An Introduction to Analysis/Sequences of numbers

The chapter begins with sequences of real and complex numbers and the question of their convergence and connection with continuity and such. Following the classical topics and the discussion of first and second countability we introduce notions of measure and semi-norms.

Groups

A group is a triple $(G, \circ, id)$ consisting of:

• (1) a set $G$
• (2) a composition law such that for every $f$ and $g$ in $G$, $g \circ f$ in $G$.
• (3) the identity $id \in G$; i.e., for every $f$ in $G$, $f \circ id = f = id \circ f$ in G.

subject to the following conditions:

• For every $f$ in $G$, $f$ is invertible; i.e., there exists some $g$ in $G$, called the inverse of f, such that $g \circ f = f = f \circ g$.
• The composition law $\circ$ is associative; that is, for every $f$, $g$, $h$ in $G$, $h \circ (g \circ f)$ = $(h \circ g) \circ f$.

A group is said to be abelian if $f \circ g = g \circ f$.

2 Theorem Let $S_n$ be the set of all bijections from ${1, 2, ... n}$ into itself. Then $S_n$ forms a group under composition.
Proof: Clear.

The group in the theorem is called a symmetric group and its elements permutations of $n$ objects. The sign of a permutation $\sigma$ is defined to be 0 if a permutation is identity, even if it can be written as the composition of an even number of permutations and odd otherwise.

We remark that there cannot be a bijection defined on a finite set that is not a permutation at the same time. For one thing, since we define that a function is single-valued, the range of a function defined on any finite set must be finite. If it is injective additionally the number of elements of its range equals to that of elements in the domain.

A ring $F$ is an abelian group $((G, +, 0), \circ, 1)$ consisting of an abelian group, the multiplication and the identity with the distribution law; i.e.,

• $h \circ (f + g) = (h \circ f) + (h \circ g)$.

A field is a ring, all of whose non-zero elements are invertible.

We assume in every field $0 \ne 1$. This is an example of a theorem that holds since a field is an abelian group.

A sequence in a set $E$ is a function $f: \mathbb{N} \to E$

2 Theorem Every infinite subset $A$ of a countable set $B$ is countable.
Proof: Since $B$ is countable, we can find a sequence in $B$. Since $A \subset B$, there is a subsequence of $B$ which is a bijection from $\mathbb{N} \to B$. $\square$

Real and complex numbers

In this chapter, we work with a number of subtleties like completeness, ordering, Archimedian property; those are usually never seriously concerned when Calculus was first conceived. I believe that the significance of those nuances becomes clear only when one starts writing proofs and learning examples that counter one's intuition. For this, I suggest a reader to simply skip materials that she does not find there is need to be concerned with; in particular, the construction of real numbers does not make much sense at first glance, in terms of argument and the need to do so. In short, one should seek rigor only when she sees the need for one. WIthout loss of continuity, one can proceed and hopefully she can find answers for those subtle questions when she search here. They are put here not for pedagogical consideration but mere for the later chapters logically relies on it.

We denote by $\mathbb{N}$ the set of natural numbers. The set $\mathbb{N}$ does not form a group, and the introduction of negative numbers fills this deficiency. We leave to the readers details of the construction of integers from natural numbers, as this is not central and menial; to do this, consider the pair of natural numbers and think about how arithmetical properties should be defined. The set of integers is denoted by $\mathbb{Z}$. It forms an integral domain; thus, we may define the quotient field $\mathbb{Q}$ of $\mathbb{Z}$, that is, the set of rational numbers, by

$\mathbb{Q} = \left \{ {a \over b} : \mathit{b} \mbox{ are integers and }\mathit{b}\mbox{ is nonzero }\right \}$.

As usual, we say for two rationals $x$ and $y$ $x < y$ if $x - y$ is positive.

We say $E \subset \mathbb{Q}$ is bounded above if there exists some $x$ in $\mathbb{Q}$ such that for any $y \in E$ $y \le x$, and is bounded below if the reversed relation holds. Notice $E$ is bounded above and below if and only if $E$ is bounded in the definition given in the chapter 1.

The reason for why we want to work on problems in analysis with $\mathbb{R}$ instead of is a quite simple one:

2 Theorem Fundamental axiom of analysis fails in $\mathbb{Q}$.
Proof: $1, 1.4, 1.41, 1.414, ... \to 2^{1/2}$. $\square$

How can we create the field satisfying this axiom? i.e., the construction of the real field. There are several ways. The quickest is to obtain the real field $\mathbb{R}$ by completing $\mathbb{Q}$.

We define the set of complex numbers $\mathbb{C} = \mathbb{R} / <i^2 + 1>$ where $i$ is just a symbol. That $i^2 + 1$ is irreducible says the ideal generated by it is maximal, and the field theory tells that $C$ is a field. Every complex number $z$ has a form:

$z = a + bi + <i^2 + 1>$.

Though the square root of -1 does not exist, $i^2$ can be thought of as -1 since $i^2 + <i^2 + 1> = -1 + 1 + i^2 + <i^2 + 1> = -1$. Accordingly, the term $<i^2 + 1>$is usually omitted.

2 Exercise Prove that there exists an irrational numbers $a, b$ such that $a^b$ is rational.

Sequences

2 Theorem Let $s_n$ be a sequence of numbers, be they real or complex. Then the following are equivalent:

• (a) $s_n$ converges.
• (b) There exists a cofinite subsequence in every open ball.

Theorem (Bolzano-Weierstrass) Every infinite bounded set has a non-isolated point.
Proof: Suppose $S$ is discrete. Then $S$ is closed; thus, $S$ is compact by Heine-Borel theorem. Since $S$ is discrete, there exists a collection of disjoint open balls $S_x$ containing $x$ for each $x \in S$. Since the collection is an open cover of $S$, there exists a finite subcover $\{ S_{x_1}, S_{x_2}, ..., S_{x_n} \}$. But

$S \cap (\{ S_{x_1} \cup S_{x_2} \cup ... S_{x_n} \} = \{ x_1, x_2, ... x_n \}$

This contradicts that $S$ is infinite. $\square$.

2 Corollary Every bounded sequence has a convergent subsequence. Proof:

The definition of convergence given in Ch 1 is general enough but is usually inconvenient in writing proofs. We thus give the next theorem, which is more convenient in showing the properties of the sequences, which we normally learn in Calculus courses.

