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--- math104-f21:hw8 [2021/10/23 06:13]
pzhou
+++ math104-f21:hw8 [2026/02/21 14:41] (current)
@@ Line 70: / Line 70: @@
 $E$ is not countable. An element of $E$ can be written as $0.a_0 a_1\cdots$, where $a_i \in \{0,2,4,6,8\}$, hence an element of $E$ is a map $\N \to \{0,2,4,6,8\}$, conversely, any such a map defines an element in $E$. Hence $E =Map( \N, \{0,2,4,6,8\})$.  Since the set $Map( \N, \{0,2\})$ is a proper subset of $E$, and $Map(\N, \{0,2\}) \cong Map (\N, \{0,1\})$, which we have shown in previous homework that is uncountable, hence $E$ is uncountable (since it contains an uncountable subset).
-$E$ is closed. For this, we only need to show that, for any $x \notin E$, there exists an $r>0$, such that $B_r(x) \cap E = \emptyset$. Suppose $x \notin E$, and let $0.a_0 a_1 \cdots$ be the decimal expansion of $x$. Then for some $n$, $a_n$ is an odd digit. We discuss two cases
+$E$ is closed. This an analog of a Cantor set. We define
-  * If $x$ has a finite decimal expansion (i.e., with a trailing 0), and let $a_N$ be the last non-zero digit, then we can let $r = 10^{-2N}$. Then for any $0<u<r$, $x+u$ has the same first $N+1$ digits $0.a_0\cdots a_N$, hence contains an odd digit; and $x-u$ has the decimal expansion $0.b_0b_1\cdots$, where $b_i = a_i$ for $i < N$, $b_N = a_N-1$, and $b_{N+1}=9$, which shows $x-u \notin E$.
+$$ E_1 = [0, 0.1] \cup [0.2, 0.3] \cup \cdots \cup [0.8, 0.9] $$
-  * If $x$ does not have a finite decimal expansion, then let $a_N$ be the digit, where $N>n$, and $a_N \neq 9$ (since we forbid the expansion with trailing 9, there must be a digit in the tail that is not 9), and $a_N \neq 0$ (since we assume $x$ does not have a finite expansion, i.e., expansion with a trailing 0).
+then $E_1$ is almost all the real numbers in $[0,1]$ with the first decimal digit being even, except those right end points $0.1, 0.3, \cdots$ in each interval. We define $E_2$ from $E_1$, by taking each closed interval $[a,b]$ in $E_1$, subdivide it into 10 closed intervals, labelled by $0, \cdots, 9$, and keep those even labelled ones. For example, $[0, 0.1]$ in $E_1$ contribute $[0, 0.01] \cup [0.02, 0.03] \cup \cdots \cup [0.08, 0.09]$ in $E_2$. More formally,
+$$ E_{n+1} = \{0, 0.2, 0.4, 0.6, 0.8\} + (1/10) E_n $$
+where for two subsets $A, B \In \R$, $A+B = \{a+b \mid a \in A, b \in B\}$ is the Minkowski sum. Thus, all $E_n$ are all closed, and $E_{n+1} \subset E_n$.
+We claim that $E = \cap_{n=1}^\infty E_n$. If the claim holds, then $E$ is the intersection of closed sets, hence $E$ is closed. Now we prove the claim. If $x \in E_1 \cap E_2$, then $x \neq 0.1, 0.3, 0.5, 0.7, 0.9$, hence the first digit (after decimal point) of $x$ is even. Similarly, if $x \in E_n \cap E_{n+1}$, then the $n$-th digit of $x$ after decimal point is even. If $x \in \cap_{n=1}^\infty E_n$, then all digits of $x$ are even, hence $E \supset \cap_{n=1}^\infty E_n$. On the other hand, it is easy to verify that $E \In E_n$ for all $n$, hence $E \subset \cap_{n=1}^\infty E_n$, thus $E = \cap_{n=1}^\infty E_n$ proving the claim.
+(whew, that's a long winding proof. It is possible to do it without using the intersection of closed set construction, but it is also long.)
+Finally, since $E$ is bounded and closed, $E$ is compact.
@@ Line 80: / Line 89: @@
+Example:
+  * $\{0, 1, 1/2, 1/3, \cdots 1/n, \cdots \}$.
+  * $\Q$, or $\Q \cap (0,1)$, or $\Q \cap [0,1]$.

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