lacunary - Theorem Index

Assume $A$ is countably infinite, $E \subset A,$ and $|E| = \infty.$ Arrange a sequence $\{x_n\}$ from the distinct elements of $A$. Let $n_1$ be the smallest positive integer such that $x_{n_1} \in E,$ and then pick $n_2, \dots, n_{k-1}$ by assigning the next $n$ in the sequence the index of the left-most entry in $\{x_n\}$ that has not yet been picked. Then, let $n_k$ be the smallest integer greater than $n_{k-1}$ such that $x_{n_k} \in E.$ Now, $\{n_k\}$ is a sequence of strictly increasing positive integers giving us the indices of the first $k$ elements of $E$ in $\{x_n\}.$

Now, define $f : \mathbb{Z}_{>0} \to E$ as $f(k) = {x_{k_n}}$, which is a bijection between the positive integers and $E$, showing that $E$ is countable.

Let $E$ be a countable subset of $A,$ and call the elements of $E$ $s_1, s_2, s_3, \dots.$ We will construct a new sequence $p$ in the following way:

$$ p_n = \neg s_{n_n}, n = 1, 2, 3, \dots $$

That is, the $n$th digit of $p$ will be the opposite of whatever the $n$th digit of $s_n$ is. So, $p$ differs from $s_1$ in the first digit, from $s_2$ in the second digit, $s_3$ in the third digit, and so on, so that it differs from all elements of $E,$ and therefore is not contained in $E.$ But, $p$ is definitely in $A$ since it its elements are the digits $0$ and $1.$ Therefore, $E$ is a proper subset of $A,$ so any countable subset of $A$ must be a proper subset of $A.$ But, $A$ can't be a proper subset of itself, and therefore $A$ must be uncountable.

I won't give a full proof here, but this can be accomplished by considering the binary representation of real numbers in the interval $[0, 1)$ consists of infinite sequences of $0$ and $1$.

Let $K$ be a compact metric space. Fix $n \in \mathbb{N}$ (any natural number will do,) then, consider the open cover $\{G_\alpha\}_n$ of balls of radius $1/n$ centered at $x \in K.$ Now, because $K$ is compact, some finite $\{V_\alpha\}_n \subset \{G_\alpha\}_n$ covers $K.$

Now, consider the set of all balls for all $n,$

$$ C = \bigcup_{n=1}^{\infty} \{V_\alpha\}_n $$

Because the union of a sequence of countable sets is countable, $C$ is countable. To show $C$ is a base for $K,$ let $G$ be an open subset of $K,$ and let $x \in G.$ Let $\epsilon > 0$ such that $B_{\epsilon}(x) \subset G.$ Pick natural $n$ such that $1/n < \epsilon/2.$ Then, $x \in \{V_\alpha\}_n$ for some $\alpha,$ (because no point in $K$ is more than $1/n$ away from the center of some $\{V_\alpha\}_n$ and therefore $C$ is a countable base of $K.$ Since we can make $\epsilon$ as close to $0$ as we like, the centers of $\{V_\alpha\}_n$ are dense as well as countable, and are therefore a countable dense subset of $K,$ so $K$ is separable.

By Let $X$ be a metric space in..., we know that $X$ is separable and by Every separable metric space has a countable..., we know that $X$ has a countable base. By the definition of base, we have that every open cover of $X$ has a countable subcover $\{G_n\}, n = 1, 2, 3, \dots.$

Suppose, for the sake of contradiction, that $\{G_n\}$ has no finite subcollection that covers $X.$ Let

$$ F_n = (G_1 \cup \cdots \cup G_n)^c. $$

Then, each $F_n$ must be nonempty (otherwise a finite subcollection of $\{G_n\}$ would cover $X.$ However, since every point in $X$ is in some $G_n,$

$$\bigcap_{n=1}^{\infty} F_n = \emptyset. $$

Now, let $E$ be a set with a point from each $F_n.$ Since there are infinitely many $F_n,$ $E$ is an infinite subset of $X,$ and therefore $E$ has a limit point, $p.$ Now, $p$ must be in some open $G_m,$ and so for some $\epsilon > 0,$ $N_{\epsilon}(p) \subset G_m.$

Now, note that each $F_{n+1} \subset F_n,$ because $F_{n+1}$ is formed by excluding all the points in $F_n$ that are also in $G_{n+1}.$ Because $p$ is a limit point of $E,$ $N_\epsilon{p}$ must contain infinitely many points of $E.$ However, only finitely many points of $E$ can be in $N_{\epsilon}(p) \subset G_m,$ because for $n >= m,$ $F_n$ contains no points in $G_m.$ Therefore, $p$ is not a limit point of $E,$ and no such limit point can exist, contradicting our hypothesis that every infinite subset of $X$ has a limit point. Therefore, $E$ must be finite, and $\{G_n\}$ must have a finite subcollection that covers $X,$ meaning $X$ is compact.

Let $X$ be a separable metric space and $E \subset X$ be closed. If $E$ is at most countable, then we are done.

Suppose that $E$ is uncountable. Note that the proof of Suppose $E \subset R^k,$ with $E$ uncountable,... only uses the property that $R^k$ is a separable metric space, and it therefore generalizes to any separable metric space. Thus $E$ is the union of a perfect set - its condensation points, $P$, and a set that is at most countable, $E \setminus P.$

Let $E$ be an open set in $R^1.$ Let $x \in E.$ Let $a = \inf \{ y : (y, x] \subset E \}$ and $b = \sup \{ y : [x, y) \subset E \}.$ Then, $a < x < b,$ because $E$ is open and thus $x$ is an interior point of $E$ and there is some neighborhood around $x$ that is entirely within $E.$ Let $I(x) = (a, b).$ By construction, $I(x)$ is connected. For all $u \in I(x),$ $I(u)$ must have the same end points as $I(x),$ since if it extended beyond, our construction of $I(x)$ would be contradicted. Also note that for all $v \in E \setminus I(x),$ $I(v) \cap I(x) = \emptyset,$ for if they intersected, they would form an open interval. Thus, each $I(x), x \in E$ is either disjoint from all others or identical to some other $I(y), y \in E,$ and $E$ is the union of all such unique $I.$

Now, in each $I,$ we can pick a rational number. Because the rationals are countable, we have at most countably many unique $I,$ and $E$ is their union.

First, for $n = 0$, $\mathbb{R}^0$ is just the empty set, so the metric axioms are vacuously satisfied for all points in the set. Now, for $n \geq 1,$

• Let $p, q \in R^n, p \neq q.$ Then $|p - q| < 0,$ so $d(p,q) > 0.$

• Let $p \in R^n.$ Now, $p - p = 0,$ so $|p - p| = 0,$ so $d(p,p) = 0.$

• Let $p, q \in R^n.$ $|p - q| = |q - p|,$ so $d(p,q) = d(q, p).$

• Let $p, q, r$ \in $R^n.$ $|p - q| \leq |p - r| + |r - p|,$ so $d(p,q) \leq d(p,r) + d(r, p).$

Therefore, $d$ is a metric on $R^n$ and $(R^n, d)$ is a metric space.

