Elements of Analysis
MATH 302

direct proof
Assume $p$ is true, then show that $q$ must also be true.

proof of contrapositive
Assume $q$ is false, then show that $p$ must also be false.

proof by contradiction
Assume $p$ is true and $q$ is false. Then show that this leads to a contradiction.
a statement is a sentence that is either true or false.
[see the comp182 logic section]
$A \rightarrow B$
 Converse: $B \rightarrow A$
 Contrapositive: $\neg B \rightarrow \neg A$
Disjoined conclusions §
[see the comp182 logic section]
$P \rightarrow (Q \vee R)$
Sets §
[see the comp182 sets section]
Indexed families of sets §
$x \in \bigcup_{n=1}^{\infty} S_n \leftrightarrow \exists n_x \in \mathbb{N} : x \in S_{n_x}$
$x \in \bigcap_{n=1}^{\infty} S_n \leftrightarrow \forall n \in \mathbb{N} : n \in S_{n_x}$
Induction §
[see the comp182 induction section]
Real vs Rational numbers §
Upper and lower bounds §
A number $u \in \mathbb{R}$ is an upper bound of $S$ if for any $x \in S$, $u \geq x$. We say that $S$ is bounded from above.
Supremum: least upper bound.
If $u$ is the supremum of $S$ and $u \in S$, $u$ is the maximum of $S$.
A number $l \in \mathbb{R}$ is a lower bound of $S$ if for any $x \in S$, $l \leq x$. We say that $S$ is bounded from below.
Infimum: greatest lower bound.
If $l$ is the infimum of $S$ and $l \in S$, $l$ is the minimum of $S$.
Let $S \subseteq \mathbb{R}$ and let $u \in \mathbb{R}$ be an upper bound of $S$. $u$ is the supremum of $S$ iff:
 $u$ is the maximum of $S$ or
 for every $\epsilon > 0$ there is an element of $S$ such that $u  \epsilon$ is less than that element
Least upper bound property (aka completeness property): Every nonempty subset of the real numbers that is bounded from above has a least upper bound that is a real number.
Bounded and nested intervals §
Archimedian property: For every real number $x$, there exists a natural number $n_x$ such that $n_x > x$.
Successor axiom: If $n \in \mathbb{N}$, $n + 1$ exists and is a natural number.
A family of sets $S_n$ indexed by the natural numbers is nested iff $S_{n + 1} \subseteq S_n \forall n \in \mathbb{N}$.
Nested intervals property: If $I_n$ is a nested family of nonempty, closed, bounded^{[1]} intervals in $\mathbb{R}$, then $\bigcap_{n = 1}^\infty I_n$ is nonempty.
Bounded Infinite sets have cluster points §
Neighbourhoods: Given $x \in \mathbb{R}$ and $\epsilon > 0$, the $\epsilon$neigbourhood is the open interval $(x  \epsilon, x + \epsilon)$.
A real number $x$ ix a cluster point ^{[2]} of a set $S \subseteq \mathbb{R}$ if $(x  \epsilon, x + \epsilon)$ contains a non$x$ element of $S$ for all $\epsilon > 0$.
Cluster points of a set need not be elements of the set.
BolsanoWeierstrass Theorem: Every bounded, infinite set of real numbers has a cluster point.
Functions §
Given sets $X$ and $Y$, a function $f$ from $X$ to $Y$ is denoted $f : X \rightarrow Y$, and it is a relation between $X$ and $Y$ such that every $x \in X$ has exactly one $y \in Y$ such that $f(x) = y$.
$X$ is the domain and $Y$ is the codomain of $f$.
$f = g$ iff they have the same domain and $f(x) = g(x)$ for all $x$ in their domain.
Let $f: X \rightarrow Y$, and $A \subseteq X$. The restriction of $f$ to $A$ is the function $f_A : A \rightarrow Y$ defined by $f_A(a) = f(a) \forall a \in A$.
Let $f: X \rightarrow Y$ be a function. Given $A \subseteq X$, the image of $A$ under $f$ is the subset $f(a) \subseteq Y$ given by $f(a) = \{y \in Y : \exists a \in A : f(a) = y\}$.
