Ch1. Measure Theory (Ex.01 ~ Ex.18)

① \(\exists N\) s.t. \(|x - y| > \frac{1}{3^N}\)

Let \(x, y \in C\) and \(x < y\). Then, there is an integer \(N \in \mathbb{N}\) such that \(|x - y| > \frac{1}{3^N}\).
If not, it means \(|x - y| \leq \frac{1}{3^k}\) for all \(k \in \mathbb{N}\).
Since \(\frac{1}{3^k} \to 0\), for every \(\varepsilon > 0\) there is \(K_{\varepsilon}\) such that \(k \geq K_{\varepsilon}\) implies \(\frac{1}{3^k} < \varepsilon\).
When \(\varepsilon \geq |x - y|\), then \(|x - y| > \frac{1}{3^k}\) for \(k \geq K_{\varepsilon}\).
This is a contradiction.

② \(\exists z \notin C\) such that \(x < z < y\)

Note that \(\frac{1}{3^N}\) is a length of a closed interval in \(C_N\).
Then, \(x\) and \(y\) are in distinct intervals.
So, there is \(z \in [0, 1]\) such that \(x < z < y\), \(z \notin C_N\), and hence \(z \notin C\).

③ \(A, B\) are separated

Let \(A = C \cap (-\infty, z)\) and \(B = C \cap (z, \infty)\).
\(\overline{A} = \overline {C \cap (-\infty, z]} \subset (-\infty, z]\) and \((-\infty, z] \cap B = \emptyset\) (\(\because b > z\) for \(\forall b \in B\)),
so \(\overline{A} \cap B = \emptyset\).
Similarly, \(\overline{B} = \overline{C \cap [z, \infty)} \subset [z, \infty)\) and \(A \cap [z, \infty) = \emptyset\) (\(\because a < z\) for \(\forall a \in A\)), so \(A \cap \overline{B} = \emptyset\).
So,\(A, B\) are separated.

④ \(A \cup B = C\), conclusion.

1. \(x \in A \cup B\): Since \(A \subset C\) and \(B \subset C\), \(x \in C\).

2. \(\nexists z \in C\): If \(y < z\), \(y \in A\).
   If \(y > z\), \(y \in B\). So, \(y \in A \cup B\).

So, \(A \cup B = C\). Since \(C\) is a union of separated sets, \(C\) is disconnected.

① \(x \in I_{k,i}\) for some \(k \in \mathbb{N}\) and \(i = 1, \cdots, 2^k\).

Choose any \(x \in C\), then \(x \in C_k\) for all \(k \in \mathbb{N}\). Each \(C_k\) has \(2^k\) closed intervals of same length (\(\forall k\)): \(C_k = \bigcup_{i=1}^{2^k} I_{k,i}\) where \(m(I_{k,i}) = \frac{1}{3^k}\) and \(I_{k,i} \cap I_{k,j} = \emptyset\) (\(i \neq j\)). Then, \(x \in I_{k,i}\) for some \(i = 1, \cdots, 2^k\).

② \(x\) is a limit point of \(C\)

Let \(y_k\) be an end point of the interval \(I_{k,i}\) such that \(y_k \neq x\) and \(|y_k - x| \leq |y_k' - x|\) (\(y_k'\) is the other end point).

1. \(r > |y_k - x|\): the open ball \(B_r(x)\) contains \(y_k\).

2. \(0 < r \leq |y_k - x|\): \(B_r(x)\) contains a point.
   Then, \((x, x + r) \subset I_{k,i} \cap B_r(x)\). So \(B_r(x)\) contains \(x_k \in (x, x + r) \subset I_{k,i}\) or \((x - r, x)\).

To sum up, \(B_r(x)\) contains a point of \(C\) not equal to \(x\). So, \(x\) is a limit point of \(C\). This holds for all \(x \in C\), so \(C\) is perfect.

① \(\delta = \frac{1}{3^N}\) with \(\frac{1}{2^N} < \varepsilon\)

For given \(\varepsilon > 0\), choose \(\delta = \frac{1}{3^N}\) with \(\frac{1}{2^N} < \varepsilon\). For \(x, y \in C\), if \(|x - y| < \delta\), this means \(x\) and \(y\) are in the same interval of \(C_N\). Let \(x = \sum a_{x,k} 3^{-k}\) and \(y = \sum a_{y,k} 3^{-k}\) with a unique binary form.