In the very first chapter, we discussed sequences of sets and their limits. Now that we have created real numbers in the previous chapter, we are ready to study real and complex-valued sequences. Throughout the chapter, by numbers we always means real or complex numbers. (In fact, we never talk about non-real and non-complex numbers in the book, anyway)

Given a sequence $a_n$ of numbers, let $E_j = \{ a_j, a_{j+1}, \cdot \}$. Then we have:

$\limsup_{j \rightarrow \infty} a_n$	$= \limsup_{j \rightarrow \infty} E_j$
	$= \bigcap_{j=1}^{\infty}\bigcup_{k=j}^{\infty}E_k$
	$= \bigcap_{j=1}^{\infty} \sup \{ a_k, a_{k+1}, ... \}$

The similar case holds for liminf as well.

Theorem Let $s_j$ be a sequence. The following are equivalent:

• (a) The sequence $s_j$ converges to $s$.
• (b) The sequence $s_j$ is Cauchy; i.e., $|s_n - s_m| \to 0$ as $n, m \to \infty$.
• (c) Every convergent subsequence of $s_j$ converges to $s$.
• (d) $\limsup_{j \to \infty} s_j = \liminf_{j \to \infty} s_j$.
• (e) For each $\epsilon > 0$, we can find some real number $N$ (i.e., N is a function of $\epsilon$) so that

  $|s_j - s| < \epsilon$ for $j \ge N$.

Proof: From the triangular inequality it follows that:

$|s_n - s_m| \le |s_n - n| + |s_m - m|$.

Letting $n, m \to \infty$ gives that (a) implies (b). Suppose (b). The Bolzano-Weierstrass theorem states that every bounded sequence has a convergent subsequence, say, $s_k$. Thus,

$|s_n - s|$	$= |s_n - s_m + s_m - s|$
	$\le |s_n - s_m| + |s_m - s|$
	$< \epsilon \ 2 + \epsilon \ 2 = \epsilon$

. Therefore, $s_j \to s$. That (c) implies (d) is obvious by definition.

2 Theorem Let $x_j$ and $y_j$ converge to $x$ and $y$, respectively. Then we have:

(a) $\lim_{j \to \infty} (x_j + y_j) = x + y$.

(b) $\lim_{j \to \infty} (x_jy_j) = xy$.

Proof: Let $\epsilon > 0$ be given. (a) From the triangular inequality it follows:

$|(x_j + y_j) - (x + y)| = |x_j - x| + |y_j - y| < {\epsilon \over 2} + {\epsilon \over 2} = \epsilon$

where the convergence of $x_j$ and $y_j$ tell that we can find some $N$ so that

$|x_j - x| < {\epsilon \over 2}$ and $|y_j - y| < {\epsilon \over 2}$ for all $j > N$.

(b) Again from the triangular inequality, it follows:

$|x_j y_j - xy|$	$= |(x_j - x)(y_j - y + y) + x(y_j - y)|$
	$\le |x_j - x|(1 + |y|) + |x| |y_j - y|$
	$\to 0$ as $j \to \infty$

where we may suppose that $|y_j - y| < 1$. $\square$

Other similar cases follows from the theorem; for example, we can have by letting $y_j = \alpha$

$\lim_{j \to \infty} (kx_j) = k \lim_{j \to \infty} x_j$.

2.7 Theorem Given $\sum a_n$ in a normed space:

• (a) $\sum \left \Vert a_n \right \Vert$ converges.
• (b) The sequence $\left \Vert a_n \right \Vert^{1 \over n}$ has the upper limit $a^*< 1$. (Archimedean property)
• (c) $\prod (a_n + 1)$ converges.

Proof: If (b) is true, then we can find a $N > 0$ and $b$ such that for all $n \ge N$:

$\left\| a_n \right\|^{1 \over n} < b < 1$. Thus (a) is true since:,

$\sum_1^{\infty} \left \Vert a_n \right \Vert = a + \sum_N^{\infty} \left \Vert a_n \right \Vert \le a + \sum_0^\infty b^n = a + {1 \over 1 - b}$ if $a = \sum_1^{N-1}$.

Continuity

Let $f:\Omega \to \mathbb{R} \cup \{\infty, -\infty\}$. We write $\{ f > a \}$ to mean the set $\{ x : x \in \Omega, f(x) > a \}$. In the same vein we also use the notations like $\{f < a\}, \{f = a\}$, etc.

We say, $u$ is upper semicontinuous if the set $\{ f < c \}$ is open for every $c$ and lower semicontinuous if $-f$ is upper semicontinuous.

2 Lemma The following are equivalent.

• (i) $u$ is upper semicontinuous.
• (ii) If $u(z) < c$, then there is a $\delta > 0$ such that $\sup_{|s-z|<\delta} f(s) < c$.
• (iii) $\limsup_{s \to z} u(s) \le u(z)$.

Proof: Suppose $u(z) < c$. Then we find a $c_2$ such that $u(z) < c_2 < c$. If (i) is true, then we can find a $\delta > 0$ so that $\sup_{|s-z|<\delta} u(s) \le c_2 < c$. Thus, the converse being clear, (i) $\iff$ (ii). Assuming (ii), for each $\epsilon > 0$, we can find a $\delta_\epsilon > 0$ such that $\sup_{|s-z|<{\delta_\epsilon}}u(s) < u(z) + \epsilon$. Taking inf over all $\epsilon$ gives (ii) $\Rightarrow$ (iii), whose converse is clear. $\square$

2 Theorem If $u$ is upper semicontinuous and $< \infty$ on a compact set $K$, then $u$ is bounded from above and attains its maximum.
Proof: Suppose $u(x) < \sup_K u$ for all $x \in K$. For each $x \in K$, we can find $g(x)$ such that $u(x) < g(x) < \sup_K u$. Since $u$ is upper semicontinuous, it follows that the sets $\{u < g(x)\}$, running over all $x \in K$, form an open cover of $K$. It admits a finite subcover since $K$ is compact. That is to say, there exist $x_1, x_2, ..., x_n \in K$ such that $K \subset \{u < g(x_1)\} \cup \{u < g(x_1)\} \cup ... \{u < g(x_n)\}$ and so:

$\sup_K u \le \max \{ g(x_1), g(x_2), ..., g(x_n) \} < \sup_K u$,

which is a contradiction. Hence, there must be some $z \in K$ such that $u(z) = \sup_K u$. $\square$

If $f$ is continuous, then both $f^{-1} ( (\alpha, \infty) )$ and $f^{-1} ( (-\infty, \beta) )$ for all $\alpha, \beta \in \mathbb{R}$ are open. Thus, the continuity of a real-valued function is the same as its semicontinuity from above and below.

