Spellchecker

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

@@ -56,7 +56,7 @@
 2. $|ab| = |a||b|$
 3. $|a + b| \leq |a| + |b|$
 
-**Corollary**. Givem $a, b \in \mathbb{R}$, then $\abs{\abs{a} - \abs{b}} \leq \abs{a - b}$.
+**Corollary**. Given $a, b \in \mathbb{R}$, then $\abs{\abs{a} - \abs{b}} \leq \abs{a - b}$.
 
 **Remark**. Every field has at least one absolute value function.
 
@@ -86,7 +86,7 @@ $$
 
 ---
 
-**Decimals**. Let $x \in \mathbb{R}$ such that $x > 0$. By the archimedian property, there exists some $b_0 \in \mathbb{N} \cup {0}$ such that $b_0 < x < b_0 + 1$. We can repeat this to see
+**Decimals**. Let $x \in \mathbb{R}$ such that $x > 0$. By the archimedean property, there exists some $b_0 \in \mathbb{N} \cup {0}$ such that $b_0 < x < b_0 + 1$. We can repeat this to see
 
 $$
 x = b_0 + \frac{b_1}{10} + \frac{b_2}{100} + \ldots + \frac{b_n}{100^n} + \ldots

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

@@ -16,7 +16,7 @@ $$
 (x_n) = (a, ar,  ar^2, ar^3, \ldots)
 $$
 
-**Example**. The *arithmatic sequence*, given base $a \in \mathbb{R}$ and distance $d \in \mathbb{R}$,
+**Example**. The *arithmetic sequence*, given base $a \in \mathbb{R}$ and distance $d \in \mathbb{R}$,
 
 $$
 (x_n) = (a, a + d, a + 2d, a + 3d, \ldots)
@@ -52,7 +52,7 @@ $$
 2. $x_n \cdot y_n \rightarrow xy$
 3. If $x_n \neq 0$ for all $n$, then $\frac{1}{x_n} \rightarrow \frac{1}{x}$
 
-**Theorem**. Suppose $(x_n)$ aand $(y_n)$ are convergent sequences and $N \in \mathbb{N}$. Then,
+**Theorem**. Suppose $(x_n)$ and $(y_n)$ are convergent sequences and $N \in \mathbb{N}$. Then,
 
 1. If $x_n \leq y_n$ for all $n \geq N$, then $\lim(x_n) \leq \lim(y_n)$
 2. If $x_n \leq a$ for all $n \geq N$, then $\lim(x_n) \leq a$
@@ -110,7 +110,7 @@ $$
 
 ## Section 3.7 - Series
 
-**Definition**. Let $(x_n)$ be a sequence in $\mathbb{R}$. Then, the *infinite series genearted by $X$* is the sequence $S = (s_n)$ with terms
+**Definition**. Let $(x_n)$ be a sequence in $\mathbb{R}$. Then, the *infinite series generated by $X$* is the sequence $S = (s_n)$ with terms
 
 $$
 s_1 = x_1; \; s_{n+1} = s_n + x_{n+1}

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

@@ -1,12 +1,12 @@
 # Chapter 4 - Limits
 
-## Secrion 4.1 - Limits of Functions
+## Section 4.1 - Limits of Functions
 
-**Definition**. Let $A \subseteq \mathbb{R}$. Then, a point $c \in \mathbb{R}$ is a *cluster point* of $A$ if for every $\delta > 0$, the $\delta$-neighborhood of $c$ contains a point $a \in A$ such thhat $a \neq c$. That is, there exists some $a$ such that $0 < |a - c| < \delta$.
+**Definition**. Let $A \subseteq \mathbb{R}$. Then, a point $c \in \mathbb{R}$ is a *cluster point* of $A$ if for every $\delta > 0$, the $\delta$-neighborhood of $c$ contains a point $a \in A$ such that $a \neq c$. That is, there exists some $a$ such that $0 < |a - c| < \delta$.
 
 **Theorem**. A real number $c$ is a cluster point for a set $A$ if and only if there exists a sequence $(a_n)$ in $A\\ \{c\}$ such that $a_n \rightarrow c$
 
-**Corollary**. A real number $c$ is a cluster point of a set $A$ if and only if every $\delta$-neighborhood conttains infinitely many points of $A$.
+**Corollary**. A real number $c$ is a cluster point of a set $A$ if and only if every $\delta$-neighborhood contains infinitely many points of $A$.
 
 **Definition**. The set of every cluster point of $A$ is called the *derived set* of $A$, and denoted $A'$.
 
@@ -18,21 +18,21 @@
 
 ---
 
-**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 goven 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 $\epsilon > 0$, there exists some $\delta > 0$ such that
 
 $$
 0 < |x-c| < \delta \Rightarrow |f(x) - L| < \epsilon
 $$
 
-**Therorem**. For a given function and cluster point, there can be at most one limit at said point.
+**Theorem**. For a given function and cluster point, there can be at most one limit at said point.
 
 **Theorem**. Let $A \subseteq \mathbb{R}$ and $f: A \rightarrow \mathbb{R}$. Then, to show that $lim_{x \rightarrow c} f(x) = L$, it suffices to show that for every sequence $(a_n)$ in $A\\ \{c\}$, the sequence $(f(a_n))$ converges tto $L$.
 