② \(b_{x,k} = b_{y,k}\) for \(k \leq N\)

Assume that there exists \(k \leq N\) such that \(a_{x,k} \neq a_{y,k}\), and let \(K\) be the first integer with the property: \(a_{x,K} = 0\), \(a_{y,K} = 2\). Then, \(\sum a_{x,k} 3^{-k} \leq \sum a_{y,k} 3^{-k}\) and \(\sum a_{y,k} 3^{-k} \geq 0\).
So,
\[ \begin{aligned} |y - x| &= \sum (a_{y,k} - a_{x,k}) 3^{-k} = \sum_{k = K}^{\infty} 2 \cdot 3^{-k} = 2 \cdot 3^{-K} \sum_{k = 0}^{\infty} 3^{-k} = 2 \cdot 3^{-K} \cdot \frac{1}{1 - \frac{1}{3}} = \frac{3}{3^K}. \end{aligned} \]
Since \(K \leq N\), \(\delta = \frac{1}{3^N} \leq \frac{1}{3^K} \leq |y - x|\), which is a contradiction.
So, \(a_{x,k} = a_{y,k}\) for all \(k \leq N\), and consequently \(b_{x,k} = b_{y,k}\) (\(k \leq N\)).

③ \(|x - y| < \delta \implies |F(x) - F(y)| < \varepsilon\)

Then,
\[ \begin{aligned} |F(x) - F(y)| &= \left| \sum (b_{x,k} - b_{y,k}) 2^{-k} \right| = \sum_{k = N + 1}^{\infty} |b_{x,k} - b_{y,k}| 2^{-k} \leq \sum_{k = N + 1}^{\infty} 2^{-k} = \frac{1}{2^N}. \end{aligned} \]
This holds for all \(\varepsilon > 0\), so \(F\) is continuous on \(C\).

① Def of \(I_n\) & \(x_n\), \(|x_n - x| \leq m(I_n)\)

Let \(J_n\) be one of the closed intervals in \(C_n\) such that \(x \in C_n\), and let \(x_n\) be a point in \(I_n\). Then \(|x_n - x| \leq m(J_n)\) because both \(x\) and \(x_n\) are in \(J_n\).

② \(\lim_{n \to \infty} |I_n| = 0\)

An open interval of length \(l_1\) is bisected from \(C_0\) of length 1, so \(l_1 < 1\). Each closed interval of \(C_1\) has length less than \(1/2\), so an open interval of length \(l_2\) is bisected from one piece of \(C_1\) with length less than \(1/2\).
Each closed interval of \(C_2\) has a length \(< 1/4\), so an open interval of length \(l_3 < 1/4\) is bisected. Repeating this countably, \(0 < l_k < \frac{1}{2^n}\). \(|I_n| = l_n\), so \(\lim_{n \to \infty} |I_n| = 0\).

③ \(\lim_{n \to \infty} x_n = x\)

\(C_n\) has \(2^n\) closed intervals including \(J_n\), so \(m(J_n) < \frac{1}{2^n}\).
This means \(\lim_{n \to \infty} m(J_n) = 0\). Since \(|x_n - x| < m(J_n)\), \(\lim_{n \to \infty} |x_n - x| = 0\).
Thus, \(\lim_{n \to \infty} x_n = x\).

① perfect

Let \(x_n\) be an end point of \(J_n\) such that \(x_n \neq x\). \(x_n\) is not removed in every iteration, so \(x_n \in \hat{C}\) for all \(n\). \(|x_n - x| \leq |J_n| \to 0\), so \(\lim_{n \to \infty} x_n = x\). Then, every neighborhood of \(x\) contains all but finitely many points of \(\{x_n\}\). So \(x\) is a limit point of \(\hat{C}\), and thus the closed \(\hat{C}\) is perfect.

② no open interval

1. If \(\exists O \subset \hat{C}\), \(\exists\) at least two points of \(\hat{C}\) in \(O\).
   Assume there exists an open interval \(O \subset \hat{C}\). Since \(\hat{C}\) is perfect, \(O\) should contain at least two distinct points of \(\hat{C}\), say \(x < y\), such that \(x, y \in O\). (If \(O\) contains only one point \(x\), \(\exists r\) s.t. \(B_r(x) \cap \hat{C} = \emptyset\), contradiction.)

