$\gdef\In{\subset}$

1. 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. $S = \{ \vec x \in \R^2 \mid |\vec x| = 1 \}$

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.

3. (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 $X$.

4. 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?

5. 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 $S$.

1. 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:

1.1 $S = \Q \cap [0,1]$ 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.

1.2 $S = \{ \vec x \in \R^2 \mid |\vec x| = 1 \}$

• $S$ 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 $\R^2$.

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 $\Q$.

3. (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.

4. 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. 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

• 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 $x-u \notin E$.
• 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).

5. 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 $S$.