2 Lemma A decreasing sequence of semicontinuous functions has the limit which is upper semicontinuous. Conversely, let $f$ be upper semicontinuous. Then there exists a decreasing sequence $f_j$ of Lipschitz continuous functions that converges to f.
Proof: The first part is obvious. To show the second part, let $f_j(x) = \inf_{y \in E} { f(x) + j^{-1} |y - x| }$. Then $f_j$ has the desired properties since the identity $|f_j(x) - f_j(y)| = j^{-1} |x - y|$. $\square$

We say a function is uniform continuous on $E$ if for each $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x, y \in E$ with $|x - y| < \delta$ we have $|f(x) - f(y)|$.

Example: The function $f(x) = x$ is uniformly continuous on $\mathbb{R}$ since

$|f(x) - f(y)| = |x - y| < \delta = \epsilon$

while the function $f(x) = x^2$ is not uniformly continuous on $\mathbb{R}$ since

$|f(x) - f(y)| = |x^2 - y^2| = |x - y||x + y|$

and $|x + y|$ can be made arbitrary large.

2 Theorem Let $f: K \to \mathbb{R}$ for $K \subset \mathbb{R}$ compact. If $f$ is continuous on $K$, then $f$ is uniformly continuous on $K$.
Proof:

The theorem has this interpretation; an uniform continuity is a global property in contrast to continuity, which is a local property.

2 Corollary A function is uniform continuous on a bounded set $E$ if and only if it has a continuos extension to $\overline{E}$.
Proof:

2 Theorem if the sequence $\{ f_j \}$ of continuos functions converges uniformly on a compact set $K$, then the sequence $\{ f_j \}$ is equicontinuous.
Let $\epsilon > 0$ be given. Since $f_j \to f$ uniformly, we can find a $N$ so that:

$|f_j(x) - f(x) | < \epsilon / 3$ for any $j > N$ and any $x \in K$.

Also, since $K$ is compact, $f$ and each $f_j$ are uniformly continuous on $K$ and so, we find a finite sequence $\{ \delta_0, \delta_1, ... \delta_N \}$ so that:

$|f(x) - f(y)| < \epsilon / 3$ whenever $|x - y| < \delta_0$ and $x, y \in E$,

and for $i = 1, 2, ... N$,

$|f_i(x) - f_i(y)| < \epsilon$ whenever $|x - y| < \delta_i$ and $x, y \in E$.

Let $\delta = \min \{ \delta_0, \delta_1, ... \delta_n \}$, and let $x, y \in K$ be given. If $j \le N$, then

$|f_j(x) - f_j(y)| < \epsilon$ whenever $|x - y| < \delta$.

If $j > N$, then

$|f_j(x) - f_j(y)|$	$\le |f_j(x) - f(x)| + |f(x) - f(y)| + |f(y) - f_j(y)|$
	$< \epsilon$

whenever $|x - y| < \delta$. $\square$

2 Theorem A sequence $f_j$ of complex-valued functions converges uniformly on $E$ if and only if $\sup_E | f_n - f_m | \to 0$as $n, m \to \infty$
Proof: Let $\| \cdot \| = \sup_E | \cdot |$. The direct part follows holds since the limit exists by assumption. To show the converse, let $f(x) = \lim_{j \to \infty} f_j(x)$ for each $x \in E$, which exists since the completeness of $\mathbb{C}$. Moreover, let $\epsilon > 0$ be given. Since from the hypothesis it follows that the sequence $\|f_j\|$ of real numbers is Cauchy, we can find some $N$ so that:

$\|f_n - f_m\| < \epsilon / 2$ for all $n, m > N$.

The theorem follows. Indeed, suppose $j > N$. For each $x$, since the pointwise convergence, we can find some $M$ so that:

$\|f_k(x) - f(x)\| < \epsilon / 2$ for all $k > M$.

Thus, if $n = \max \{ N, M \} + 1$,

$|f_j(x) - f(x)| \le |f_j(x) - f_n(x)| + |f_n(x) - f(x)| < \epsilon.$ $\square$

2 Corollary A numerical sequence $a_j$ converges if and only if $|a_n - a_m| \to 0$ as $n, m \to \infty$.
Proof: Let $f_j$ in the theorem be constant functions. $\square$.

2 Theorem (Arzela) Let $K$ be compact and metric. If a family $\mathcal{F}$ of real-valued functions on K bestowed with $\| \cdot \| = \sup_K | \cdot |$ is pointwise bounded and equicontinuous on a compact set $K$, then:

• (i) $\mathcal{F}$ is uniformly bounded.
• (ii) $\mathcal{F}$ is totally bounded.

Proof: Let $\epsilon > 0$ be given. Then by equicontinuity we find a $\delta > 0$ so that: for any $x, y \in K$ with $x \in B(\delta, y)$

$|f(x) - f(y)| < \epsilon / 3$

Since $K$ is compact, it admits a finite subset $K_2$ such that:

$K \subset \bigcup_{x \in K_2} B(\delta, x)$.

Since the finite union of totally and uniformly bounded sets is again totally and uniformly bounded, we may suppose that $K_2$ is a singleton $\{ y \}$. Since the pointwise boundedness we have:

$|f(x)| \le |f(x) - f(y)| + |f(y)| \le \epsilon / 3 + |f(y)| < \infty$ for any $x \in K$ and $f \in \mathcal{F}$,

showing that (i) holds. To show (ii) let $A = \cup_{f \in \mathcal{F}} \{ f(y) \}$. Then since $\overline{A}$ is compact, $A$ is totally bounded; hence, it admits a finite subset $A_2$ such that: for each $f \in \mathcal{F}$, we can find some $g(y) \in A_2$ so that:

$|f(y) - g(y)| < \epsilon / 3$.

Now suppose $x \in K$ and $f \in \mathcal{F}$. Finding some $g(y)$ in $A_2$ we have:

$|f(x) - g(x)| \le |f(x) - f(y)| + |f(y) - g(y)| + |g(y) - g(x)| < \epsilon / 3$.