2. contradiction
   There exists \(N \in \mathbb{N}\) such that \(l_N < \frac{1}{2^N} < \frac{1}{4} |x - y|\). Then, \(x, y \in J_N\), \(x \in I_{x,N}\), and \(y \in I_{y,N}\). Let \(x_N\) be a right end point of \(J_{x,N}\), \(y_N\) be a left end point of \(J_{y,N}\), and \(z = \frac{1}{2}(x_N + y_N)\). \(x < z < y\) implies \(z \in O\). Also, \(z\) is in a bisected open interval, so \(z \notin \hat{C}\) and thus \(z \notin O\). This is a contradiction.

① \(x \in \overline{A}\) iff \(d(x, A) = 0\)

1. \(\Rightarrow\)
   It is trivial when \(x \in A\), so the following proof covers the case \(x \in A'\). \(x\) is a limit point of \(A\), so for given \(\varepsilon > 0\) there is a point \(a_{\varepsilon} \in A\) such that \(a_{\varepsilon} \neq x\) and \(d(x, a_{\varepsilon}) < \varepsilon\). By the def of \(d(x, A) = \inf \{ d(x, a) : a \in A \}\), \(d(x, A) \leq d(x, a_{\varepsilon}) < \varepsilon\), so \(d(x, A) < \varepsilon\). This holds for all \(\varepsilon > 0\), so \(d(x, A) = 0\).

2. \(\Leftarrow\)
   \(d(x, A) = 0\) so \(\inf \{ d(x, a) : a \in A \} = 0\). Then, for given \(\varepsilon > 0\), there is \(a_{\varepsilon} \in A\) such that \(d(x, a_{\varepsilon}) < 0 + \varepsilon = \varepsilon\).
   

1. \(x \notin A\): \(B_{\varepsilon}(x)\) contains a point of \(A\) for all \(\varepsilon > 0\), so \(x \in A'\).
   
2. \(x \in A\): nothing to prove.
   So, \(x \in \overline{A}\).

② \(\overline{E} = \bigcap O_n\)

1. \(x \in E\): \(d(x, E) = 0\) so \(x \in O_n\) for all \(n\). So, \(x \in \bigcap O_n\).
   
2. \(x \in O_n\) for all \(n\), so \(d(x, E) < \frac{1}{n}\) \(\forall n\). This means \(d(x, E) = 0\), so \(x \in \overline{E}\).

③ \(m(E) = \lim_{n \to \infty} m(O_n)\)

1. \(E = \overline{E}\), \(O_{n+1} \subset O_n\).
   \(E\) is closed, so \(\overline{E} = E\). It is obvious that \(O_n \supset O_{n+1}\) for all \(n\).

2. \(m(O_1) < \infty\), compact.
   \(E\) is bounded, so there is an open ball \(B_R(o) \supset E\). For \(x \in O_1\) and \(e \in E\), \(d(x, o) \leq d(x, e) + d(e, o) < 1 + d(e, o)\) (\(\because e \in E'\)) \(\leq 1 + R\) (\(\because e \in B_R(o)\)). So \(|x| < R + 1\) and \(x \in B_{R+1}(o)\). This holds for all \(x \in O_1\), so \(m(O_1) < \infty\).

3. \(\therefore m(E) = \lim_{n \to \infty} m(O_n)\) (Cor 1.3.3)

① \(E\) is unbounded & closed

Let \(E = \mathbb{Z}\). For \(k \in \mathbb{Z}\), \(O_n = \bigcup_{k \in \mathbb{Z}} B_{1/k}(k)\), where \(m(B_{1/k}(k)) = \frac{2}{k}\). So \(m(O_1) = \sum_{k \in \mathbb{Z}} \frac{2}{k} = \infty\), and \(\lim_{n \to \infty} m(O_n) = \infty\). However, \(m(E) = 0\). (Note that \(E\) is closed)

② \(E\) is open & bounded

1. \(E = \bigcup_{k=1}^{\infty} \left( q_k - \frac{1}{k}, q_k + \frac{1}{k} \right)\), \(m(E)\)
   Let \(\{q_k\}\) be an enumeration of \(\mathbb{Q}\) in \([0,1]\) and \(E = \bigcup_{k=1}^{\infty} \left( q_k - \frac{1}{k}, q_k + \frac{1}{k} \right)\). Each \(\left( q_k - \frac{1}{k}, q_k + \frac{1}{k} \right)\) is open, so \(E\) (countable union of open sets) is open. Since \(q_k \in [0,1]\) for all \(k \in \mathbb{N}\), \(E \subset \left( -1, 1 + \frac{1}{1} \right)\).