Let $\vec{y}, \vec{z}$ be points in a ball with center $\vec{x}$ and radius $r$. Then, by definition, $|\vec{y} - \vec{x}| < r,$ and $|\vec{z} - \vec{x}| < r.$ Suppose $\vec{p} \in \{ \lambda \vec{y} + (1 - \lambda)\vec{z} | 0 < \lambda < 1 \}.$ We will show that $|\vec{x} - \vec{p}| < r.$

$$ \begin{aligned} |\vec{x}-\vec{p}| &= |\vec{x} - (\lambda\vec{y} + (1-\lambda)\vec{z})| \quad(\text{substitute definition of } \vec{p})\\ &= |\vec{x} - \lambda\vec{y} - \vec{z} + \lambda\vec{z}| \quad(\text{expand})\\ &= |\vec{x} - \lambda\vec{y} - \vec{z} + \lambda\vec{z} + \lambda\vec{x} - \lambda\vec{x}| \quad(\text{add and subtract } \lambda\vec{x})\\ &= |\lambda(\vec{x}-\vec{y}) + (1-\lambda)(\vec{x}-\vec{z})| \quad(\text{factor } \lambda \text{ and } 1-\lambda)\\ &\le \lambda\,|\vec{x}-\vec{y}| + (1-\lambda)\,|\vec{x}-\vec{z}| \quad(\text{triangle inequality})\\ & < \lambda r + (1-\lambda)r \quad(\text{since } |\vec{x}-\vec{y}|,|\vec{x}-\vec{z}| < r)\\ &= r \quad(\text{because } \lambda + (1-\lambda)=1). \end{aligned} $$

So, $\vec{p}$ is within our ball and therefore all balls are convex.

Suppose $x \in \left ( \bigcup_{\alpha} E_\alpha \right )^c.$ Then, $x \notin \bigcup_{\alpha} E_\alpha,$ so $x$ is not in any $E_\alpha.$ Therefore, for every $E_\alpha,$ $x \in E_\alpha^c,$ and thus $x \in \bigcap_{\alpha} \left ( E_{\alpha}^c \right ).$ Conversely, suppose $x \in \bigcap_{\alpha} \left ( E_{\alpha}^c \right ).$ Then, $x$ is in every $E_\alpha^c,$ that is, $x$ is not in any $E_\alpha.$ Therefore, $x \in \left ( \bigcup_{\alpha} E_\alpha \right )^c.$

Let $p \in X, p \notin E'.$ Then some neighborhood $N$ of $p$ contains no points in $E,$ other than possibly $p$ itself. If $N$ contains only $p,$ then $p$ is not a limit point of $E'.$ Suppose, for the sake of contradiction, some point $q \in N, q \neq p$ is a limit point of $E'.$ Then, every neighborhood of $q$ contains some point $m$ in $E'.$ Let $M \subset N, p \notin M,$ be such a neighborhood, and let $m \in M, m \neq q$ be a limit point of $E$. Now, $m$ has neighborhoods wholly in $N,$ and such neighborhoods can have no point in $E,$ so we have a contradiction, and therefore $q$ is not a limit point of $E'.$ Hence, $p$ is an interior point of $E'^c,$ and $E'^c$ is open, and therefore $E'$ is closed.

If $E$ is closed, then we are done, because a set equals its closure if it is closed.

Suppose $p$ is a limit point of $E.$ Then every neighborhood of $p$ contains some $q \in E, q \neq p.$ Since $q \in E, ~ q \in \overline{E}$, so $p$ is a limit point of $\overline{E}.$

Conversely, suppose $p$ is a limit point of $\overline{E}.$ Then, every neighborhood $N$ of $p$ contains a point of $q \in \overline{E}, q \neq p.$ If $q \in E,$ then $N$ clearly contains a point in $E.$ Otherwise, $q \in E', q \not in E.$ Now, since $q \in N$ and every neighborhood is an open set, $q$ has some neighborhood $M \subset N.$ Since $q \in E',$ $M$ contains some point $s \in E.$ Since $M \subset N,$ $s \in N,$ and therefore $N$ contains a point in $E.$ Thus, all neighborhoods of $p$ contain some point in $E,$ and $p$ is a limit point of $E.$

Consider $P = \{\frac{1}{n} | n \in \mathbb{N}\}.$ Then $P$ has one limit point, $0,$ but $P' = \{0\}$ has no limit points, since the only number that contains $0$ in all its neighborhoods is $0$ itself (see From this, it's evident that a finite....)

Let $p \in \bigcup_{i=1}^n \overline{A_i}.$ Then, $p \in \overline{A_i}$ for some $i.$ If $p \in A_i,$ then $p \in \bigcup_{i=i}^n A_i = B_n,$ so $p \in \overline{B_n}.$ If $p$ is in some $A_{i}',$ then every neighborhood of $p$ contains some point $q \in A_i, q \neq p,$ and since $A_i \subset B_n,$ $q \in B_n,$ so $p \in \overline{B_n}.$

Conversely, let $p \in \overline{B_n}.$ If $p \in B_n,$ $p \in A_i$ for some $i,$ and thus $p \in \overline{A_i} \subset \bigcup_{i=1}^n \overline{A_i}.$ If $p$ is only in $B_{n}',$ suppose for contradiction that $p \notin \bigcup_{i=1}^n \overline{A_i}.$ For each $i,$ let $N_i$ be a neighborhood centered at $p$ with $N_i \cap A_i = \emptyset.$ Then, $N = \bigcap_{i=1}^n N_i$ is a neighborhood of $p$ (see (a) - For any collection $\{Ga\}$ of...,) but $N \cap N_i = \emptyset$ for all $i,$ so $N \cap B_n = \emptyset.$ But, this contradicts our hypothesis that $p \in \overline{B_n},$ so our contradictory assumption must be invalid, and $p \in \bigcup_{i=1}^n \overline{A_i}.$

Clearly, $P$ is bounded, for it lies within $[0, 1].$ Each $E_n$ is composed of the union of $2^n$ closed intervals, and the union of finitely many closed intervals is also closed. $P$ is then the intersection of infinitely many closed intervals, which is again closed. Therefore, $P$ is closed and bounded, and by heine-borel, is compact.

Suppose, for the sake of contradiction, that some segment $(\alpha, \beta)$ \subset $P$ and let $L = \beta - \alpha.$ Pick some $n \in \mathbb{N}$ such that $1/3^n < L.$ Now, $E_n$ is the union of $2^n$ intervals of length $1/3^n,$ and since $(\alpha, \beta) \subset P,$ it must be the case that $(\alpha, \beta)$ is a subset of some interval of length $1/3^n.$ However, this can't be the case, since $L > 1/3^n,$ by construction. Therefore, our provision assumption is incorrect, and $P$ contains no segment.