Since there can be finitely many such $g$, this shows (ii). $\square$

2 Corollary (Ascoli's theorem) If a sequence $f_j$ of real-valued functions is equicontinuous and pointwise bounded on every compact subset of $\Omega$, then $f_j$ admits a subsequence converging normally on $\Omega$.
Proof: Let $K \subset \Omega$ be compact. By Arzela's theorem the sequence $f_j$ is totally bounded with $\| \cdot \| = \sup_K | \cdot |$. It follows that it has an accumulation point and has a subsequence converging to it on $K$. The application of Cantor's diagonal process on an exhaustion by compact subsets of $\Omega$ yields the desired subsequence. $\square$

2 Corollary If a sequence $f_j$ of real-valued $\mathcal{C}^1$ functions on $\Omega$ obeys: for some $c \in \Omega$,

$\sup_j |f_j(c) | < \infty$ and $\sup_{x \in \Omega, j} |f_j'(x)| < \infty$,

then $f_j$ has a uniformly convergent subsequence.
Proof: We want to show that the sequence satisfies the conditions in Ascoli's theorem. Let $M$ be the second sup in the condition. Since by the mean value theorem we have:

$|f_j(x)| \le |f_j(x) - f_j(c)| + |f_j(c)| \le |x - c|M + |f_j(c)|$,

and the hypothesis, the sup taken all over the sequence $f_j$ is finite. Using the mean value theorem again we also have: for any $j$ and $x, y \in \Omega$,

$|f_j(x) - f_j(y)| \le |x - y| M$

showing the equicontinuity. $\square$

First and second countability

2 Theorem (first countability) Let $E$ have a countable base at each point of $E$. Then we have the following:

• (i) For $A \subset E$, $x \in \overline{A}$ if and only if there is a sequence $x_j \in A$ such that $x_j \to x$.
• (ii) A function is continuous on $E$ if and only if $f(x_j) \to f(x)$ whenever $x_j \to x$.

Proof: (i) Let $\{B_j\}$ be a base at $x$ such that $B_{j + 1} \subset B_j$. If $x \in \overline{A}$, then every $B_j$ intersects A</math>. Let $x_j \in B_j \cap E$. It now follows: if $x \in G$, then we find some $B_N \subset G$. Then $x_k \in B_k \subset B_N$ for $k \ge N$. Hence, $x_j \to x$. Conversely, if $x \not\in \overline{A}$, then $E \backslash \overline{E}$ is an open set containing $x$ and no $x_j \subset E$. Hence, no sequence in $A$ converges to $x$. That (ii) is valid follows since $f(x_j)$ is the composition of continuous functions $f$ and $x$, which is again continuous. In other words, (ii) is essentially the same as (i). $\square$.

2. 2 Theorem $\Omega$ has a countable basis consisting of open sets with compact closure.
Proof: Suppose the collection of interior of closed balls $G_{\alpha}$ in $\Omega$ for a rational coordinate $\alpha$. It is countable since $\mathbb{Q}$ is. It is also a basis of $\Omega$; since if not, there exists an interior point that is isolated, and this is absurd.

2.5 Theorem In $\mathbb{R}^k$ there exists a set consisting of uncountably many components.

Usual properties we expect from calculus courses for numerical sequences to have are met like the uniqueness of limit.

2 Theorem (uncountability of the reals) The set of all real numbers is never a sequence.
Proof: See [1].

Example: Let f(x) = 0 if x is rational, f(x) = x^2 if x is irrational. Then f is continuous at 0 and nowhere else but f' exists at 0.

2 Theorem (continuous extension) Let $f, g: \overline{E} \to \mathbb{C}$ be continuous. If $f = g$ on $E$, then $f = g on \overline{E}$.
Proof: Let $u = f - g$. Then clearly $u$ is continuous on $\overline{E}$. Since ${0}$ is closed in $\mathbb{C}$, $u^{-1}(0)$ is also closed. Hence, $u = 0$ on $\overline{E}$. $\square$

, and for each $j$, $B_j = \overline{ \{ x_k : k \ge j \} }$. Since for any subindex $j(1), j(2), ..., j(n)$, we have:

$x_{j(n)} \in B_{j(1)} \cap B_{j(2)} ... B_{j(n)}$.

That is to say the sequence $B_j$ has the finite intersection property. Since $E$ is coun

Since $E$ is first countable, $E$ is also sequentially countable. Hence, (a) $\to$ (b).

Suppose $E$ is not bounded, then there exists some $\epsilon > 0$ such that: for any finite set $\{ x_1, x_2, ... x_n \} \subset E$,

$E \not \subset B(\epsilon, x_1) \cup ... B(\epsilon, x_n)$.

Let recursively $x_1 \in E$ and $x_j \in E \backslash \bigcup_1^{j-1} B(\epsilon, x_k)$. Since $d(x_n, x_m) > \epsilon$ for any n, m, $x_j$ is not Cauchy. Hence, (a) implies that $E$ is totally bounded. Also,

separated by neighborhoods

3 Theorem the following are equivalent:

• (a) $\| x \| = 0$ implies that $x = 0$.
• (b) Two points can be separated by disjoint open sets.
• (c) Every limit is unique.

3 Theorem The separation by neighborhoods implies that a compact set $K$ is closed in E.
Proof [2]: If $K = E$, then $K$ is closed. If not, there exists a $x \in E \backslash K$. For each $y \in K$, the hypothesis says there exists two disjoint open sets $A(y)$ containing $x$ and $B(y)$ containing $y$. The collection $\{ B(y) : y \in E \}$ is an open cover of $K$. Since the compactness, there exists a finite subcover $\{ B(y_1), B(y_2), ... B(y_n) \}$. Let $A(x)$ = $A(y_1) \cap A(y_2) ... A(y_n)$. If $z \in K$, then $z \in B(y_k)$ for some $k$. Hence, $z \not\in A(y_k) \subset A$. Hence, $A(x) \subset E \backslash K$. Since the finite intersection of open sets is again open, $A(x)$ is open. Since the union of open sets is open,

$E \backslash K = \bigcup_{x \not\in K} A(x)$ is open. $\square$

Seminorm

Illustrations of unit circles in different norms.

Let $G$ be a linear space. The seminorm of an element in $G$ is a nonnegative number, denoted by $\| \cdot \|$, such that: for any $x, y \in G$,

1. $\|x\| \ge 0$.
2. $\| \alpha x \| = |\alpha| \|x\|$.
3. $\|x + y\| \le \|x\| + \|y\|$. (triangular inequality)

Clearly, an absolute value is an example of a norm. Most of the times, the verification of the first two properties while whether or not the triangular inequality holds may not be so obvious. If every element in a linear space has the defined norm which is scalar in the space, then the space is said to be "normed linear space".

