Set theory
What even is a set?¶
A set is a well defined collection of objects, a set could contain the two shoes on you feet, or the \(5\) pieces of cheese on this cutting board (that I'm going to pretend exists), sets can even contain other sets, but sets can not contain themselves, because it would lead to a paradox: would the set that contains every set that doesn't contain itself contain itself?" This also means that there isn't a set that contains everything.
But the thing is, using some symbols, you can describe almost all of math. These symbols can just be pronounced as words, and it would make a sentence, such as " \(¬ \exists (x): |x| < 0\) " as " there does not exist \(x\) such that the absolute value of \(x\) is strictly less than \(0\) ". Time to rapidfire through each one's pronunciation and meaning.
meanings of things¶
\[ ∀ \text{ Is pronounced "for any" or "for all" (but I prefer "for any") and means what it says. It than has an open parentheses, a thing (or sometimes multiple things separated by a comma) (} x, y, z, \text{ or a set) that I will call } x \text{ for now, a closed parentheses (parenthese is not a word), a } \cdot \text{, a statement that implies something about } x \text{, a colon, and finish it off with a statement including } x. \]
\[ \text{left and right parentheses are not pronounced.} \]
\[ \cdot \text{ Is pronounced "such that" and it's only used in two contexts: "for any } x \text{ such that..." and "there exists } x \text{ such that...".} \]
\[ : \text{ Is pronounced however a colon is pronounced.} \]
\[ , \text{ Is pronounced however a comma is pronounced.} \]
\[ \exists \text{ Is pronounced "there exists" and I don't think I need to explain that.} \]
\[ ¬ \text{ Is pronounced "is not" or "does not" as in "there does not exist } x \text{".} \]
\[ \in \text{ Is pronounced "is an element of" where an element of a set is a singular object that is contained in that set.} \]
\[ x, y, \text{ And } z \text{ are pronounced "} x, y, \text{ And } z \text{" and they all mean "a thing that could be an element of a set".} \]
\[ \text{capital letters are sets.} \]
\[ \iff \text{ Is pronounced "if and only if" as in "if statement } a \text{ is true, statement } b \text{ is true, and if statement } a \text{ is false, statement } b \text{ is false".} \]
\[ ∩ \text{ Is pronounced "and" and means "} a ∩ b \text{ is true if and only if statement } a \text{ is true and } b \text{ is true", it can also mean the intersection of two sets, in that case, it is pronounced "intersectioned with", but I'll get to its formal meaning in the next chapter.} \]
\[ = : \text{ Is pronounced "equals by definition" and means "define the thing on the left as the thing on the right", or was it the other way around?} \]
\[ ∨ \text{ Is pronounced "or" and means "} a ∨ b \text{ is true if statement } a \text{ is true or } b \text{ is true... Or both!", it can also mean the union of two sets, in that case, it is pronounced "unioned with", but I'll get to its formal meaning in the next chapter.} \]
\[ \text{succ Is pronounced "the immediate successor of" and means "that number } + 1 \text{".} \]
\[ → \text{ Is technically called the if then sign, but it is pronounced "implies" and means "statement } a → b \text{ is true if statement } a \text{ being true implies statement } b \text{ is true", so } a → b \text{ is true if statement } a \text{ is true and statement } b \text{ is true, or if statement } a \text{ is false and statement } b \text{ is false, or if statement } a \text{ is true and statement } b \text{ is false, but not if statement } a \text{ is false and statement } b \text{ is true. Also, if there was an element sign two spaces behind, pronounce it "being an element of" as opposed to "is an element of".} \]
\[ \text{I don't think that I need to explain the } < \text{sign.} \]
\[ ℝ \text{ Is pronounced "the set of all real numbers" and means, well, the set of all real numbers.} \]
\[ ℕ \text{ Is pronounced "the set of all natural numbers" and means "the set of all positive integers", it's debatable weather or not it includes } 0 \text{.} \]
definitions¶
\[ Ø \text{ Is pronounced "the empty set" and means "the set of which is empty inside", but here that is in set theory:} \]
\[ ¬ \exists (x) \cdot x \in Ø \]
\[ ⊆ \text{ Is pronounced "is a subset of", and the meaning of that is:} \]
\[ A ⊆ B \iff ∀(x) \cdot x \in A: x \in B \]
\[ pow \text{ Is pronounced "the power set of" as in "} pow(S) \text{", and the meaning of that is:} \]
\[ ∀(P) \cdot ∀(U) \cdot U ⊆ S: U \in P ∩ ∀(T) \cdot T ¬ ⊆ S: T ¬ \in P: P = : pow(S) \]
\[ = \text{ Is pronounced "is the same as" or "is equal to", and the meaning of that is:} \]
\[ A = B \iff A ⊆ B ∩ B ⊆ A \]
\[ ¬ \exists (S) \cdot S \in S \]
\[ x \in \in S \iff \exists (U) \cdot U \in S ∩ x \in U \]
\[ x \in \in S \text{ can also be written as } \in^2 \]
\[ x \in \in \in S \iff \exists (U) \cdot U \in S ∩ x \in^2 U \]
\[ x \in \in \in S \text{ can also be written as } \in^3 \]
\[ x \in \in \in \in S \iff \exists (U) \cdot U \in S ∩ x \in^3 U \]
\[ x \in \in \in \in S \text{ can also be written as } \in^4 \]
\[ \vdots \]
\[ x \in^{a + b} S \iff \exists (U) \cdot U \in^a S ∩ x \in^b U \]
\[ x \in^{a + b} S \text{ can also be written as } x \in^a \in^b S \]
\[ \in^S \text{ Is pronounced "is a super element of" (} S \text{ for super), and the definition is:} \]
\[ x \in^S S \iff x \in S ∨ \exists (U) \cdot U \in S ∩ x \in^S U \]
.¶
Was recursion in the rule book? I guess so.