Suppose $K$ is a finite set in metric space $X$ and that $\{G_\alpha\}, \alpha \in A$ is an open cover of $K.$ Since $K$ is finite, we can enumerate its points as $\{k_1, \dots, k_n\},$ for some $n \geq 0.$ Then, for each $i = 1, \dots, n$ (there are none when $n = 0$,) pick an $\alpha(i) \in A$ with $x_i \in G_\alpha(i).$ Define the index set

$$ A_0 = \{\alpha(1), \dots, \alpha(n)\} \subset A. $$

Because $n$ is finite, $A_0$ is finite as well, and

$$ K = \{k_1, \dots, k_n\} \subset \bigcup_{\alpha \in A_0} G_\alpha, $$

so $\{G_\alpha\}, \alpha \in A_0$ is a finite sub-cover of the original open cover. Therefore, every open cover of $K$ has a finite sub-cover, and $K$ is compact.

Suppose $K$ is compact relative to $X$ and that $\{V_\alpha\}$ is an open cover of $K$ relative to $Y,$ such that $K \subset \bigcup_\alpha V_\alpha.$ We need to show that a finite subset of $\{V_\alpha\}$ covers $K.$ Because $\{V_\alpha\}$ is open relative to $Y$ and $Y \subset X$ (see Suppose $Y \subset X.$ A subset $E$...) there are sets $G_\alpha,$ open relative to $X,$ such that $V_a = Y \cap G_\alpha,$ for each $\alpha.$ Now, since $K$ is compact relative to $X,$ we have

$$ K \subset G_{\alpha_1} \cup \cdots \cup G_{\alpha_n}, \tag{a} $$

for some finite set of indices $a_1, \dots, a_n.$ Now, since $K \subset Y,$

$$ K \subset V_{\alpha_1} \cup \cdots \cup V_{\alpha_n}, \tag{b} $$

which shows $K$ is compact relative to $Y.$

Conversely, suppose $K$ is compact relative to $Y$ and let $\{G_\alpha\}$ be a cover of $K$ open relative to $X.$ We need to show there is a finite subset of $\{G_\alpha\}$ that covers $K.$ Let $V_\alpha = Y \cap G_\alpha,$ for each $\alpha.$ Then (b) will hold for some set of indicies, $\alpha_1, \dots, \alpha_n,$ and since each $V_\alpha \subset G_\alpha,$ (a) is implied by (b) and we've shown $K$ is compact relative to $X.$

We will proceed both sides of the implication by proving the contrapositive, i.e., that if the interval property doesn't hold, then the set isn't connected, and conversely, that if the set isn't connected, the interval property doesn't hold.

Suppose $x, y \in E$ and $z \in (x, y), z \notin E.$ Then $E = A_z \cup B_z,$ where

$$ A_z = E \cap (- \infty, z), \quad B_z = E \cap (z, \infty). $$

Since $x \in A_z$ and $y \in B_z,$ they are nonempty, and since $A_z \subset (- \infty, z)$ and $B_z \subset (z, \infty),$ they are separated. Therefore, $E$ is not connected.

Conversely, suppose, for the sake of contradiction, that $E$ is not connected. Then there are nonempty separated sets $A$ an $B$ such that $A \cup B = E.$ Let $x \in A, y \in B$ and assume $x < y.$ Define

$$ z = \sup{(A \cap [x, y])}. $$

By Let $E$ be a nonempty set of..., $z \in \overline{A},$ and because $A$ and $B$ are separated, $z \notin B.$ Therefore $x \leq z < y.$

If $z \notin A,$ it follows that $x < z < y,$ and $z \notin E.$

If $z \in A,$ then $z \notin \overline{B},$ hence there exists $z_1$ such that $z < z_1 < y$ and $z_1 \notin B$ (because $z \notin \overline{B}$ means there is a neighborhood of $z$ that contains no points of $B$.) Thus, $x < z_1 < y$ and $z_1 \notin E.$

Suppose $g \in G$ with $\text{ord}(g) = n$ and let $m \in \mathbb{N}$. Then, $\langle g^m \rangle = \langle g^{\gcd{(m,n)}} \rangle$.

So, $ \langle g \rangle = \langle g^m \rangle \iff \gcd{(m,n)} = 1$.

In more detail, say $G$ can be partitioned into $n = (G:H)$ cosets of $H$, where $n$ is a natural number. Let $A$ be a set of coset representatives $A \subseteq G$ such that $\mathcal{P} = \bigcup_{a_i \in A} a_i H$ is a partition of $G$ formed by the left cosets of $H$.

Then, $\mathcal{P}$ is

$$ \begin{align} \mathcal{P} = \{ & a_1 H, \\ & a_2 H, \\ & a_3 H, \\ & \cdots, \\ & a_n H\} \end{align}, \tag{a} $$

and each $a_i H$ is a disjoint coset of of $H$ in $G$.

Now, $H$ can be partitioned into $m = (H:K)$ cosets of $K$, where $m$ is a natural number. Let $B$ be a set of coset representatives $B \subseteq H$ such that $\mathcal{Q} = \bigcup_{b_i \in B} b_i K$ is a partition of $H$ formed by the left cosets of $K$.

Then, $\mathcal{Q} = \{ b_1 K, b_2 K, b_3 K, \cdots, b_m K\}$. Now, we can recover $H$ by taking the union of the elements of $\mathcal{Q}$, that is, $H = \bigcup_{q \in Q} q$, so the elements of $\mathcal{Q}$, after a round of flattening, are the same as those of $H.$ If we replace $H$ in (a) with $\mathcal{Q}$, distribute each $a_n$ group action over the elements of $\mathcal{Q}$, and then take the union across each resulting set we end up with

$$ \begin{align} \mathcal{R} = \{ & a_1 b_1 K, a_1 b_2 K, a_1 b_3 K, \cdots, a_1 b_m K, \\ & a_2 b_1 K, a_2 b_2 K, a_2 b_3 K, \cdots, a_2 b_m K, \\ & \cdots, \\ & a_3 b_1 K, a_3 b_2 K, a_3 b_3 K, \cdots, a_3 b_m K, \\ & a_n b_1 K, a_n b_2 K, a_n b_3 K, \cdots, a_n b_m K\}, \end{align} $$

which is $G$ partitioned into the left cosets of $K$. Because we constructed each element in $\mathcal{R}$ by partitioning the elements of a partition of $G$ into cosets of $K$ (then taking their union), we know they are disjoint and cover all of $G$. There are $nm = (G:H)(H:K) = (G:K)$ elements in $\mathcal{R}$, and since $n$ and $m$ are finite, so is $(G:K)$.

Suppose $X$ and $Y$ are sets with $X \subset Y$ and $Y \subset X.$ Now, suppose $x \in X.$ Then, $x \in Y.$ Conversely, suppose $y \in Y.$ Then $y \in X.$ Thus, $(\forall x)(x \in X \iff x \in Y),$ and $X = Y$.