A liner space is a pure algebraic concept; defining norms, hence, induces topology with which we can work on problems in analysis. (In case you haven't noticed this is a book about analysis not linear algebra per se.) In fact, a normed linear space is one of the simplest and most important topological space. See the addendum for the remark on semi-norms.

An example of a normed linear space that we will be using quite often is a sequence space $l^p$, the set of sequences $x_j$ such that

For $1 \le p < \infty$, $\sum_1^{\infty} |x_j|^p < \infty$

For $p = \infty$, $x_j$ is a bounded sequence.

In particular, a subspace of $l^p$ is said to have dimension n if in every sequence the terms after nth are all zero, and if $n < \infty$, then the subspace is said to be an Euclidean space.

An open ball centered at $a$ of radius $r$ is the set $\{ z : \| z - a \| < r \}$. An open set is then the union of open balls.

Continuity and convergence has close and reveling connection.

3 Theorem Let $L$ be a seminormed space and $f: \Omega \to L$. The following are equivalent:

• (a) $f$ is continuous.
• (b) Let $x \in \Omega$. If $f(x) \in G$, then there exists a $\omega \subset \Omega$ such that $x \in \omega$ and $f(\omega) \subset G$.
• (c) Let $x \in \Omega$. For each $\epsilon > 0$, there exists a $\delta > 0$ such that

$|f(y) - f(x)| < \epsilon$ whenever $y \in \Omega$ and $|y - x|< \delta$

• (e) Let

Proof: Suppose (c). Since continuity, there exists a $\delta > 0$ such that:

$|f(x_j) - f(x)| < \epsilon$ whenever $|x_j - x| < \delta$

The following special case is often useful; in particular, showing the sequence's failure to converge.

2 Theorem Let $x_j \in \Omega$ be a sequence that converges to $x \in \Omega$. Then a function $f$ is continuous at $x$ if and only if the sequence $f(x_j)$ converges to $f(x)$.
Proof: The function $x$ of $j$ is continuous on $\mathbb{N}$. The theorem then follows from Theorem 1.4, which says that the composition of continuous functions is again continuous. $\square$.

2 Theorem Let $u_j$ be continuous and suppose $u_j$ converges uniformly to a function $u$ (i.e., the sequence $\|u_j - u\| \to 0$ as $j \to \infty$. Then $u$ is continuous.
Proof: In short, the theorem holds since

$\lim_{y \to x}u(y) = \lim_{y \to x} \lim_{j \to \infty} u_j(y) = \lim_{j \to \infty} \lim_{y \to x} u_j(y) = \lim_{j \to \infty} u_j(x) = u(x)$.

But more rigorously, let $\epsilon > 0$. Since the convergence is uniform, we find a $N$ so that

$\|u_N(x) - u(x)\| \le \|u_N - u\| < \epsilon / 3$ for any $x$.

Also, since $u_N$ is continuous, we find a $\delta > 0$ so that

$\|u_N(y) - u_N(x)\| < \epsilon / 3$ whenever $|y - x| < \delta$

It then follows that $u$ is continuous since:

$\| u(y) - u(x) \| \le \| u(y) - u_N(y) \| + \| u_N(y) - u_N(x) \| + \| u_N(x) - u(x) \| < \epsilon$

whenever $|y - x| < \delta$ $\square$

2 Theorem The set of all continuous linear operators from $A$ to $B$ is complete if and only if $B$ is complete.
Proof: Let $u_j$ be a Cauchy sequence of continuous linear operators from $A$ to $B$. That is, if $x \in A$ and $\|x\| = 1$, then

$\|u_n(x) - u_m(x)\| = \| (u_n - u_m)(x) \| \to 0$ as $n, m \to \infty$

Thus, $u_j(x)$ is a Cauchy sequence in $B$. Since B is complete, let

$u(x) = \lim_{j \to \infty} u_j(x)$ for each $x \in A$.

Then $u$ is linear since the limit operator is and also $u$ is continuous since the sequence of continuous functions converges, if it does, to a continuous function. (FIXME: the converse needs to be shown.) $\square$

Metric spaces

We say a topological space is metric or metrizable if every open set in it is the union of open balls; that is, the set of the form $\{ x; d(x, y) < r \}$ with radius $r$ and center $y$ where $d$, called metric, is a real-valued function satisfying the axioms: for all $x, y, z$

1. $d(x, y) = d(y, x)$
2. $d(x, y) + d(y, z) \ge d(x, z)$
3. $d(x, y) = 0$ if and only if $x = y$. (Identity of indiscernibles)

It follows immediately that $|d(x, y) - d(z, y)| \le d(x, y)$ for every $x, y, z$. In particular, $d$ is never negative. While it is clear that every subset of a metric space is again a metric space with the metric restricted to the set, the converse is not necessarily true. For a counterexample, see the next chapter.

2 Theorem Let $K, d)$ be a compact metric space. If $\Gamma$ is an open cover of $K$, then there exists a $\delta > 0$ such that for any $E \subset K$ with $\sup_{x, y \in E} d(x, y) < \delta$ $E \subset$ some member of $\Gamma$
Proof: Let $x, y$ be in $K$, and suppose $x$ is in some member $S$ of $\Gamma$ and $y$ is in the complement of $S$. If we define a function $\delta$ by for every $x$

$\delta(x) = \inf \{ d(x, z); z \in A \in \Gamma, x \ne A \}$,

then

$\delta(x) \le d(x, y)$

The theorem thus follows if we show the inf of $\delta$ over $K$ is positive. $\square$

2 Lemma Let $K$ be a metric space. Then every open cover of $K$ admits a countable subcover if and only if $K$ is separable.
Proof: To show the direct part, fix $n = 1, 2, ...$ and let $\Gamma$ be the collection of all open balls of radius ${1 \over n}$. Then $\Gamma$ is an open cover of $K$ and admits a countable subcover $\gamma$. Let $E_n$ be the centers of the members of $\gamma$. Then $\cup_1^\infty E_n$ is a countable dense subset. Conversely, let $\Gamma$ be an open cover of $K$ and $E$ be a countable dense subset of $K$. Since open balls of radii $1, {1 \over 2}, {1 \over 3}, ...$, the centers lying in $E$, form a countable base for $K$, each member of $\Gamma$ is the union of some subsets of countably many open sets $B_1, B_2, ...$. Since we may suppose that for each $n$ $B_n$ is contained in some $G_n \in \Gamma$ (if not remove it from the sequence) the sequence $G_1, ...$ is a countable subcover of $\Gamma$ of $K$. $\square$

Remark: For more of this with relation to cardinality see [3].