2. Then,
   \[ m(E) \leq \sum_{k=1}^{\infty} m \left( \left( q_k - \frac{1}{k}, q_k + \frac{1}{k} \right) \right) = \sum_{k=1}^{\infty} \frac{2}{k} = 2 \sum_{k=1}^{\infty} \frac{1}{k} = \frac{1}{2}. \]

3. \(\lim_{n \to \infty} m(O_n) \geq 1\)
   If \(x \in \mathbb{Q} \cap [0,1]\), it is obvious that \(x \in E\) and \(d(x, E) = 0\). Consider the case \(x \in (\mathbb{R} - \mathbb{Q}) \cap [0,1]\), which is irrational in \((0,1)\). Since \(E\) contains all rationals in \([0,1]\) and \(\mathbb{Q}\) is dense in \(\mathbb{R}\), \(x \in E\) and \(d(x, E) = 0\) (\(\because \mathbb{Q}\) dense). Since \(d(x, E) = 0\) for all \(x \in [0,1]\), \(O_n\) contains \([0,1]\) for all \(n \in \mathbb{N}\). This means \(m(O_n) \geq 1\) for all \(n \in \mathbb{N}\), so \(\lim_{n \to \infty} m(O_n) \geq 1\).

4. \(\therefore m(E) \neq \lim_{n \to \infty} m(O_n)\)

① \(\| L(x) \| \leq M |x|\) for some \(M > 0\), \(x \in \mathbb{R}^d\)

Let \(x = x_1 e_1 + x_2 e_2 + \cdots + x_d e_d\), \(x_i \in \mathbb{R}\) and \(e_i\) is a unit vector. Then \(L(x) = x_1 L(e_1) + \cdots + x_d L(e_d)\) and \[\| L(x) \| \leq |x_1| \| L(e_1) \| + \cdots + |x_d| \| L(e_d) \| \leq (x_1^2 + \cdots + x_d^2)^{1/2} ( \| L(e_1) \|^2 + \cdots + \| L(e_d) \|^2 )^{1/2}\] (\(\because\) Cauchy-Schwartz ineq) \(\triangleq M |x|\).

② \(L\) is continuous

For \(x, y \in \mathbb{R}^d\) with \(|x - y| < \delta = \varepsilon / M\), \[\| L(x) - L(y) \| = \| L(x - y) \| \leq M |x - y| < \varepsilon.\] This holds for all \(\varepsilon > 0\), so \(L\) is (uniformly) continuous.

③ \(L(E)\) is compact, conclusion

Since \(L\) is continuous, \(L(E)\) is compact. (PMA 4.1A)

① \(F = \bigcup_{n=1}^{\infty} K_n\), where \(F\) is closed set and \(K_n\) is a compact set in \(\mathbb{R}^d\)

\(\overline{B_n(0)}\) is closed and bounded, so it is compact. Let \(K_n = F \cap \overline{B_n(0)}\), which is an intersection of closed and compact sets. So, \(K_n\) is compact.
When \(x \in \bigcup K_n\), there is \(N \in \mathbb{N}\) such that \(x \in K_N\). Since \(K_N \subset F\), \(x \in F\). \(\forall x \in F\) : there is \(N \in \mathbb{N}\) with \(N \geq |x|\), which means \(x \in \overline{B_N(0)}\). Since \(x \in K_N\), \(x \in \bigcup K_n\). Thus, \(F = \bigcup K_n\).

② \(L(E)\) is \(F_\sigma\)

Let \(E = \bigcup_{n=1}^{\infty} F_n\), where \(F_n\) is closed. Then, \(L(E) = L( \bigcup F_n ) = \bigcup L(F_n)\). Each \(F_n\) can be written as \(F_n = \bigcup_{k=1}^{\infty} K_{n,k}\) where \(K_{n,k}\) is compact. Then, \(L(F_n) = \bigcup_{k=1}^{\infty} L(K_{n,k})\) and thus, \(L(E) = \bigcup_{n=1}^{\infty} \bigcup_{k=1}^{\infty} L(K_{n,k})\). \(L(K_{n,k})\) is compact, which is closed as well. Thus, \(L(E)\) is an \(F_\sigma\) set.