Suppose $E$ is a bounded infinite subset of $R^k.$ Then, it is a subset of a $k$-cell $I \subset R^k,$ and because every $k$-cell is compact, $I$ is compact. Since @{infinite subsets of a compact set $K$ have a limit point in $K$, $E$ has a limit point in $I$ and therefore in $R^k.$

Suppose $E_n$ has $[\alpha, \beta]$ as an interval and thus $\alpha, \beta \in E_n.$ Then, by definition, $E_{n+1}$ will contain $[\alpha, \frac{\beta - \alpha}{3}]$ and $\frac{2(\beta - \alpha)}{3}, \beta]$ as intervals, so $\alpha, \beta \in E_{n+1}.$ Note that $E_0$ has $[0, 1]$ as an interval. By induction, all $E_n$ contain $0$ and $1$, and therefore so does their intersection $P,$ and $P$ is nonempty.

Let $x \in P.$ Let $r > 0,$ and pick $n \in \mathbb{N}$ to be large enough that $1/3^n < r.$ Then, $x$ lies in one of the $2^n$ intervals of length $1/3^n$ in $E_n;$ call it $I_n.$ The endpoints of $I_n$ are also in $P$ (see The Cantor set is not empty.,) and at least one of them is not $x.$ Since the endpoints are contained in a neighborhood of $x$ with radius $r > 0,$ all neighborhoods of $x$ are limit points of $P,$ and therefore $P$ is perfect.

Suppose $F \subset K \subset X,$ with $F$ closed relative to $X,$ and $K$ compact. Let $\{V_\alpha\}$ be an open cover of $F.$ Since $F^c$ is open relative to $X$ (see A set $E$ is open iff its...), if we add it to $\{V_\alpha\},$ we obtain an open cover of $K;$ let's call it $\Omega.$ Since $K$ is compact, we can obtain a finite subcover of $K$ by discarding all but a finite number of sets from $\Omega;$ let's call it $\Phi.$ Since $F \subset K,$ $\Phi$ is also a finite subcover of $F,$ and therefore $F$ is compact.

Suppose $K$ is compact relative to metric space $X.$ Let $p \in K^c.$ For each $q \in K,$ we can define $r_q = \frac{1}{2} d(p,q),$ and let $P_q$ and $Q_q$ be neighborhoods of radius $r_q$ around $p$ and $q,$ respectively. Note that $P_q$ and $Q_q$ are disjoint, because we defined their radii to be half the distance between them, and they are open.) Now, since $K$ is compact, we can pick a finite number of points in $K,$ $q_1, \dots, q_n,$ such that $K \subset Q_{q_1} \cup \cdots \cup Q_{q_n} = Q$ ($Q$ is a finite subcover of $K.$) Using the same set of points as reference, let $P_{q_1} \cap \cdots \cap P_{q_n} = P.$ Note that $P \cap Q = \{\},$ since each $P_q$ is disjoint with its paired $Q_q$ (to be in $P$, a point must be in all $P_q,$ but any point in $Q$ is not in at least one $P_q.$) $P$ is open, since it is the intersection of finitely many open sets (see (a) - For any collection $\{Ga\}$ of...) and obviously contains $p,$ since each $P_q$ contains $p$. Therefore, $p$ has a neigborhood $P$ that is disjoint with $K$ (since $P \cap K \subset P \cap Q = \{\}$), and is therefore an interior point of $K^c.$ It follows that $K^c$ is open, and that $K$ is closed.

Let $A$ and $B$ be sets. We want to show that

$$ (A \cap B)^c = A^c \cup B^c. $$

Suppose $x \in (A \cap B)^c.$ Then, $x$ is not in $A \cap B,$ that is, $x$ is either not in $A$ or it is not in $B$ or it is in neither. If $x$ is not in $A,$ then it is in $A^c,$ and therefore it is in $A^c \cup B^c.$ The same approach works with $B,$ and therefore $x \in A^c \cup B^c,$ and we have shown $(A \cap B)^c = A^c \cup B^c.$

Let $A$ and $B$ be sets. We want to show that

$$ (A \cup B)^c = A^c \cap B^c. $$

Suppose $x \in (A \cup B)^c.$ Then, if $x \in A$ or $x \in B,$ then $x \in A \cup B$ and $x \notin (A \cup B)^c,$ a contradiction. Therefore, $x \notin A$ and $x \notin B.$ That is, $x \in A^c$ and $x \in B^c,$ therefore $x \in A^c \cap B^c.$

Let $\{V_n\}$ be a countable base of $R^k$ (see $R^k$ is separable. and Every separable metric space has a countable...,) and let $W$ be the union of those $V_n$ for which $E \cap V_n$ is at most countable. We will show that $P = W^c.$

Suppose $p \in W^c.$ Then $p$ is in no $V_n$ for which $V_n \cap E$ is at most countable, that is, every neighborhood of $p$ has uncountably many points in $E,$ and thus $p \in P.$

Conversely, suppose $p \in P.$ Suppose, for the sake of contradiction, that $p \in W.$ Then $p \in V_n$ for some $V_n$ where $V_n \cap E$ is at most countable. But, since $p$ is an interior point of this $V_n,$ there is a neighborhood $N(p) \subset V_n,$ and since every neighborhood of $p$ has uncountably many points in $E,$ we have a contradiction, and thus our assumption that $p \in W$ must be incorrect, and therefore $p \in W^c,$ and $P = W^c.$ Furthermore, since $W$ is a union of open sets, $W$ is open, and $W^c = P$ is closed.

Since $W$ is open, only countably many $V_n$ are required to cover it. Each of these $V_n$ has at most countably many points in $E,$ so $W = P^c$ has at most countably many points in $E,$ that is, there are at most countably many points of $E$ that are not in $P.$

Now, to show all points in $P$ are limit points of $P,$ suppose $p \in P.$ Let $N_{r}(p), r > 0$ be a neighborhood of $p.$ Then, $N_{r}(p) \cap E$ is uncountable. Now, since there are at must countably many points in $E$ that are not in $P,$ there are at most countably many points in $(N_{r}(p) \cap E) \setminus P,$ and therefore there must be uncountably many points in $N_{r}(p) \cap E \cap P.$ Therefore, every neighborhood of $p$ contains infinitely many points in $P$ other than $p,$ $p$ is a limit point of $P,$ and $P$ is perfect.

The points of $R^k$ that have only rational coordinates are a subset of $R^k,$ we'll call it $Q^k,$ countable and dense.

We know that the rationals are countable, and because $n$-tuples of countable elements are countable, $Q^k$ is countable.