2 Theorem Let $K$ be a metric space. Then the following are equivalent:

• (i) $K$ is compact.
• (ii) Every sequence of $K$ admits a convergent subsequence. (sequentially compact)
• (iii) Every countable open cover of $K$ admits a finite subcover. (countably compact)

Proof (from [4]): Throughout the proof we may suppose that $K$ is infinite. Lemma 1.somthing says that every sequence of a infinite compact set has a limit point. Thus, the first countability shows (i) $\Rightarrow$ (ii). Supposing that (iii) is false, let $G_1, G_2, ...$ be an open cover of $K$ such that for each $n = 1, 2, ...$ we can find a point $x_n \in (\cup_1^n G_j)^c = \cap_1^n {G_j}^c$. It follows that $x_n$ does not have a convergent subsequence, proving (ii) $\Rightarrow$ (iii). Indeed, suppose it does. Then the sequence $x_n$ has a limit point $p$. Thus, $p \in \cap_1^\infty {G_j}^c$, a contradiction. To show (iii) $\Rightarrow$ (i), in view of the preceding lemma, it suffices to show that $K$ is separable. But this follows since if $K$ is not separable, we can find a countable discrete subset of $K$, violating (iii). $\square$

Remark: The proof for (ii) $\Rightarrow$ (iii) does not use the fact that the space is metric.

2 Theorem A subset $E$ of a metric space is precompact (i.e., its completion is compact) if and only if there exists a $\delta > 0$ such that $E$ is contained in the union of finitely many open balls of radius $\delta$ with the centers lying in E.

Thus, the notion of relatively compactness and precompactness coincides for complete metric spaces.

2 Theorem (Heine-Borel) If $E$ is a subset of $\mathbb{R}^n$, then the following are equivalent.

• (i) $E$ is closed and bounded.
• (ii) $E$ is compact.
• (iii) Every infinite subset of $E$ contains some of its limit points.

2 Theorem Every metric space is perfectly and fully normal.

2 Corollary Every metric space is normal and Hausdorff.

2 Theorem If $x_j \to x$ in a metric space $X$ implies $f(x_n) \to f(x)$, then $f$ is continuous.
Proof: Let $E \subset X$ and $x \in$ the closure of $E$. Then there exists a $x_j \in E \to x$ as $j \to \infty$ as $j \to \infty$. Thus, by assumption $f(x_j) \to f(x)$, and this means that $f(x)$ is in the closure of $f(E)$. We conclude: $f(\overline{E}) \subset \overline{f(E)}$. $\square$

2 Theorem Suppose that $f: (X_1, d_1) \to (X_2, d_2)$ is continuous and satisfies

$d_2(f(x), f(y)) \ge d_1(x, y)$ for all $x, y$

If $F \subset X_2$ and $(F, d_2)$ is a complete metric space, then $f(F)$ is closed.
Proof: Let $x_n$ be a sequence in $F$ such that $\lim_{n, m \to \infty} d_2(f(x_n), f(x_m)) = 0$. Then by hypothesis $\lim_{n, m \to \infty} d_1(x_n, x_m) = 0$. Since $F$ is complete in $d_2$, the limit $y = \lim_{n \to \infty} x_n$ exists. Finally, since $f$ is continuous, $\lim_{n\to \infty} d_2(f(x_n), f(y)) = 0$, completing the proof. $\square$

Measure

A measure, which we shall denote by $\mu$ throughout this section, is a function from a delta-ring $\mathcal{F}$ to the set of nonnegative real numbers and $\infty$ such that $\mu$ is countably additive; i.e., $\mu (\coprod_1^\infty E_j) = \sum_1^{\infty} \mu (E_j)$ for $E_j$ distinct. We shall define an integral in terms of a measure.

4.1 Theorem A measure $\mu$ has the following properties: given measurable sets $A$ and $B$ such that $A \subset B$,

• (1) $\mu (B \backslash A) = \mu (B) - \mu (A)$.
• (2) $\mu (\varnothing) = 0$.
• (3) $\mu (A) \le \mu (B)$ where the equality holds for $A = B$. (monotonicity)

Proof: (1) $\mu (B) - \mu (A) = \mu (B \backslash A \cup A) - \mu (A) = \mu (B \backslash A) + \mu (A) - \mu (A)$. (2) $\mu (\varnothing) = \mu (A \backslash A) = \mu (A) - \mu (A). (3) \mu (B) - \mu (A) = \mu (B \backslash A) \ge 0$ since measures are always nonnegative. Also, $\mu (B) - \mu (A) = 0$ if $A = B$ since (2).

One example would be a counting measure; i.e., $\mu E$ = the number of elements in a finite set $E$. Indeed, every finite set is measurable since the countable union and countable intersection of finite sets is again finite. To give another example, let $x_0 \in \mathcal{F}$ be fixed, and $\mu(E) = 1$ if $x_0$ is in $E$ and 0 otherwise. The measure $\mu$ is indeed countably additive since for the sequence ${E_j}$ arbitrary, $\mu(\coprod_0^{\infty} E_j) = 1 = \mu(E_j)$ for some $j$ if $x_0 \in \bigcup E_j$ and 0 otherwise.

We now study the notion of geometric convexity.

2 Lemma

${x+y \over 2} = a \left( a {x+y \over 2} + (1-a)y \right) + (1-a) \left( ax + (1-a){x+y \over 2} \right)$ for any $x, y$.

2 Theorem Let $D$ be a convex subset of a vector space. If $0 < a < 1$ and $f(ax + (1-a)y) \le af(x) + (1-a)f(y)$ for $x, y \in D$, then

$f({x+y \over 2}) \le {f(x) + f(y) \over 2}$ for $x, y \in D$

Proof: From the lemma we have:

$2a(1-a)f({x+y \over 2}) \le a(1-a) (f(x) + f(y))$ $\square$

Let $\mathcal{F}$ be a collection of subsets of a set $G$. We say $\mathcal{F}$ is a delta-ring if $\mathcal{F}$ has the empty set, G and every countable union and countable intersection of members of $\mathcal{F}$. a member of $\mathcal{F}$ is called a measurable set and G a measure space, by analogy to a topological space and open sets in it.

