{Q} \times E $ to $ B $ in the obvious way, and the former is the Cartesian product of two countable sets.
+
+For any point $ x \in G \subset X $ where $ G $ is open, $ x $ must be an interior point, so there exists $ N_r(x) $ such that $ N_r(x) \subset G $. Since $ E $ is dense, $ x $ is either a point of $ E $ or is a limit point of $ E $, in the former case, $ N_r(x) $ is a neighborhood of $ x $ with rational coordinates and with a center in $ E $, therefore $ N_r(x) $ is the $ V_{\alpha} $ we needed.
+
+Otherwise $ x $ is a limit point of $ E $. By the Archimedean property, there exists a rational number $ q $ such that $ 0 < q < \frac{r}{3} $. Since $ x $ is a limit point of $ E $, there exist a point in $ E $ such that $ e \in N_q(x) $, consider the neighborhood $ N_q(e) $. First of all $ x \in N_q(e) $, and for each point $ f \in N_q(e) $, $ d(x, f) < d(x, e) + d(e, f) < q + q < \frac{2r}{3} < r $, therefore $ N_q(e) \subset N_r(x) \subset G $, and so $ N_q(e) $ is the $ V_{\alpha} $ we needed.
+
+In either case, we have shown that for any point $ x \in G $, there exists a neighborhood $ V_{\alpha} $ such that $ x \in V_{\alpha} \subset G $, so $ B $ is a base, and it is a countable base.

Books/BabyRudin/Chapter02/ex24.cecilia.tex

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+\subsection*{Exercise 24 (Cecilia)}
+Suppose the process described in the hint doesn't stop, there exists an infinite sequence $ S_{\delta} $ of points $ x_1, x_2, \cdots $ such that there pairwise distances are at least $ \delta $. Let's say that (for the sake of contradiction) this subset has a limit point $ p $. Consider a neighborhood $ N_{\frac{\delta}{3}}(p) $, the neighhood cannot contain more than one point of $ S_{\delta} $, but then that contradicts theorem 2.20.
+
+Therefore the process must stop, and so the set $ S_{\delta} $ is finite for any $ \delta > 0 $.
+
+Because the process stopped, the neighborhoods $ N_{\delta}(s) $ for all $ s \in S_{\delta} $ covers $ X $. Therefore for every point $ x \in X $, there exists a point $ s \in S_{\delta} $ such that $ x \in N_{\delta}(s) $.
+
+Consider the set $ S = S_1 \cup S_{\frac{1}{2}} \cup S_{\frac{1}{3}} \cup \cdots $, the set is countable because it is a countable union of finite sets. For any point $ x \in X $, and for any neighborhood $ N_r(x) $ where $ r > 0 $, there exists a rational number $ 0 < \frac{p}{q} < r $ by the Archimedean property, and so $ 0 < \frac{1}{q} < r $. Therefore there exists a point $ s \in S_{\frac{1}{q}} $ such that $ x \in N_{\frac{1}{q}}(s) \subset N_r(x) $, so every neighborhood of $ x $ contains a point $ s \in S $, in other words, it is a limit point of $ S $, and so $ S $ is dense in $ X $.
+
+Therefore $ X $ is separable.

Books/BabyRudin/Chapter02/ex25.cecilia.tex

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+\subsection*{Exercise 25 (Cecilia)}
+Since $ K $ is compact, consider these open covers of the space as follows:
+
+For each $ n \in \mathbb{N} $, define the open cover of the space as $ \{ N_{\frac{1}{n}}(k) \mid k \in K \} $.
+
+Each of these open cover has a finite subcover by compactness of $ K $, so the union of these finite subcovers is a countable set.
+
+We claim that the centers of these finite subcovers $ C $ form a countable dense subset of $ K $.
+
+To see this, consider any point $ k \in K $ and any neighborhood $ N_r(k) $ where $ r > 0 $, there exists a rational number $ 0 < \frac{p}{q} < r $ by the Archimedean property, and so $ 0 < \frac{1}{q} < r $. The point $ k $ must be covered by an open set in the finite subcover of radius $ \frac{1}{3q} $, and that must be a subset of $ N_r(k) $. In other words, there exists a point $ c \in C $ such that $ c \in N_r(k) $, meaning $ k $ is a limit point of $ C $, and so $ C $ is dense in $ K $.
+
+Therefore $ K $ is separable.
+
+We claim that these finite subcovers forms a base of $ K $.
+
+To see this, consider any point $ k \in K $ and any neighborhood $ N_r(k) $ where $ r > 0 $, there exists a rational number $ 0 < \frac{p}{q} < r $ by the Archimedean property, and so $ 0 < \frac{1}{q} < r $. The point $ k $ must be covered by an open set in the finite subcover of radius $ \frac{1}{3q} $, and that must be a subset of $ N_r(k)$. Therefore the finite subcovers forms a base of $ K $.
+
+Therefore $ K $ has a countable base.

Books/BabyRudin/Chapter02/ex26.cecilia.tex

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+\subsection*{Exercise 26 (Cecilia)}
+Just to clarify the hints:
+$ F_n $ is non empty because the finite collection does not cover $ X $.
+$ \cap F_n $ is empty because $ \cap F_n $ exclude every $ G $ but $ G $ covers $ X $.
+
+Consider a limit point $ l $, $ l \in X $, so $ l $ is covered by $ G_i $ for some $ i $. Consider a neighborhood $ N_r(l) \subset G_i $, this neighborhood cannot contain any point $ e_j $ where $ j > i $, so such a neighborhood can contain at most a finite number of points, and that contradicts theorem 2.20.