To show $Q^k$ is dense, consider an arbitrary point $p$ in $R^k.$ Now, let $\epsilon > 0.$ Since the the rationals are dense in the reals, we can pick a $q \in Q^k$ with $d(p, q) < \epsilon,$ by picking rational approximations of the coordinates of $p$ and forming $q$ such that $q$ is within $\epsilon$ of $p,$ i.e.

$$ |p_i - q_i| < \frac{\epsilon}{\sqrt{k}} \implies |p - q| < \sqrt{\sum_{i=1}^k (p_i - q_i)^2 } < \sqrt{k} \cdot \frac{\epsilon}{\sqrt{k}} = \epsilon. $$

Let $I$ be a $k$-cell, consisting of all points $\vec{x} = (x_1, \dots, x_k)$ such that $a_j \leq x_j \leq b_j, 1 \leq j \leq k.$ Let

$$ \delta = { \sqrt{\sum_{j=1}^k (b_j - a_j)^2} }, $$

i.e., the maximum distance between any two points in $I$ (the diagonal). Then for any points $\vec{x}, \vec{y} \in I, |\vec{x} - \vec{y}| \leq \delta.$

Suppose, for the sake of contradiction, that there is an open cover $\{G_\alpha\}$ of $I$ that contains no finite subcover of $I.$ Now, let $c_j = (a_j + b_j)/2,$ i.e. $c_j$ is the midpoint of $[a_j, b_j].$ We can subdivide $I$ into $2^k$ $k$-cells $Q_i,$ determined by the intervals $[a_j, c_j]$ and $[c_j, b_j].$ At least one $Q_i,$ call it $I_1,$ cannot be covered by any finite subcollection of $\{G_\alpha\},$ or else $I$ would have a finite subcover in $\{G_\alpha\}.$ We then can subdivide $I_1$ and so on, obtaining a sequence $\{I_n\}$ with the following properties:

(a) $I_{n+1} \subset I_{n}$ (each $k$-cell in the sequence is nested in the previous.)

(b) $I_n$ is not covered by any finite subset of $\{G_\alpha\}.$

(c) If $\vec{x}, \vec{y} \in I_n,$ then $|\vec{x} - \vec{y}| \leq 2^{-n} \delta.$

From (a) and If ${In}$ is a sequence of intervals..., there is some point $\vec{x}^*$ that is in every $I_n.$ For some $\alpha, x^* \in G_\alpha,$ since $\{G_\alpha\}$ is a cover of all of $I.$ $G_\alpha$ is open, so for some $r > 0, |\vec{y} - \vec{x^*}| < r$ implies that $y \in G_\alpha,$ that is, $\vec{x}$ has a neighborhood that lies entirely within $G_\alpha.$ If we make $n$ big enough, we have that $2^{-n} \delta < r,$ so by (c), $I_n \subset G_\alpha,$ that is, $I_n$ is entirely covered by $G_\alpha.$ But, this contradicts (b), so our provisional assumption is incorrect, and $\{G_\alpha\}$ must have a finite subcover that covers $I,$ and therefore $I$ is compact.

Let $\delta > 0,$ and pick $x_1 \in X.$ Now, continue picking $x_{j+1}$ such that $d(x_i, x_{j+1}) \geq \delta$ i.e. so that each new point is at least $\delta$ away from each existing point. Suppose this process continues infinitely; then the points $\{x_i\}$ are an infinite subset of $X,$ and thus must have a limit point $p.$ Now, because every neighborhood of $p$ must contain infinitely many points in ${x_i}$, we can let $r = \delta / 2,$ $N_r{p}$ be a neighborhood of $p,$ and pick $x, y \in \{x_i\}, N_r{p}.$ But, by the triangle inequality

$$ \begin{aligned} d(x, y) & \leq d(x, p) + d(p, y) \\ & < \delta / 2 + \delta / 2 \\ & = \delta \end{aligned} $$

so $d(x,y) < \epsilon,$ which contradicts our assumption that $\{x_i\}$ could be an infinite set where all points were at least $\delta$ apart. Therefore, $\{x_i\}$ has only finitely many points, and $X$ can be covered with finitely many open balls of radius $\delta$ (for if it couldn't be, we could always fit another point into $\{x_i\}.$)

Now, if we let $\delta = 1/n, n=1,2,3,\dots,$ we can consider the set of points $\{x_i\}_n$ as the finite set of points at least $1/n$ apart in $X.$ The collection of all such points can be called $\{x_{i_n}\}$ and is dense in $X:$ let $p$ be a point in $X.$ and $N_r{p}, r>0$ a neighborhood of $p.$ If we pick $n$ such that $1/n < r,$ then some $\{x_i\}_n$ will be in $N_r{p},$ because if not, we would have a contradiction with the fact that shown above that no point in $X$ is more than $1/n$ away from a point in $\{x_i\}_n.$

Since each $\{x_i\}_n$ is finite, and there are countably many $\{x_i\}_n,$ the $\{x_{i_n}\}$ is a countably dense subset of $X$ and therefore $X$ is separable.

Suppose $N_r(p)$ is a neighborhood in $X.$ Let $q \in N_r(p).$ We need to show that $q$ is an interior point of $N_r(p).$ Let $s = r - d(p, q);$ because $d(p, q) < r$, we have $s > 0.$ Now let $N_s(q)$ be the neighborhood of radius $s$ around $q.$ We need to show that $N_s(q) \subset N_r(p).$ Suppose $x \in N_s(q).$ First note that because $s = r - d(p, q),$ $d(p,q) = r - s.$ Now,

$$ \begin{aligned} d(p, x) & \leq d(p,q) + d(q, x) \\ & < r - s + s \\ & = r. \end{aligned} $$

Therefore, $N_s(q) \subset N_r(p),$ so $q$ is an interior point of $N_r(p),$ and since $q$ was arbitrary, every point of $N_r(p)$ is interior. Hence, $N_r(p)$ is open.

Let $X$ be a separable metric space, and let $x \in G \subset X,$ with $G$ open. Now, because $X$ is separable, it by definition has a countable dense subset $E.$ If $x \in E,$ we can pick a rational $\delta > 0$ and let $N_{\delta}(x)$ be a neighborhood such that $N_{\delta}(x) \subset G$ ($\delta$ must also be small enough such that this neighborhood is within $G,$ which is possible because $x$ is an interior point of $G$ and we can pick a rational as close to any real as we'd like.) Now we have that $x \in N_{\delta}(x) \subset G,$ and since there are countably many $x \in E$ and countably many neighborhoods with rational radius around each $x \in E,$ there are countably many such neighborhoods in $G.$ On the other hand, if $x \notin E,$ then $x$ is a limit point of $E,$ and thus there is $p \in E$ as close as we'd like to $x.$ Pick $p \in E$ such that $d(p,x) < \delta$ for some rational $\delta$ such that $x \subset N_{\delta}(p) \subset G.$ Again, since there are countably many such $p \in E,$ with countably many neighborhoods of rational radius each, there are countably many such neighborhoods in $G.$

Now, if we let $G$ be the union of all open sets $G_\beta \subset X,$ then $G$ is open, and let ${V_\alpha}$ be the union of all the $N_r(q), q \in G \cap E,$ with $r \in Q,$ then every $x \in G$ is in some ${V_\alpha},$ and there are countably many ${V_\alpha},$ so ${V_\alpha}$ is a countable base for $X.$

If (a) holds, then $E \subset I$ for some $k$-cell $I,$ and (b) follows from the facts that every $k$-cell is compact and closed subsets of compact sets are compact. Then, (c) follows from the fact that any infinite subset $K$ of a compact set $E$ has a limit point in $E$. To complete the cycle of implication, we must now show that (c) implies (a).