2 Theorem (Hölder's inequality) For $1 \le p \le \infty$, if $1/p + 1/q = 1$, then

$\int |fg|d\mu \le \left( \int |f|^p d\mu \right)^{1/p} \left( \int |g|^q d\mu \right)^{1/q}$

Proof: By replacing $f$ and $g$ with $|f|$ and $|g|$, respectively, we may assume that $f$ and $g$ are non-negative. Let $C = \int g^q d\mu$. If $C = 0$, then the inequality is obvious. Suppose not, and let $d\nu = {g^q d\mu \over C}$. Then, since $x^p$ is increasing and convex for $x \ge 0$, by Jensen's inequality,

$\int fg d\mu = \int fg^{(1-q)} d\nu C \le \left( \int f^p g^{-q} d\nu \right)^{(1/p)} C$.

Since $1/q = 1 - 1/p$, this is the desired inequality. $\square$

4 Corollary (Minkowski's inequality) Let $p \ge 1$ and $\| \cdot \|_p = \left( \int | \cdot |^p \right)^{1/p}$. If $f \ge 0$ and $\int \int f(x, y) dx dy < \infty$, then

$\| \int f(\cdot, y) dy \|_p \le \int \| f(\cdot, y) \|_p dy$

Proof (from [5]): If $p = 1$, then the inequality is the same as Fubini's theorem. If $p > 1$,

$\int |\int f(x, y)dy|^p dx$	$= \int |\int f(x, y)dy|^{p-1} \int f(x, z)dz dx$
(Fubini)	$= \int \int |\int f(x, y)dy|^{p-1} f(x, z)dx dz$
(Hölder)	$\le \left( \int |\int f(x, y)dy|^{(p-1)q}dx \right)^{1/q} \int \| f(\cdot, z) \|_p dz$

By division we get the desired inequality, noting that $(p - 1)q = p$ and $1 - {1 \over q} = {1 \over p}$. $\square$

TODO: We can probably simplify the proof by appealing to Jensen's inequality instead of Hölder's.

Remark: by replacing $\int dy$ by a counting measure we get:

$\|f+g\|_p \le \|f\|_p + \|g\|_p$

under the same assumption and notation.

2 Lemma For $1 \le p \le \infty$ and $a, b > 0$, if $1/p + 1/q = 1$,

$a^{1 \over p}b^{1 \over q} = \inf_{t > 0} \left( {1 \over p} t^{-{1 \over q}}a + {1 \over q} t^{1 \over p}b \right)$

Proof: Let $f(t) = {1 \over p} t^{-{1 \over q}}a + {1 \over q} t^{1 \over p}b$ for $t > 0$. Then the derivative

$f'(t) = {1 \over pq} \left( t^{-{1 \over p}} a + t^{1 \over q} b \right)$

becomes zero only when $t = a/b$. Thus, the minimum of $f$ is attained there and is equal to $a^{1 \over p}b^{1 \over q}$. $\square$

When $t$ is taken to 1, the inequality is known as Young's inequality.

2 Theorem (Hölder's inequality) For $1 \le p \le \infty$, if $1/p + 1/q = 1$, then

$\int |fg| \le \left( \int |f|^p \right)^{1/p} \left( \int |f|^q \right)^{1/q}$

Proof: If $p = \infty$, the inequality is clear. By the application of the preceding lemma we have: for any $t > 0$

$\int |f|^{1 \over p} |g|^{1 \over q} \le {1 \over p} t^{-{1 \over q}}\int |f| + {1 \over q} t^{1 \over p} \int |g|$

Taking the infimum we see that the right-hand side becomes:

$\left( \int |f| \right)^{1/p} \left( \int |g| \right)^{1/q}$ $\square$

The convex hull in $\Omega$ of a compact set K, denoted by $\hat K$ is:

$\left \{ z \in \Omega : |f(z)| \le \sup_K|f|\mbox{ for }f\mbox{ analytic in }\Omega \right \}$.

When $f$ is linear (besides being analytic), we say the $\hat K$ is geometrically convex hull. That $K$ is compact ensures that the definition is meaningful.

Theorem The closure of the convex hull of $E$ in $G$ is the intersection if all half-spaces containing $E$.
Proof: Let F = the collection of half-spaces containing $E$. Then $\overline{co}(E) \subset \cap F$ since each-half space in $F$ is closed and convex. Yet, if $x \not\in \overline{co}(E)$, then there exists a half-space $H$ containing $E$ which however does not contain x. Hence, $\overline{co}(E) = \cap F$. $\square$

Let $\Omega \subset \mathbb{R}^n$. A function $f:\Omega \to \mathbb{R}^n$ is said to be convex if

$\sup_K |f - h| = \sup_{\mbox{b}K} | f - h |$.

for some $K \subset \Omega$ compact and any $h$ harmonic on $K$ and continuous on $\overline{K}$.

Theorem The following are equivalent:

• (a) $f$ is convex on some $K \subset \mathbb{R}$ compact.
• (b) if $[a, b] \subset K$, then

  $f(\lambda a + (1 - \lambda) b) \le \lambda f(a) + (1 - \lambda) f(b)$ for $\lambda \in [0, 1]$.

• (c) The difference quotient

  ${f(x + h) - f(x) \over h}$ increases as $h$ does.

• (d) $f$ is measurable and we have:

  $f \left( {a + b \over 2} \right) \le {a + b \over 2}$ for any $[a, b] \subset K$.

• (e) The set $\{ (x, y) : x \in [a, b], f(x) \le y \}$ is convex for $[a, b] \subset K$.

Proof: Suppose (a). For each $x \in [a, b]$, there exists some $\lambda \in [0, 1]$ such that $x = \lambda a + (1 - \lambda) b$. Let $A(x) = A(\lambda a + (1-\lambda) b) = \lambda f(a) + (1-\lambda) f(b)$, and then since $\mbox{b}[a, b] = \{a, b\}$ and $(f - A)(a) = 0 = (f - A)(b)$,

$|f(x)| = |f(x) - A(x) + A(x)|$	$\le \sup \{|f(a) - A(a)|, |f(b) - A(b)|\} + A(x)$
	$=\lambda f(a) + (1-\lambda) f(b)$

Thus, (a) $\Rightarrow$ (b). Now suppose (b). Since $\lambda = {x - a \over b - a}$, for $\lambda \in (0, 1)$, (b) says:

$(b - x)f(x) + (x - a)f(x)$	$\le (b -x)f(a) + (x - a)f(b)$
${f(a) - f(x) \over a - x}$	$\le {f(b) - f(x) \over b - x}$

Since $a - x < b - x$, we conclude (b) $\Rightarrow$ (c). Suppose (c). The continuity follows since we have:

$\lim_{h > 0, h \to 0}f(x + h) = f(x) + \lim_{h > 0, h \to 0}h{f(x + h) - f(x) \over h}$.

