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docs/math/real-analysis/11-metric-spaces.md

@@ -24,19 +24,19 @@ $$
 
 ---
 
-**Definition**. Let $(S, d)$ be a metric space. Then, for each $\epsilon > 0$, the *$\epsilon$-neighborhood* or *$\epsilon$-ball* of a point $a \in S$ is the set
+**Definition**. Let $(S, d)$ be a metric space. Then, for each $\varepsilon > 0$, the *$\varepsilon$-neighborhood* or *$\varepsilon$-ball* of a point $a \in S$ is the set
 
 $$
-V_\epsilon(a) = {x \in S | d(a, x) < \epsilon}
+V_\varepsilon(a) = {x \in S | d(a, x) < \varepsilon}
 $$
 
-**Definition**. Let $(S, d)$ be a metric space. Then, a subset $G \subseteq S$ is *open* if for each $x \in G$, there exists some $\epsilon > 0$ so that $V_\epsilon(x) \subseteq G$.
+**Definition**. Let $(S, d)$ be a metric space. Then, a subset $G \subseteq S$ is *open* if for each $x \in G$, there exists some $\varepsilon > 0$ so that $V_\varepsilon(x) \subseteq G$.
 
 **Definition**. Let $(S, d)$ be a metric space. Then, a subset $G \subseteq S$ is *closed* if its complement $C(G) = S - G = S \ F$ is closed.
 
-**Definition**. Let $(S, d)$ be a metric space. A point $c \in S$ is a *cluster point$ of a set $A \subseteq S$ if every $\epsilon$-neighborhood of $c$ contains some point $a \in A$ such that $a \neq c$.
+**Definition**. Let $(S, d)$ be a metric space. A point $c \in S$ is a *cluster point$ of a set $A \subseteq S$ if every $\varepsilon$-neighborhood of $c$ contains some point $a \in A$ such that $a \neq c$.
 
-**Theorem**. Every $\epsilon$-neighborhood of a point is an open set.
+**Theorem**. Every $\varepsilon$-neighborhood of a point is an open set.
 
 **Theorem**. The union of an arbitrary collection of open sets is open.
 
@@ -50,22 +50,22 @@ $$
 
 ---
 
-**Definition**. A *sequence* $(x_n)$ in a metric space $(S, d)$ converges to a point $x \in S$ if given any $\epsilon > 0$, there exists a $K \in \mathbb{N}$ such that given $n \in \mathbb{N}$,
+**Definition**. A *sequence* $(x_n)$ in a metric space $(S, d)$ converges to a point $x \in S$ if given any $\varepsilon > 0$, there exists a $K \in \mathbb{N}$ such that given $n \in \mathbb{N}$,
 
 $$
-n \geq K \Rightarrow d(x_n, x) \leq \epsilon
+n \geq K \Rightarrow d(x_n, x) \leq \varepsilon
 $$
 
 **Theorem**. Let $(x_n)$ be a sequence in metric space $(S, d)$. Then,
 
-- $(x_n)$ converges to $x$ if and only if every $\epsilon$-neighborhood of $x$ contains all but finitely many terms of $(x_n)$.
+- $(x_n)$ converges to $x$ if and only if every $\varepsilon$-neighborhood of $x$ contains all but finitely many terms of $(x_n)$.
 - If $(x_n) \rightarrow x$ and $(x_n) \rightarrow x'$, then $x = x'$.
 - If $(x_n)$ converges, then $(x_n)$ is bound.
 