Assume, for the sake of contradiction, that $E$ is not bounded. Then, $E$ must contain an indexed set of points $\{x_n\}, n = 1, 2, 3, \dots$ where each $x_n$ must satisfy $|x_n| > n.$ $\{x_n\}$ is obviously infinite, but we will show it has no limit points in $E.$ Let $p \in E.$ Then for some positive integer $N,$ $|p| < N.$ Since $N$ is finite, there can only be finitely many points $\{q_\alpha\}$ in $\{x_n\}$ with $|x_n| < N.$ If $\{q_\alpha\}$ is empty, then $p$ is obviously not a limit point of $\{x_n\}.$ Otherwise, let $r = \min\{d(p, q_\alpha)\}.$ Then $r > 0$ and let $N_r\{p\}$ be a neighborhood of $p.$ Since no point in $\{x_n\},$ other than perhaps $p,$ lies in $\{x_n\},$ $p$ is clearly not a limit point of $\{x_n\}.$ Thus, our provisional assumption is invalid and (c) implies $E$ is bounded.

To show that (c) implies $E$ is closed, assume for the sake of contradiction that $E$ is not closed. Then, there is a point $x_0 \in R^k$ which is a limit point of $E$ but is not in $E.$ We will construct an infinite subset of $E$ and show that it has no limit point in $E.$ For $n = 1, 2, 3, \dots,$ let the point $x_n$ be some point in $E$ such that $|x_n - x_0| < 1/n;$ let $\{x_n\}$ be the set of such points. $\{x_n\}$ is certainly infinite, because $|x_n - x_0|$ will eventually be bigger than $1/n$ for some $n$ if we keep reusing the same $x_n$ infinitely many times. Now, $\{x_n\}$ has $x_0$ as a limit point, and we will show it is its only limit point in $R^k.$ Assume $y \in R^k, y \neq x_0.$ Then, via the triangle inequality,

$$ \begin{aligned} |x_n - y| & \geq |x_0 - y| - |x_n - x_0| \\ & \geq |x_0 - y| - \frac{1}{n} \\ & \geq \frac{1}{2} |x_0 - y| \end{aligned} $$

for all but finitely many $n,$ and thus $y$ is not a limit point of $\{x_n\}$ because its its neighborhoods do not contain infinitely many points of $\{x_n\}$. Thus, $\{x_n\}$ has no limit point in $E,$ which contradicts (c), and therefore our provisional assumption that $E$ is not closed is incorrect, and (c) implies that $E$ is closed.

Assume, for the sake of contradiction, that no point in $K$ is a limit point of $E.$ Then any point $q$ in $K$ has a neighborhood with at most one point in $E;$ $q,$ if $q \in E.$ Since $E$ is infinite, an infinite number of these singleton neighborhoods would be required to cover it, and therefore to cover $K,$ since $E \subset K.$ But, this contradicts our hypothesis that $K$ is compact. Therefore, our provisional assumption must be false, and $K$ must contain a limit point of $E.$

Let $I_n = [a_n, b_n],$ and let $E$ be the set of all $a_n.$ Then, $E$ is nonempty, because even if $a_n = b_n$ for all $n,$ it at least contains a single point. It is also bounded above by $b_1,$ since any $b_n$ is in $[a_1, b_1].$ Let $x = \sup E.$ Let $m$ and $n$ be positive integers and we have that

$$ a_n \leq a_{m+n} \leq b_{m+n} \leq b_m, $$

so that $x \leq b_m$ for each $m.$ Since $a_m \leq x,$ we have that $a_m \leq x \leq b_m,$ that is, $x \in I_m$ for all $m = 1, 2, 3, \dots,$ so $x \in \bigcap_{i=1}^n I_n$ and thus $\bigcap_{i=1}^n I_n$ is not empty.

We will proceed using proof by induction. First, for the base case, note that $B_1$ is the set of $1$-tuples formed by elements of $A$, so $B_1 = A$ and is thus countable. Now, for the inductive step, assume $B_{n-1}$ is countable $(n = 2, 3, 4, \dots).$ Then we have that

$$ B_n = \{(b,a) | b \in B_{n-1}, a \in A\} = \bigcup_{b \in B_{n-1}} ({b} \times A). $$

So, for any given $n-1$-tuple $b$, we form $n$-tuples by appending each element of $a$ to it, and so the set of pairs $(b,a)$ has the same cardinality as $A,$ and is thus countable. $B_n$ is thus the union of the countable set of countable sets (the set of sets formed by appending each element of $A$ to each element of $B_{n-1}$) and is therefore countable itself, by a theorem proved above. Therefore, by induction, every $B_n$ is countable.

Let $A$ and $B$ be boolean variables. We want to show that

$$ \neg (A \land B) \leftrightarrow \neg A \lor \neg B. $$

First, assume $\neg (A \land B)$ is true. Then, both $A \land B$ must be false, so either $A$ must be false or $B$ must be false, or both must be false. If $A$ is false, then $\neg A$ is true, and so is $\neg A \lor \neg B.$ The same is true if $B$ is false, so $\neg (A \land B) \rightarrow \neg A \lor \neg B.$

Now, assume $\neg A \lor \neg B$ is true. Then, either $\neg A$ or $\neg B$ must be true, so either $A$ or $B$ or both must be false. Now, if $A$ is false, then $A \land B$ is false. The same holds if $B$ is false, and thus $\neg (A \land B)$ is true. Therefore $\neg A \lor \neg B \rightarrow \neg (A \land B)$ and we have shown $ \neg (A \land B) \leftrightarrow \neg A \lor \neg B. $

Let $A$ and $B$ be boolean variables. We want to show that

$$ \neg (A \lor B) \leftrightarrow \neg A \land \neg B. $$

First, assume $\neg (A \lor B)$ is true. Then, $A \lor B$ is false. If $A$ were true, then we'd have a contradiction, and similarly with $B,$ so both $A$ and $B$ must be false, that is, $\neg A$ and $\neg B$ most both be true, and $\neg (A \lor B) \rightarrow \neg A \land \neg B.$

Now, assume $\neg A \land \neg B.$ Here, both $\neg A$ and $\neg B$ must be true, so both $A$ and $B$ must be false. Therefore, $A \lor B$ is false, so $\neg A \land \neg B \rightarrow \neg (A \lor B)$ and we have shown $\neg (A \lor B) \leftrightarrow \neg A \land \neg B.$

Let $p$ be a limit point of $E$ and let $N_r(p)$ be a neighborhood of $p.$ Suppose that $N_r(p)$ contains only finitely many points of $E.$ Since we have finitely many points, we can inspect each and find the minimum distance from $p$ to any point in $N_r(p) \bigcup E\setminus\{p\}$ and call it $s.$ Now, we can make a new neighborhood $N_s(p),$ which contains none of the points in $N_r(p) \bigcup E\setminus\{p\}$ since they're all at least $s$ away from $p,$ by construction. But then, $p$ is not a limit point of $E,$ since it has a neighborhood that contains no points of $E\setminus\{p\}$ Therefore, we have a contradiction, and $N_r(p)$ must therefore contain infinitely many points.