Also, let $x = 2^{-1}(a + b)$ such that $a < b$, for $a, b \in K$. Then we have:

${f(a) - f(x) \over a - x}$	$\le {f(b) - f(x) \over b - x}$
$f(x) - f(a)$	$\le f(b) - f(x)$
$f \left( {a + b \over 2} \right)$	$\le {f(a) + f(b) \over 2}$

Thus, (c) $\Rightarrow$ (d). Now suppose (d), and let $E = \{ (x, y) : x \in [a, b], y \ge f(x) \}$. First we want to show

$f\left({1 \over 2^n} \sum_1^{2^n} x_j \right) \le {1 \over 2^n} \sum_1^{2^n} f(x_j)$.

If $n = 0$, then the inequality holds trivially. if the inequality holds for some $n - 1$, then

$f \left( {1 \over 2^n} \sum_1^{2^n} x_j \right)$	$=f \left( {1 \over 2} \left( {1 \over 2^{n - 1}} \sum_1^{2^{n - 1}} x_j + {1 \over 2^{n - 1}} \sum_1^{2^{n - 1}} x_{2^{n - 1} + j} \right) \right)$
	$\le {1 \over 2} f \left( {1 \over 2^{n - 1}} \sum_1^{2^{n - 1}} x_j \right) + {1 \over 2} f \left( {1 \over 2^{n - 1}} \sum_1^{2^{n - 1}} x_{2^{n - 1} + j} \right)$
	$\le {1 \over 2^n} \sum_1^{2^{n - 1}} f(x_j) + {1 \over 2^n} \sum_1^{2^{n - 1}} f(x_{2^{n - 1} + j})$
	$= {1 \over 2^n} \sum_1^{2^n}f(x_j)$

Let $x_1, x_2 \in [a, b]$ and $\lambda \in [0, 1]$. There exists a sequence of rationals number such that:

$\lim_{j \to \infty} {p_j \over 2q_j} = \lambda$.

It then follows that:

$f(\lambda x_1 + (1 - \lambda) x_2)$	$\le \lim_{j \to \infty} {p_j \over 2q_j} f(x_1) + \left( 1 - {p_j \over 2q_j} \right) f(x_2)$
	$= \lambda f(x_1) + (1 - \lambda) f(x_2)$
	$\le \lambda y_1 + (1 - \lambda) y_2$.

Thus, (d) $\Rightarrow$ (e). Finally, suppose (e); that is, $E$ is convex. Also suppose $K$ is an interval for a moment. Then

$f(\lambda a + (1 - \lambda) b) \le \lambda f(a) + (1 - \lambda) f(b)$. $\square$

4. Corollary (inequality between geometric and arithmetic means)

$\prod_1^n a_j^{\lambda_j} \le \sum_1^n \lambda_j a_j$ if $a_j \ge 0$, $\lambda_j \ge 0$ and $\sum_1^n \lambda_j = 1$.

Proof: If some $a_j = 0$, then the inequality holds trivially; so, suppose $a_j > 0$. The function $e^x$ is convex since its second derivative, again $e^x$, is $> 0$. It thus follows:

$\prod_1^n a_j^{\lambda_j} =e^{\sum_1^n \lambda_j \log(a_j)} \le \sum_1^n \lambda_j e^{\log(a_j)} = \sum_1^n \lambda_j a_j$. $\square$

The convex hull of a finite set $E$ is said to be a convex polyhedron. Clearly, the set of the extremely points is the subset of $E$.

4 Theorem (general Hölder's inequality) If $a_{jk} > 0$ and $p_j > 1$ for $j = 1, ... n_j$ and $k = 1, ... n_k$ and $\sum_1^{n_j} p_j = 1$, then:

$\sum_{k=1}^{n_k} \prod_{j=1}^{n_j} a_{jk} \le \prod_{j=1}^{n_k} \left( \sum_{k=1}^{n_k} a_{jk}^{p_j} \right)^{1/p}$ (Hörmander 11)

Proof: Let $A_j = \left( \sum_{k} a_{jk}^{p_j} \right)^{1/p_j}$.

4. Theorem A convex polyhedron is the intersection of a finite number of closed half-spaces.
Proof: Use induction. $\square$

4. Theorem The convex hull of a compact set is compact.
Proof: Let $f(\lambda_1, \lambda_2, ... \lambda_n, x_1, x_2, ... x_n) = \sum_1^n \lambda_j x_j$. Then $f$ is continuous since it is the finite sum of continuous functions $\lambda_j x_j$. Since the intersection of compact sets is compact and

$\mbox{co}(K) = \bigcap_{n = 1, 2, ...} f(\lambda_1, \lambda_2, ... \lambda_n, x_1, x_2, ... x_n)$,

$\mbox{co}(K)$ is compact. $\square$

Example: Let $p(z) = (z - s_1)(z - s_2) ... (z - s_n)$. Then the derivative of $p$ has zeros in $\mbox{co}(\{s_1, s_2, ... s_n\})$.

Addendum

1. Show the following are equivalent with assuming that $K$ is second countable.
   • (1) $K$ is compact.
   • (2) Every countable open cover of $K$ admits a finite subcover (countably compact)
   • (3) $K$ is sequentially compact.

Nets are, so to speak, generalized sequences.

1. Show the following are equivalent with assuming Axiom of Choice:
   • (1) $K$ is compact; i.e, every open cover admits a finite subcover.
   • (2) In $K$ every net has a convergent subnet.
   • (3) In $K$ every ultrafilter has an accumulation point (Bourbaki compact)
   • (4) There is a subbase $\tau$ for $K$ such that every open cover that is a subcollection of $\tau$ admits a finite subcover. (subbase compact)
   • (5) Every nest of non-empty closed sets has a non-empty intersection (linearly compact) (Hint: use the contrapositive of this instead)
   • (6) Every infinite subset of $K$ has a complete accumulation point (Alexandroff-Urysohn compact)