-**Definition**. A sequence $(x_n)$ in metric space $(S, d)$ is a *Cauchy sequence* if for every $\epsilon > 0$, there exists some $H \in \mathbb{N}$ such that for any $m, n \in \mathbb{N}$,
+**Definition**. A sequence $(x_n)$ in metric space $(S, d)$ is a *Cauchy sequence* if for every $\varepsilon > 0$, there exists some $H \in \mathbb{N}$ such that for any $m, n \in \mathbb{N}$,
 
 $$
-m, n \geq H \Rightarrow d(x_n, x_m) < \epsilon
+m, n \geq H \Rightarrow d(x_n, x_m) < \varepsilon
 $$
 
 **Definition**. A metric space in which every Cauchy sequence converges is said to be *complete*.

docs/math/real-analysis/2-reals.md

@@ -56,32 +56,32 @@
 2. $|ab| = |a||b|$
 3. $|a + b| \leq |a| + |b|$
 
-**Corollary**. Given $a, b \in \mathbb{R}$, then $\abs{\abs{a} - \abs{b}} \leq \abs{a - b}$.
+**Corollary**. Given $a, b \in \mathbb{R}$, then $||a| - |b|| \leq |a - b|$.
 
 **Remark**. Every field has at least one absolute value function.
 
 **Theorem**. In an ordered field $F$, for any $r > 0$, we know that
 
-1. $\abs{x = r}$ if and only if $x = r$ or $x = -r$
-2. $\abs{x < r}$ if and only if $-r < x < r$
-3. $\abs{x > r}$ if either $x > r$ or $x < -r$
+1. $|x = r$ if and only if $x = r$ or $x = -r$
+2. $|x < r$ if and only if $-r < x < r$
+3. $|x > r$ if either $x > r$ or $x < -r$
 
 ---
 
-**Definition**. The *standard distance function* or *metric* on the real numbers $\mathbb{R}$ given $a, b$ is $\abs{a - b}$.
+**Definition**. The *standard distance function* or *metric* on the real numbers $\mathbb{R}$ given $a, b$ is $|a - b|$.
 
 **Theorem**. For any real numbers $a, b, c$,
 
-1. $\abs{a - b} > 0$ if and only if $a \neq b$ and $\abs{a - b} = 0$ if and only if $a = b$
-2. $\abs{a - b} = \abs{b - a}$
-3. $\abs{a - c} \leq \abs{a - b} + \abs{b + c}$
+1. $|a - b| > 0$ if and only if $a \neq b$ and $|a - b| = 0$ if and only if $a = b$
+2. $|a - b| = |b - a|$
+3. $|a - c| \leq |a - b| + |b + c|$
 
 **Definition** A set together with a function satisfying these three properties is known as a *metric space*.
 
-**Definition** The $\epsilon$-neighborhood of $a \in \mathbb{R}$, denoted $V_\epsilon(a)$ is the set of all real numbers $x \in \mathbb{R}$ such that $\abs{x - a} < \epsilon$. That is,
+**Definition** The $\varepsilon$-neighborhood of $a \in \mathbb{R}$, denoted $V_\varepsilon(a)$ is the set of all real numbers $x \in \mathbb{R}$ such that $|x - a| < \varepsilon$. That is,
 
 $$
-V_\epsilon(a) = (a - \epsilon, a + \epsilon)
+V_\varepsilon(a) = (a - \varepsilon, a + \varepsilon)
 $$
 
 ---

docs/math/real-analysis/3-sequences-series.md

@@ -26,10 +26,10 @@ $$
 
 ---
 
-**Definition**. A sequence $X = (x_n)$ is said to *converge* to a number $x \in \mathbb{R}$ if when given any $\epsilon > 0$, there exists some $K \in \mathbb{N}$ such that for every $n \in \mathbb{N}$ with $n \geq K$,
+**Definition**. A sequence $X = (x_n)$ is said to *converge* to a number $x \in \mathbb{R}$ if when given any $\varepsilon > 0$, there exists some $K \in \mathbb{N}$ such that for every $n \in \mathbb{N}$ with $n \geq K$,
 
 $$
-\abs{x_n - x} < \epsilon
+|x_n - x| < \varepsilon
 $$
 
 If this is the case, we say that $X$ converges to  $x$, and $x$ is a *limit* of X. This can be written as
@@ -66,9 +66,9 @@ $$
 