Recursion? Get it?
numbers¶
\[ 0 = Ø \]
\[ \text{succ} (n) \text{ (Which mathematically equals } n + 1 \text{) Is how you would usually define numbers, so I'll define numbers that way, I'll say that succ} (n) \text{ is the set that contains all numbers } 0 \text{-} n \text{. But first: the union of two sets, denoted as an } ∨ \text{ sign.} \]
\[ x \in A ∨ B \iff x \in A ∨ x \in B \]
\[ set \text{ Is pronounced "the set containing" as in "} set(S) \text{", and} \]
\[ ∀(S) \cdot E \in S ∩ ∀(T) \cdot T ¬= E: T ¬ \in S: S = : set(E) \]
\[ \text{Around } 100 \text{ lines?? (I might add or remove another definition, but at the time of typing this, this is on } 95 \text{ lines.)} \]
\[ \text{succ} (n) = : set(n) ∨ n \]
back to definitions¶
\[ ∩ \text{ Is pronounced "and" and means "} a ∩ b \text{ is true if and only if statement } a \text{ is true and } b \text{ is true", it can also mean the intersection of two sets, in that case, it is pronounced "intersectioned with", but} \]
\[ x \in A ∩ B \iff x \in A ∩ x \in B \]
\[ \text{Here's another definition of the subset: } ∀(A, B) \cdot ¬ \exists (x) \cdot x \in A ¬ → x \in B: A ⊆ B. \]
\[ \text{And another one! } A ⊆ B \iff ¬ \exists (x) \cdot x \in A ∩ x ¬ \in B \]
\[ n_1 < n_2 \iff n_1 \in n_2 \]
\[ \text{WARNING! The next statement is the axiom of choice, kinda controversial.} \]
\[ ∀(S) \cdot S ¬= Ø: \exists (x) \cdot x \in S \]
group theory¶
A group (call it \(G\)) is a certain type of set, including an addition like thing represented with a \(+\) sign (this addition like thing could also be multiplication), let's start with the set of numbers \(0\) - \(4\) under modular addition. To be a group, it has to follow \(4\) different rules.
\[ 1 \text{, Closure} \]
\[ ∀(a, b) \cdot a \in G ∩ b \in G: a + b \in G \]
Because it is modular, this holds true for modular addition.
\[ 2 \text{, Associativity} \]
\[ ∀(a, b, c) \cdot a \in G ∩ b \in G ∩ c \in G: (a + b) + c = a + (b + c) \]
Because addition is associative, this holds true for modular addition.
\[ 3 \text{, Identity} \]
\[ \exists (e) \cdot e \in G ∩ ∀(a) \cdot a \in G: a + e = a ∩ ∀(b) \cdot b \in G: e + b = b \]
Because of \(0\), this holds true for modular addition.
\[ 4 \text{, Inverses} \]
\[ ∀(a) \cdot a \in G: \exists (b) \cdot b \in G ∩ a + b = e ∩ b + a = e \]
Because of negatives and them looping back around, this holds true for modular addition. (Also, modular multiplication almost works, but it fails at this step because there is no \(\frac{1}{0}\).)
Thus, the set of numbers \(0\) - \(4\) under modular addition is a group.
limits¶
\[ \lim_{n \to \infty} f(n) = x \iff ∀(y) \cdot y \in ℝ ∩ y \ne x: \exists (n) \cdot n \in ℕ ∩ ∀(k) \cdot k \in ℕ ∩ k ≥ n: |f(k) - x| < |f(k) - y| \]
\[ \lim_{n \to \infty} f(n) \to \infty \iff ∀(y) \cdot y \in ℝ: \exists (n) \cdot n \in ℕ ∩ ∀(k) \cdot k \in ℕ ∩ k ≥ n: f(k) > y \]
\[ \lim_{n \to \infty} f(n) \to -\infty \iff ∀(y) \cdot y \in ℝ: \exists (n) \cdot n \in ℕ ∩ ∀(k) \cdot k \in ℕ ∩ k ≥ n: f(k) < y \]
\[ \lim_{n \to \infty} f(n) = 0 \iff ∀(y) \cdot y \in ℝ ∩ y \ne 0: \exists (n) \cdot n \in ℕ ∩ ∀(k) \cdot k \in ℕ ∩ k ≥ n: |f(k)| < |y| \]
\[ \lim_{n \to \infty} f(n) = x^{+} \iff ∀(y) \cdot y \in ℝ ∩ y \ne x: \exists (n) \cdot n \in ℕ ∩ ∀(k) \cdot k \in ℕ ∩ k ≥ n: |f(k) - x| < |f(k) - y| ∩ f(k) > x \]
\[ \lim_{n \to \infty} f(n) = x^{-} \iff ∀(y) \cdot y \in ℝ ∩ y \ne x: \exists (n) \cdot n \in ℕ ∩ ∀(k) \cdot k \in ℕ ∩ k ≥ n: |f(k) - x| < |f(k) - y| ∩ f(k) < x \]
set theory proofs?¶
No.