We know that $P$ is infinite, because by definition, all points in perfect sets are limit points, and only infinite sets have limit points.

Suppose, for the sake of contradiction, that $P$ is countable. Label the points of $P$ as $x_1, x_2, \dots.$ We will construct a sequence of ${V_n}$ of neighborhoods.

As a base step, let $V_1$ be any neighborhood of $x_1;$ let $V_1 = \{ y \in R^k | ~ |y - x_1| < r \}$ (note: subsequent $V_{n+1}$ aren't required to be neighborhoods of $x_{n+1}.$) Then the closure $\overline{V_1}$ of $V_1$ is $\overline{V_1} = \{ y \in R^k | ~ |y - x_1| \leq r \}.$

For the inductive step, suppose as an induction hypothesis that we have some $V_n$ that's been constructed such that $V_n \cap P$ is not empty. Since every point of $P$ is a limit point of $P,$ we can make a neighborhood $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_n,$ (ii) $x_n \notin \overline{V_{n+1}},$ (iii) $V_{n+1} \cap P$ is not empty. Now, $V_{n+1}$ satisfies our induction hypothesis, and since $V_1$ does too, we have ${V_n}$ defined for all $n = 1, 2, 3, \dots.$

For each $n$, let $K_n = \overline{V_n} \cap P.$ Since $\overline{V_n}$ is closed and bounded, $\overline{V_n}$ is compact. Since $x_n \notin \overline{V_{n+1}},$ no point of $P$ lies in $\bigcap_{n=1}^\infty K_n.$ Since $K_n \subset P,$ this implies that $\bigcap_{n=1}^\infty K_n$ is empty. But, each $K_n$ is nonempty, by (iii), and $K_{n+1} \subset K$, by (i). But the intersection of nonempty compact nested sets is nonempty, so we have a contradiction, so our provisional assumption that $P$ is countable must be incorrect. Therefore, $P$ is uncountable.

Let $G_\alpha = K_\alpha^c$ for each $\alpha,$ and note that since $K_\alpha$ is compact and therefore closed, $G_\alpha$ is open. Then, fix a member $K_1$ of $\{K_\alpha\}.$ Assume, for contradiction's sake, that no point of $K_1$ is in all $K_\alpha,$ that is, that $\bigcap K_\alpha = \emptyset.$ Then, any point $x \in K_1$ is in some $K_\alpha^c = G_\alpha,$ so $\{G_\alpha\}$ forms an open cover of $K_1.$ Since $K_1$ is compact, some finite subset $G_{\alpha_1}, \dots, G_{\alpha_n}$ of $\{G_\alpha\}$ forms a finite subcover of $K_1$ such that $K_1 \subset G_{\alpha_1} \cup \cdots \cup G_{\alpha_n} = \left ( K_{\alpha_1} \cap \cdots \cap K_{\alpha_n} \right )^c$ (by De Morgan's.) Therefore $K_1 \cap K_{\alpha_1} \cap \cdots \cap K_{\alpha_n} = \emptyset.$ This is an empty intersection of a finite subcollection of $\{K_\alpha\},$ which contradicts our hypothesis that all finite intersections are nonempty. Therefore, our assumption that no point in $K_1$ is in all $K_\alpha$ is incorrect, and some point in $K_1$ is in all $K_\alpha,$ and therefore $\bigcap K_\alpha$ is not empty.

First, consider the case that $E^c$ is empty, and therefore open. If $E^c$ has no limit points, it is vacuously closed. Suppose $E^c$ has a limit point $x.$ Since $E$ is empty, $x$ must be in $E^c,$ therefore $x$ is closed. Now, consider the case that $E^c$ is empty, and therefore closed. If $E$ is empty, it is open, and the theorem is satisfied. If $E$ is not empty, a point $x$ in $E$ has only points in $E$ in any neighborhood, since all points are in $E,$ and therefore $E$ is open.

Now we deal with the cases where neither $E$ nor $E^c$ are empty.

Now, let $E^c$ be closed. Let $x \in E.$ Since $E^c$ is closed, $x$ is not a limit point of $E^c,$ that is $x$ has some neighborhood that doesn't contain a point in $E^c$ and must therefore be a subset of $E.$ Therefore, $x$ is an interior point of $E,$ and $E$ is open.

Conversely, assume $E$ is open. Let $x$ be a limit point of $E^c.$ Suppose, for the sake of contradiction, that $x \in E.$ Then, since $E$ is open, $x$ is an interior point of $E$ and has some neighborhood that is a subset of $E.$ This is a contradiction, since every neighborhood of $x$ must contain at least one point of $E^c$ to be a limit point of $E^c.$ Therefore, $x$ must be in $E^c,$ and it follows that $E^c$ is closed.

Suppose $E$ is open relative to $Y.$ Then, for every $p \in E,$ there is some $r_p > 0$ such that $d(p, q) < r_p,$ $q \in X$ implies that $q \in Y.$ Let $V_p$ be the set of all $q \in X$ where $d(p,q) < r_p$ (and thus $q \in Y$) and let

$$ G = \bigcup_{p \in E} V_p. $$

Then, since each $V_p$ is an open subset of $X,$ so is $G.$ Now, since $p \in V_p$ for each $p \in E,$ $E \subset G \cap Y.$ Also, since $V_p \cap Y \subset E$ for every $p \in E,$ $G \cap Y \subset E,$ and $E = G \cap Y.$

Conversely, suppose $E = Y \cap G$ for some open subset $G$ of $X.$ Now, suppose $p \in E.$ Then, $p \in G,$ and there is some neighborhood $V_p \subset G.$ Then, $V_p \cap Y \subset E,$ so $E$ is open relative to $Y.$

Let $x \in \mathbb{R}$ and $\epsilon > 0.$ Pick $n \in \mathbb{N}$ such that $1/n < \epsilon.$ Now, because the reals are Archimedean, for some $m \in \mathbb{Z}$ we have that $m = \lfloor nx \rfloor.$ Then,

$$ m \leq nx \leq m+1, \quad \frac{m}{n} \leq x \leq \frac{m+1}{n}. $$

Now, if we let $q = \frac{m}{n},$ we have

$$ 0 \leq x - q < \frac{1}{n} < \epsilon, $$

so $|x - q| < \epsilon.$ Therefore, every neighborhood of $x$ contains some $q \in \mathbb{Q},$ and so $x$ is either a limit point of $\mathbb{Q}$ or else $x \in \mathbb{Q},$ and so the rationals are therefore dense in the reals.