 ---
 
-**Definition**. The *extended real numbers* are $\hat{\mathbb{R}} = \mathbb{R} \cup \{ \infty, -\infty \}$ are a totally-ordered set witth supremum and infimum. Note that this set is no longer a field.
+**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 limitt of $f$ at $c$ is infinite if given some $\alpha$,  there exists some $V_\delta(c)$ such that forr 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 $x \in V_\epsilon(c)$, 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 $x \in A$,
 
@@ -78,4 +78,4 @@ $$
 f(x) \leq g(x) \leq h(x) \; \text{ for all } x \in A, x \neq c
 $$
 
-Then, $\lim_{x \rightarrow c} g(x) = L$.
+Then, $\lim_{x \rightarrow c} g(x) = L$.

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

@@ -1,4 +1,4 @@
-# Chapter 5 - Continuiy
+# Chapter 5 - Continuity
 
 ## Section 5.1 - Continuous Functions
 
@@ -8,7 +8,7 @@ $$
 |x - a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon
 $$
 
-Note that if $a$ is an *isolateed point* of $A$, that is, not a cluster point, then $a$ is automatically continuous.
+Note that if $a$ is an *isolated point* of $A$, that is, not a cluster point, then $a$ is automatically continuous.
 
 If $a$ is a cluster point of $A$, then this definition collapses to the definition of $\lim_{x \rightarrow a} f(x) = f(a)$.
 
@@ -20,17 +20,17 @@ 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 thaat 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 $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x \in S$,
 
 $$
 d_S(x, a) < \delta \Rightarrow d_T(f(x),  f(a)) < \epsilon
 $$
 
-**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 xists some $U(a) \in A$ such that $f(U) \subseteq V$.
+**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 $f(U) \subseteq V$.
 
-## Section 5.2 - Combinations of continuous Functions.
+## Section 5.2 - Combinations of continuous Functions
 
-**Theorem**. Let $f, g: A \rightarrow \mathbb{R}$ be continuous at $a \in A$. Then, 
+**Theorem**. Let $f, g: A \rightarrow \mathbb{R}$ be continuous at $a \in A$. Then,
 
 - $f + g$ and $fg$ are continuous at $a$
 - If $g(x) \neq 0$ for all $x \in A$, then $\frac{f}{g}$ is continuous at $a$.
@@ -51,15 +51,15 @@ $$
 
 **Corollary**. Let $f: A \rightarrow \mathbb{R}$ be a continuous function, with $A$ being a compact subset of metric space $S$. Then, $f(A)$ is closed and bounded. Moreover, there exists a $p, q \in A$ such that $f(p)$ and $f(q)$ are the supremum and infimum of $f(A)$.
 
-**Corollary**. Maximum-Minimum Theorem. If $I = [a, b]$ is a closed and bounded interval and $f: I \rightarrow \mathbb{R}$ is continuous on $I$, then $f$ has an absolute minumum and maximum on $I$.
+**Corollary**. Maximum-Minimum Theorem. If $I = [a, b]$ is a closed and bounded interval and $f: I \rightarrow \mathbb{R}$ is continuous on $I$, then $f$ has an absolute minimum and maximum on $I$.
 
 ---
 
 **Theorem**. Let $S, T$ be metric spaces and $A \subseteq S$. Then, if $f: A \rightarrow T$ is continuous on $A$, and $A$ is a connected subset of $S$, then $f(A)$ is a connected subset of $T$.
 
-**Corollary**. Suppose that $I$ is an interval. Let $f: I \rightarrow \mathbb{R}$ be continuous on $I$. Then, $f(I)$ is an intterval.
+**Corollary**. Suppose that $I$ is an interval. Let $f: I \rightarrow \mathbb{R}$ be continuous on $I$. Then, $f(I)$ is an interval.
 
-**Theorem**. (Bolzano's) Invermediate Value Theorem. Suppose $f: [a, b] \rightarrow \mathbb{R}$ is continuous on $[a, b]$ with $a \neq b$. Then, given some $k$ such that $f(a) < k < f(b)$, there exists some $c \in (a, b)$ such that $k = f(c)$.
+**Theorem**. (Bolzano's) Intermediate Value Theorem. Suppose $f: [a, b] \rightarrow \mathbb{R}$ is continuous on $[a, b]$ with $a \neq b$. Then, given some $k$ such that $f(a) < k < f(b)$, there exists some $c \in (a, b)$ such that $k = f(c)$.
 
 ---
 
@@ -77,4 +77,4 @@ Note that if $f$ is uniformly continuous, it must be continuous on $A$.
 
 **Theorem**. Suppose $A \subseteq \mathbb{R}$ and $f: A \rightarrow \mathbb{R}$ is uniformly continuous. Then, if $(x_n)$ is a Cauchy sequence in $A$, $(f(x_n))$ is a Cauchy sequence in $\mathbb{R}$.
 
-**Remark**. Suppose $S, T$ are metric spaces and $f: S \rightarrow T$ is uniformly continuous. Then, if $(x_n)$ is a Cauchy sequence in $S$, $(f(x_n))$ is a Cauchy sequence in $T$.
+**Remark**. Suppose $S, T$ are metric spaces and $f: S \rightarrow T$ is uniformly continuous. Then, if $(x_n)$ is a Cauchy sequence in $S$, $(f(x_n))$ is a Cauchy sequence in $T$.