Open sets §
A set $U \subseteq \mathbb{R}$ is open if for every $x \in U$, there exists $\epsilon_x > 0$ such that $(x  \epsilon_x, x + \epsilon_x) \subseteq u$.
The intersection of any finite family of open sets is open.
Closed sets §
A set $F \subseteq \mathbb{R}$ is closed if $F$ contains all of its cluster points.
If a set has no cluster points, it is closed.
 the union of open sets is open
 the intersection of finitely many open sets is open
 the intersection of closed sets is closed
 the union of finitely many closed sets is closed
Continuity §
$f$ is continuous if and only if for every $x_0$ in the domain: for every $\epsilon > 0$, there exists $\delta > 0$ such that for all $x \in \mathbb{R}$, if $x  x_0 < \delta$, then $f(x)  f(x_0) < \epsilon$.
A function $f : \mathbb{R} \rightarrow \mathbb{R}$ is continuous if for every open set $u$ in the codomain of $f$, its preimage $f^{1}(u)$ is also open.
Sequences §
a sequence is a function $\mathbb{N} \rightarrow \mathbb{R}$.
a sequence converges to a real number $L$ if for every $\epsilon$neighbourhood of $L$, the sequence eventually remains in the $\epsilon$neighbourhood.
$(\forall \epsilon > 0, \exists N_\epsilon \in \mathbb{N} : \forall n \geq N_\epsilon, x_n  L < \epsilon) \Leftrightarrow x_n \in (L  \epsilon, L + \epsilon)$
a sequence can have at most one limit.
Squeeze theorem: if $\lim l_n = L$, $\lim u_n = L$, and $l_n \leq x_n \leq u_n$, then $\lim x_n = L$.
 $\lim (x_n + y_n) = \lim x_n + \lim y_n$
 $\lim (c \cdot x_n) = c (\lim x_n)$ for any constant $c$
 $\lim (x_n \cdot y_n) = (\lim x_n) (\lim y_n)$
 $\lim (\frac{x_n}{y_n}) = \frac{\lim x_n}{\lim y_n}$ if $\lim y_n \neq 0$
A real number $L$ is a cluster point of a set $S$ of real numbers iff there exists a sequence of elements of $S$, all different from $L$, that converges to $L$.
A set $F \subseteq \mathbb{R}$ is closed iff for an arbitrary convergent sequence $x_n$ in $F$, the limit $\lim x_n$ is an element of $F$.
A sequence is monotone if it is either weakly increasing, or weakly decreasing.
Every bounded, monotone sequence of real numbers converges.
Given a sequence $(x_n)$, a subsequence is a sequence $(y_n)$ defined by $y_k = x_{n_k}$ for some choice of natural numbers $n_1 < n_2 < \cdots$.
BolzanoWeierstrass Theorem: Every bounded sequence of real numbers has a convergent subsequence.
A sequence is a cauchy sequence if the terms of the sequence are “eventually” close to each other.
$\forall \epsilon > 0, \exists N \in \mathbb{N} : \forall m, n \geq N, x_m  x_n < \epsilon$
A sequence of real numbers is convergent if and only if it is a Cauchy sequence.
A set $S \subseteq \mathbb{R}$ is disconnected if there exist sets $U, V \subseteq S$ such that:
 $U \cup V = S$
 $U \neq \varnothing$ and $V \neq \varnothing$
 $U \cap V = \varnothing$
 $U$ and $V$ are open in $S$
Given $D \subseteq \mathbb{R}$, if $f : D \rightarrow \mathbb{R}$ is continuous, then the image of any connected set is continuous.
Extreme value theorem: Suppose $f: [a, b] \rightarrow \mathbb{R}$ is continuous. Then $f([a, b]) = \{f(x) : x \in [a, b]\}$ has a maximum and a minimum.
A set $K \subseteq \mathbb{R}$ is compact if it is closed and bounded.
Given $D \subseteq \mathbb{R}$, if $f : D \rightarrow \mathbb{R}$ is continuous, then the image of every compact set is compact.
Covering property (aka HeineBorel theorem)" $K$ is closed and bounded if and only if every open cover of $K$ has a finite subcover.