(a) Suppose $p \in X$ and $p \notin \overline{E}.$ Then $p$ is not in $E$ and is not in $E',$ and is in fact in $\overline{E}^c.$ Now, since $p$ is not a limit point of $E,$ it has some neighborhood $N_r(p)$ that does not intersect $E.$ Any point $x$ in $N_r(p)$ is an interior point of $N_r(p)$, and therefore has its own neighborhood $N_\epsilon{(x)}$ that does not intersect $E,$ and therefore $x$ is not a limit point of $E.$ Thus, any point in $\overline{E}^c$ is an interior point of $\overline{E}^c,$ and $\overline{E}^c$ is therefore open and its complement, $\overline{E},$ is closed (since a set is open iff its complement is closed.)

(b) Suppose $E$ is closed. Then it contains its limit points, so $E = E \cup E' = \overline{E}.$ Conversely, suppose $E = \overline{E}.$ By (a), $E$ is closed.

(c) Suppose that $E \subset F \subset X,$ and that $F$ is closed. Suppose $p \in \overline{E}.$ If $p \in E,$ then $p \in F$ because $E \subset F.$ If $p \in E',$ then it must be in $F$ also, since $F$ contains all points of $E,$ and is closed, and thus must contain the limit points of $E.$

First, note that since $R^k = \bigcup_{n=1}^{\infty}F_n,$ every point in $R^k$ is in some $F_n,$ and so $\bigcap_{n=1}^\infty F_n^c$ must be empty.

Suppose, for the sake of contradiction, that no $F_n$ has a non-empty interior, that is, every $F_n$ is closed with an empty interior. Then, each $F_n^c$ is open. Moreover, since $F_n$ has an empty interior, it has no points for which there exists a neighborhood that contains only points in $F_n,$ that is, every neighborhood of each point of $F_n$ contains a point in $F_n^c,$ so every point in $F_n$ is a limit point of $F_n^c,$ and every point in $R^k$ is either in $F_n^c$ or is a limit point of $F_n^c,$ so each $F_n^c$ is non-empty and dense in $R^k.$

As a base step, let $x_1$ be some point in $R^k,$ and let $B_1$ be an open ball around $x_1.$ Since the interior of $F_1$ is empty, $B_1 \cap F_1^c$ is not empty, and is open.

For the inductive step, suppose we have some $B_n$ that's constructed such that $B_n \cap F_n^c$ is non-empty and open. Then, we can pick a point $x_{n+1} \in B_n \cap F_n^c, x_{n+1} \neq x_n,$ and make a ball $B_{n+1}$ around it such that $B_{n+1} \subset B_n \cap F_{n}^c,$ $x_n \notin B_{n+1},$ and $B_{n+1} \cap F_{n+1}^c$ is open and not empty. Now, since $B_{n+1}$ satisfies our induction hypothesis, and since $B_1$ does as well, we have $B_n$ defined for all $n = 1, 2, 3, \dots.$

Since the set of points $\{x_n\}$ is infinite (by induction) and bounded (all points are within $B_1$,) it has a limit point $p$ in $R^k.$. Now, suppose, for contradiction, that $p$ is not in every $F_n^c.$ Then, for some $N,$ $p$ is not in $F_N^c.$ Since $p \notin F_N^c$ and $B_{N+1} \subseteq F_N^c$, we have $p \notin B_{N+1}$. Since $B_{N+1}$ is open and $p$ is outside it, $p$ is at some positive distance $\varepsilon$ from any point $q \in B_{N+1}$. But $x_n \in B_n \subseteq B_{N+1}$ for all $n > N$, so all these infinitely many points are at distance at least $\varepsilon$ from $p$. Thus any neighborhood of $p$ with radius less than $\varepsilon$ contains at most finitely many points of $\{x_n\}$, contradicting that $p$ is a limit point.

Now, this means that $p \in \bigcap_{n=1}^\infty F_n^c,$ meaning $\bigcap_{n=1}^\infty F_n^c \neq \emptyset,$ a contradiction! Therefore, our supposition that every $F_n$ has an empty interior must be incorrect, and some $F_n$ must have a non-empty interior.

Suppose $y \in E.$ Then $y \in \overline{E}.$ Suppose $y \notin E.$ Now, by hypothesis, for every $h > 0,$ there is some $x$ such that $y - h < x < y,$ because otherwise, $x$ would be an upper bound on $E.$ Therefore, every neighborhood $N_h(y)$ contains some $x \in E,$ and thus $y$ is a limit point of $E$ and $y \in \overline{E}.$

Let $G = \bigcup_{\alpha} G_\alpha, x \in G.$ Then $x$ is in some $G_\alpha$ for some $\alpha,$ and is an interior point of that $G_\alpha,$ since $G_\alpha$ is open. Therefore, $x$ has some neighborhood that is a subset of $G_\alpha$ and therefore of $G,$ so $x$ is an interior point of $G$ and $G$ is open - this shows (a).

Note that

$$ (\bigcap_{\alpha} F_\alpha)^c = \bigcup_{\alpha} F_\alpha^c, \tag{e} $$

and by (a) above, (e) is open. Then its complement, $\bigcap{\alpha} F_\alpha$ is closed, and we've shown (b).

Now, let $x$ be in $G = \bigcap_{i=1}^n G_i,$ so $x$ is in every $G_i,$ and has a neighborhood $N_i$ in every $G_i$ with radius $r_i > 0.$ Let $r$ be $\min\{r_1, \dots, r_n\}.$ Then, $x$ has a neighborhood $N$ of radius $r$ in every $G_i,$ and thus in $G,$ so $x$ is an interior point of $G,$ and we've shown (c).

Now, $(\bigcup_{i=1}^n F_i)^c = \bigcap_{i=1}^n F_i^c$ is open by (c), so its complement, $\bigcup_{i=1}^n F_i$ is closed, and we've shown (d).

We can construct an infinite array where the rows are sequence constructed by the sets that make up the entries of $\{E_n\}.$ Then, we can create a single sequence from all the entries of the sets of $\{E_n\}$ by iterating over them in the following order:

sequence = []
for i in range(1, k):
    for j in range(0, i):
        n = i - j
        m = j + 1
        sequence.append(f"E_{n},{m}")

Which, for $k = 5,$ yields

$$ E_{1,1}, E_{2,1}, E_{1,2}, E_{3,1}, E_{2,2}, E_{1,3}, E_{4,1}, E_{3,2}, E_{2,3}, E_{1,4}. $$

This sequence may contain duplicates, so some indices may need to be skipped in constructing a subset $T$ of the positive integers such that $T ~ S,$ but we've now shown that $S$ is at most countable. To show $S$ is infinite and therefore countable, note that the infinite set $E_1$ is a subset of $S,$, and therefore $S$ is infinite and countable.