 **Theorem**. Suppose $(x_n)$ is a sequence if real numbers. Then,
 
-1. If $x_n \rightarrow x$, then $\abs{x_n} \rightarrow \abs{x}$
-2. If $\abs{x_n} \rightarrow 0$, then $x_n \rightarrow 0$
-3. $x_n \rightarrow x$ if and only if $\abs{x_n - n} \rightarrow 0$
+1. If $x_n \rightarrow x$, then $|x_n| \rightarrow |x|$
+2. If $|x_n| \rightarrow 0$, then $x_n \rightarrow 0$
+3. $x_n \rightarrow x$ if and only if $|x_n - n| \rightarrow 0$
 
 **Theorem**. Suppose $(x_n)$ is a sequence if real numbers, with each $x_n \geq 0$. Then, given some $k \in \mathbb{N}$, if $x_n \rightarrow x$, then $\sqrt[k]{x_n} \rightarrow \sqrt[k]{x}$.
 
@@ -100,11 +100,9 @@ is a *subsequence* of $X$,
 
 ## Section 3.5 - The Cauchy Criterion
 
-**Definition**. A sequence $(x_n)$ is said to be a *Cauchy sequence* such that for any given $\epsilon$, there exists a natural number $H$ such that all natural numbers $m, n \geq H$, then
+**Definition**. A sequence $(x_n)$ is said to be a *Cauchy sequence* such that for any given $\varepsilon$, there exists a natural number $H$ such that all natural numbers $m, n \geq H$, then
 
-$$
-\abs{x_m - x_n} \leq \epsilon
-$$
+$$|x_m - x_n \leq \varepsilon$$
 
 **Theorem**. If $(x_n)$ is a Cauchy sequence, then $(x_n)$ is convergent.
 
@@ -130,15 +128,15 @@ $$
 
 In particular, $s_n - s_{n - 1} = x^n$. Thus, the Cauchy criteria takes the form
 
-**Theorem**. Cauchy Criteria for Series. The series $\sum x_n$ converges if and only if, for a given $\epsilon$, there exists some natural number $H$ such that when $m > n > H$,
+**Theorem**. Cauchy Criteria for Series. The series $\sum x_n$ converges if and only if, for a given $\varepsilon$, there exists some natural number $H$ such that when $m > n > H$,
 
 $$
-\abs{s_m - s_n} = \abs{\sum_{i = m + 1}^n x_i} < \epsilon
+|s_m - s_n| = |\sum_{i = m + 1}^n x_i| < \varepsilon
 $$
 
 **Corollary**. $n$-th Term Test. If $\sum x_n$ converges, then $x_n \rightarrow 0$.
 
-**Corollary**. Absolute Convergence Test. If $\sum \abs{x_n}$ converges, then $\sum x_n$ converges.
+**Corollary**. Absolute Convergence Test. If $\sum |x_n|$ converges, then $\sum x_n$ converges.
 
 ---
 

docs/math/real-analysis/4-limits.md

@@ -18,10 +18,10 @@
 
 ---
 
-**Definition**. Suppose $f: A \rightarrow \mathbb{R}$ is a function with domain $A \subseteq \mathbb{R}$, and let $c \in A$ be a cluster point of $A$. then, a real number $L$ is a *limit of $f$ at $c$* if given any $\epsilon > 0$, there exists some $\delta > 0$ such that
+**Definition**. Suppose $f: A \rightarrow \mathbb{R}$ is a function with domain $A \subseteq \mathbb{R}$, and let $c \in A$ be a cluster point of $A$. then, a real number $L$ is a *limit of $f$ at $c$* if given any $\varepsilon > 0$, there exists some $\delta > 0$ such that
 
 $$
-0 < |x-c| < \delta \Rightarrow |f(x) - L| < \epsilon
+0 < |x-c| < \delta \Rightarrow |f(x) - L| < \varepsilon
 $$
 
 **Theorem**. For a given function and cluster point, there can be at most one limit at said point.
@@ -32,12 +32,12 @@ $$
 
 **Definition**. The *extended real numbers* are $\hat{\mathbb{R}} = \mathbb{R} \cup \{ \infty, -\infty \}$ are a totally-ordered set with supremum and infimum. Note that this set is no longer a field.
 
