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--- math104-f21:hw8 [2021/10/15 22:40]
pzhou
+++ math104-f21:hw8 [2022/01/11 08:36] (current)
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   * An (infinite) countable subset $S \In \R$, such that $S \In S'$, ie, every point of $S$ is a limit point of $S$.
+===== Solution =====
+. Determine whether following subset $S$ of metric space $X$ is (a) open or not (b) closed or not (c ) bounded or not (d) compact or not. (You may use Heine-Borel theorem for $\R^k$)
+  * $X = \R$ with usual metric. $S = \Q \cap [0,1]$.
+  * $X = \R^2$ with Euclidean metric. $S = \{ \vec x \in \R^2 \mid |\vec x| = 1 \}$
+Answer:
+.1 $S = \Q \cap [0,1]$ in $\R$.
+  * $S$ is not open, for any $x \in S$, and any $r > 0$, $B_r(x) \not \subset S$, since $B_r(x)$ contains irrational numbers, and $S$ only contains rational numbers.
+  * $S$ is not closed. Since for $x \in (0,1)$ and $x \notin S$, for any $r > 0$, $B_r(x) \cap S \neq \emptyset$, since $B_r(x)$ would contain rational numbers.
+  * $S$ is bounded.
+  * $S$ is not compact, since by Heine-Borel theorem, $S$ is compact if and only if $S$ is closed and bounded. But $S$ is not closed.
+.2 $S = \{ \vec x \in \R^2 \mid |\vec x| = 1 \}$
+  * $S$ is not open. For example, consider the point $(0,1) \in S$, for any $r>0$, $B_r( (0,1) )$ contains a point $(0, 1+r/2)$, which is not in $S$.
+  * $S$ is closed. Since for any $x \notin S$, we can let $r = |d(0, x) - 1 | /2$, then $B_r(x) \cap S = \emptyset$.
+  * $S$ is bounded, since $S$ is contained in a bounded set $B_{2}(0)$.
+  * $S$ is compact, since $S$ is closed and bounded subset of $\R^2$.
+. True or False, give your reasoning or give an counter-example.
+  * Let $(X, d)$ be a metric space, then any finite subset $S \In X$ is closed.
+  * Let $X = \Q$ with the usual metric. Then, any closed bounded subset $S \In X$ is compact.
+Answer:
+(a) True. We proved in class that, in a metric space, a singleton is closed. And any finite union of closed set is closed.
+(b) False. Since $[0, 1] \cap \Q$ is not compact (see problem 1), but it is closed and bounded in $\Q$.
+. (Open and closed subset are relative notion) Let $(X, d)$ be a metric space. $U \In Y \In X$ any subset. Prove that
+  * If $Y$ is open relative to $X$, then $U$ is open relative to $Y$ if and only if $U$ is open relative to $X$.
+  * (Optional) If $Y$ is closed relative to $X$, then $U$ is closed relative to $Y$ if and only if $U$ is closed relative to $X$.
+Proof: We first prove the direction that:  $U$ is open in $Y$ implies $U$ is open in $X$. For any point $p \in U$, since $U$ is open in $Y$, there exists $r>0$, such that $B_r^Y(p) = \{ q \in Y \mid d(p,q) < r \}$ is contained in $U$. On the other hand, $B_r^Y(p) = B_r^X(p) \cap Y$, where $B_r^X(p) = \{ q \in X \mid d(p,q) < r \}$. Since $Y$ is open in $X$, and finite intersection of open sets are open, hence $B_r^Y(p)$ is open in $X$. Since for each $p \in U$, the above constructed $B_p = B_r^Y(p)$ is contained in $U$ and is open in $X$, hence $U = \cup_{p \in U} B_p$ is a union of open sets in $X$, hence $U$ is open in $X$.
+Now we prove that other direction.  Suppose $U$ is open in $X$, then $U \cap Y = U$ is open in $Y$ by definition of induced topology on $Y$. We are done.
+. Let $E \In [0,1]$ consist of those real numbers, such that the decimal expansion only contains even digits $0,2, \cdots, 8$. Is $E$ countable? Is $E$ closed in $\R$? Is $E$ compact?
+Caveat: here I made a mistake when I say "the" decimal expansion of a real number. There are some real numbers that admits exactly two decimal expansions, where one of the expansion is finite (ie with a trailing 0), and one of the expansion has a trailing 9s. For example, $0.1 = 0.09999\cdots$. Here I should have said: $E \In [0,1]$ consist of $x \in \R$, such that $x$ **admit** an decimal expansion that only contains even digits. This bug affects the answer, if $E$ is closed and if $E$ is compact. If you are very careful, and noticed this bug, then you don't lose any points.
+Answer: In the following, when we say the decimal expansion, we use the unique decimal expansion that does not have a trailing 9.
+$E$ 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. This an analog of a Cantor set. We define
+$$ E_1 = [0, 0.1] \cup [0.2, 0.3] \cup \cdots \cup [0.8, 0.9] $$
+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.
+. Give examples.
+  * An (infinite) countable subset in $\R$ that is compact.
+  * An (infinite) countable subset $S \In \R$, such that $S \In S'$, ie, every point of $S$ is a limit point of $S$.
+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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