① \(\|L(x) - L(y)\| \leq \sqrt{d} M \ell\)

Let \(Q\) be a cube of length \(\ell\) in \(\mathbb{R}^d\) and let \(x\) be a corner of the \(Q\). For \(y \in Q\), \(|y - x| \leq \sqrt{d} \ell\). So, \(\|L(x) - L(y)\| \leq M |x - y| \leq \sqrt{d} M \ell\).

② \(L(Q) \subset Q'\) with length \(2 \sqrt{d} M \ell\)

Let \(Q'\) be a cube in \(\mathbb{R}^n\) with length \(2 \sqrt{d} M \ell\) centered at \(L(x)\). The minimum distance between \(L(x)\) and the exterior of the \(Q'\) is \(\sqrt{d} M \ell\). Since \(\|L(x) - L(y)\| \leq \sqrt{d} M \ell\) for all \(y \in Q\), \(L(Q) \subset Q'\). Let \(|Q'| = (2 \sqrt{d} M)^n |Q| \triangleq c |Q|\).

\(\Leftarrow\): \(R = R \cup \emptyset\). Let \(\Omega = R\).

\(\Rightarrow\) (\(\Omega = R_1\))

\(\Omega = \bigcup_{i=1}^{\infty} R_i\), where \(R_i\) is open rectangle and \(R_i \cap R_j = \emptyset\) for all \(i \neq j\). Let \(R_1\) be a non-empty subset (wlog): \(R_1 \subset \Omega\).

Assume that \(R_1 \neq \Omega\) (\(\exists j > 1\) s.t. \(R_j \neq \emptyset\)).
\(\Omega = R_1 \cup \left( \bigcup_{j=2}^{\infty} R_j \right)\).
Since \(R_i \cap R_j = \emptyset\) (\(\forall i \neq j\)), \(R_1 \cap \left( \bigcup_{j=2}^{\infty} R_j \right) = \emptyset\).
Then \(\Omega\) is a union of two non-empty disjoint open sets, so \(\Omega\) is disconnected.
This is a contradiction.
Thus, \(R_1 = \Omega\).

① \(O_n = \{ x : d(x, F) < \frac{1}{n} \}\) is open

Let \(y \in O_n\) and \(\varepsilon = \frac{1}{n} - d(y, F) > 0\)
For \(z \in B_\varepsilon(y)\), \(d(z, F) \leq d(z, y) + d(y, F) < \varepsilon + d(y, F) = \frac{1}{n}\)
This means \(z \in O_n\), and this holds for all \(z \in B_\varepsilon(y)\)
So, \(B_\varepsilon(y) \subset O_n\)
Since this holds for all \(y \in O_n\), \(O_n\) is open

② \(F = \bigcap_{n=1}^{\infty} O_n \sim G_\delta\)

\(d(y, F) = 0\) for \(y \in F\), so \(y \in O_n\) for all \(n\) and \(y \in \bigcap_{n=1}^{\infty} O_n\)
So, \(F \subset \bigcap O_n\)
If \(z \in \bigcap O_n\): \(d(z, F) < \frac{1}{n}\) for all \(n\)
Since \(d(z, F) = 0\), \(z \in F = F\)
Thus, \(\bigcap O_n = F\)

③ \(U = F_\sigma\) for open set \(U\) \(\therefore F^c = \bigcup O_n^c\)

① \(\mathbb{R} - \mathbb{Q}\)

The set of all irrationals is \(\mathbb{R} - \mathbb{Q} = \mathbb{R} - \bigcup \{ q_i \}\)
\(= \mathbb{R} \cap \bigcap ( \{ q_i \}^c ) = \mathbb{R} \cap \bigcap ( \mathbb{R} \setminus \{ q_i \} )\)
Each \(\mathbb{R} \setminus \{ q_i \}^c\) is dense in \(\mathbb{R}\) (\(\because q_i\) is a limit point of \(\mathbb{R} \setminus \{ q_i \}^c\))
So \(\mathbb{R} - \mathbb{Q}\) is a countable intersection of open and dense subsets

② \(\mathbb{Q}\)

Assume that \(\mathbb{Q} = \bigcap O_n\) for some open sets \(O_n\)
Naturally, \(\mathbb{Q} \subset O_n\) for all \(n\)
The point \(x \in \mathbb{R} \setminus \mathbb{Q}\) is a limit point of \(\mathbb{Q}\)
Since \(O_n^c \cap \mathbb{Q} = \emptyset\), \(x\) is a limit point of \(\mathbb{Q} \subset \mathbb{R}\)
This implies \(O_n\) is dense in \(\mathbb{R}\), and \(\mathbb{Q}\) is a countable intersection of open & dense sets, which isn’t empty

③ concl.