-**Definition**. At any point $c$, the limit of $f$ at $c$ is infinite if given some $\alpha$,  there exists some $V_\delta(c)$ such that for all $x \in V_\epsilon(c)$, then $f(x) \in V_\alpha(\infty)$.
+**Definition**. At any point $c$, the limit of $f$ at $c$ is infinite if given some $\alpha$,  there exists some $V_\delta(c)$ such that for all $x \in V_\varepsilon(c)$, then $f(x) \in V_\alpha(\infty)$.
 
-**Definition**. The limit of a function at infinity is defined if for a given $\epsilon$, there exists some $\alpha$ so that there exists some $V_\delta(c)$ such that for all $x \in A$,
+**Definition**. The limit of a function at infinity is defined if for a given $\varepsilon$, there exists some $\alpha$ so that there exists some $V_\delta(c)$ such that for all $x \in A$,
 
 $$
-x > \alpha \Rightarrow |f(x) - L| < \epsilon
+x > \alpha \Rightarrow |f(x) - L| < \varepsilon
 $$
 
 ## Section 4.2 - Limit Theorems

docs/math/real-analysis/5-continuity.md

@@ -2,10 +2,10 @@
 
 ## Section 5.1 - Continuous Functions
 
-**Definition**. Let $A \subseteq \mathbb{R}$, and $f: A \rightarrow \mathbb{R}$. Then, if $a \in A$, $f$ is *continuous at $a$* if, given any $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x \in A$,
+**Definition**. Let $A \subseteq \mathbb{R}$, and $f: A \rightarrow \mathbb{R}$. Then, if $a \in A$, $f$ is *continuous at $a$* if, given any $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x \in A$,
 
 $$
-|x - a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon
+|x - a| < \delta \Rightarrow |f(x) - f(a)| < \varepsilon
 $$
 
 Note that if $a$ is an *isolated point* of $A$, that is, not a cluster point, then $a$ is automatically continuous.
@@ -20,10 +20,10 @@ Note that a function cannot be continuous at a point outside of its domain, even
 
 ---
 
-**Definition**. Let $(S, d_S)$ and $(T, d_T)$ be metric spaces. A function $f: S \rightarrow T$ is continuous at a point $a \in S$ if given any $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x \in S$,
+**Definition**. Let $(S, d_S)$ and $(T, d_T)$ be metric spaces. A function $f: S \rightarrow T$ is continuous at a point $a \in S$ if given any $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x \in S$,
 
 $$
-d_S(x, a) < \delta \Rightarrow d_T(f(x),  f(a)) < \epsilon
+d_S(x, a) < \delta \Rightarrow d_T(f(x),  f(a)) < \varepsilon
 $$
 
 **Theorem**. A function $f: S \rightarrow T$ is continuous at a point $a \in A$ if and only if given some neighborhood $V(f(a)) \in B$, there exists some $U(a) \in A$ such that $f(U) \subseteq V$.
@@ -63,10 +63,10 @@ $$
 
 ---
 
-**Definition**. Let $A \subseteq R$. Then, a function $f: A \rightarrow \mathbb{R}$ is *uniformly continuous* if given any $\epsilon > 0$, there exists some $\delta > 0$ depending only on $\epsilon$ such that for any $x, y \in A$,
+**Definition**. Let $A \subseteq R$. Then, a function $f: A \rightarrow \mathbb{R}$ is *uniformly continuous* if given any $\varepsilon > 0$, there exists some $\delta > 0$ depending only on $\varepsilon$ such that for any $x, y \in A$,
 
 $$
-|x - y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon
+|x - y| < \delta \Rightarrow |f(x) - f(y)| < \varepsilon
 $$
 
 Note that if $f$ is uniformly continuous, it must be continuous on $A$.