Then, Baire’s thm (PMA ex 3-22) says \((\mathbb{R} - \mathbb{Q}) \cap \mathbb{Q}\) is not empty
However, \((\mathbb{R} - \mathbb{Q}) \cap \mathbb{Q} = \emptyset\), which is a contradiction

① \(\mathbb{Q} \cup (\mathbb{R} - \mathbb{Q})\) is a Borel set

Since a closed subset is a Borel set, \(\{ q_i \}\) is a Borel set for all \(q_i \in \mathbb{Q}\)
The Borel \(\sigma\)-algebra is closed under countable union, so \(\bigcup \{ q_i \}\) is a Borel set
Since Borel \(\sigma\)-algebra is closed under complement, \(\mathbb{Q}^c = \mathbb{R} - \mathbb{Q}\) is a Borel set

② \(A = (\mathbb{Q} \cap (0,1)) \cup ((\mathbb{R} - \mathbb{Q}) \cap (2,3))\) is not \(G_\delta\), \(F_\sigma\)

① \(J_*(E) \leq J_*(\overline{E})\) for \(E \subset \overline{E}\) and thus \(J_*(E) \leq J_*(\overline{E})\)

Let \(\widetilde{C_1} = \{ \sum |I_i| : E \subset \bigcup I_i \}\) and \(\widetilde{C_2} = \{ \sum |I_i| : \overline{E} \subset \bigcup I_i \}\)
Then \(J_*(E) = \inf \widetilde{C_1}\) and \(J_*(\overline{E}) = \inf \widetilde{C_2}\)
Since \(E \subset \overline{E}\), the covering of \(\overline{E}\) also covers \(E\)
If \(C \in \widetilde{C_2}\), then \(C \in \widetilde{C_1}\)
This means \(\widetilde{C_2} \subset \widetilde{C_1}\)
Thus, \(\inf \widetilde{C_2} = J_*(\overline{E}) \geq \inf \widetilde{C_1} = J_*(E)\)

② \(J_*(E) \geq J_*(\overline{E})\)

Let \(I = \bigcup_{i=1}^{n} I_i\) covers \(E\): \(E \subset I\)
Since \(E \subset I\) implies \(\overline{E} \subset \overline{I} = \bigcup \overline{I_i}\)
Let \(S_E = \sum_{i=1}^{n} |I_i|: E \subset \bigcup I_i\)
Since \(|\overline{I_i}| = |I_i|\), \(\sum |I_i| = \sum |\overline{I_i}|\) and thus \(S_E \leq S_{\overline{E}}\)
So, \(\inf S_E = J_*(E) \leq J_*(\overline{E}) = \inf S_{\overline{E}}\) (\(\because \sum |I_i| \leq S_{\overline{E}}\))

① (let \(B_n = \bigcup_{k=n}^{\infty} E_k\)) \(\bigcap_{n=1}^{\infty} B_n \subset E\)

\(x \in E^c\), then \(x \in E_k\) for (at most) finitely many \(k\)
Let \(K-1\) be the maximum of such \(k\)’s, then \(x \in E_k^c\) for all \(k \geq K\)
Then, \(x \in \bigcap_{k=K}^{\infty} E_k^c = (\bigcup_{k=K}^{\infty} E_k)^{c} = B_K^{c}\)
Since \(x \in B_K^c\), \(x \in \bigcap_{n=1}^{\infty} B_n^c\)
Thus, in reverse, \(\bigcap_{n=1}^{\infty} B_n \subset E\)

② \(E \subset \bigcap_{n=1}^{\infty} B_n\)

\(x \in (\bigcap_{n=1}^{\infty} B_n)^{c} = \bigcup_{n=1}^{\infty} B_n^c\)
Then, there is a positive integer \(M\) such that \(x \in B_M^c = \bigcap_{k=M}^{\infty} E_k^c\)
Since \(x \in E_k^c\) for all \(k \geq M\), \(x \in E_k\) for at most finitely many \(k\)
Thus, \(x \notin E\) and \(E \subset \bigcap_{n=1}^{\infty} B_n\)

③ \(\therefore E = \bigcap_{n=1}^{\infty} B_n\)