In this thread we learn Mathematical Analysis 1. Textbook: I'll

1.7 Definition: Suppose S is an ordered set, and [math]E \subset S[/math]. If there exists a [math]\beta \subset S[/math] such that [math]x \leq \beta[/math] for every [math]x \in E[/math], we say that E is bounded above, and call [math]\beta[/math] an upper bound of E.

Just testing.

>>8433376
1.8 Definition: Suppose S is an ordered set, [math]E \subset S[/math] and [math]E[/math] is bounded above. Suppose there exists an [math]\alpha \in S[/math] with the following properties:

(i) [math]\alpha[/math] is an upper bound of E.
(ii) If [math]\gamma < \alpha[/math] then [math]\gamma[/math] is not an upper bound of E.

Then [math]\alpha[/math] is called the least upper bound of [math]E[/math] or supremum.
We write: [math]\alpha = \sup E[/math]

The greatest lower bound, or infimum, of a set which is bounded below is defined in the same manner and written as:
[math]\alpha = \inf E[/math]

1.10 Definition: An ordered set [math]S[/math] is said to have the least-upper-bound property if the following is true:
If [math]E \subset S[/math], [math]E[/math] is not empty, and [math]E[/math] is bounded above, then [math]\sup E[/math] exists in [math]S[/math]

1.11 Theorem: Suppose [math]S[/math] is an ordered set with the least-upper-bound property, [math]B \subset S[/math], [math]B[/math] is not empty, and [math]B[/math] is bounded below. Let [math]L[/math] be the set of all lower bounds of [math]B[/math]. Then
[eqn]\alpha = \sup L[/eqn]

exists in [math]S[/math], and [math]\alpha = \inf B[/math].
In particular, [math]\inf B[/math] exists in [math]S[/math].

The proof is in the textbook.

Why isn't this thread being bumped when I post something?

>>8433412
I'm a slow learner. I need time. Keep up the good work though.

>>8433365
what is the point of this thread? to practice latex? do that on you own time and don't clutter up the board with these simple definitions and theorems from a textbook that is freely available by google

>>8433406
So now we're moving on to fields. >>8433415
Continue here after you have a general understanding of the previous definitions and perhaps look at the proof of 1.11.

Fields:
1.12 Definition: A field is a set [math]F[/math] with two operations, called addition and multiplication, which satisfy the following so-called "field axioms" (A), (M) and (D):

(A) Axioms for addition
(A1) If [math]x \in F[/math], then their sum [math][/math]x+y[math][/math] is in [math]F[/math].
(A2)Addition is commutative: [math]x+y=y+x[/math] for all [math]x,y \in F[/math].
(A3)Addition is associative: [math](x+y)+z=x+(y+z)[/math] for all [math]x,y,z \in F[/math].
(A4)[math]F[/math] contains an element [math]0[/math] such that [math]0 + x = x[/math] for every [math]x \in F[/math].
(A5)To every [math]x \in F[/math] corresponds an element [math]-x \in F[/math] such that [math]x + (-x) = 0[/math].

(M) Axioms for multiplication
(M1) If [math]x \in F[/math] and [math]y \in F[/math], then their product [math]xy[/math] is in [math]F[/math].
(M2) Multiplication is commutative: [math]xy = yx[/math] for all [math]x,y \in F[/math].
(M3)Multiplication is associative: [math](xy)z = x(yz)[/math] for all [math]x,y,z \in F[/math].
(M4)[math]F[/math] contains an element [math]1 \neq 0[/math] such that [math]1x = x[/math] for every [math]x \in F[/math].
(M5)If [math]x \in F[/math] and [math]x \neq 0[/math] then there exists an element [math]1/x \in F[/math] such that [math]x \cdot (1/x) = 1 [/math].

(D) The distributive law
[eqn]x(y+x) = xy + xz[/eqn]
holds for all [math]x,y,z \in F[/math].

>>8433452
"Why do Asians look like aliens?"
"Redpilled Professors (general)"
"How to not get emotional when learning?"
"Why are brainlets allowed to study in universities? They are nothing but trouble. "

These are all from page 1, the rest is pop-sci. If this thread is clutter I say we need more of it. But okay.

>>8433465
>there is already clutter so it is okay for me to add to it
in any case, those threads give way to discussions. you're just copy--pasting from a textbook, and i'm not sure why.

does rudin ever go into the Archimedean property of a field and so on?

>>8433507
he mentions the archimedian property, yes

>>8433457
What book is this?
Not sure what this course is, but we had this shit in our pre-calc course where the lecturer was giving out his own material (lots of errata).

Either way, I want to note the similarity between A5 and one of the algebraic axiom.

(just looking at the reals)
For every [math]x \neq 0[/math] exists the inverse [math]-x[/math] and [math]x^{-1}[/math] such that
[math]x+(-x)=0[/math]
[math]x \cdot x^{-1} = 1[/math]

>>8433365
Here's a question to you.

It is known that the set lR is a complete ordered field. Is there a field that is ordered, but not complete? If so, give an example. Is there a field that is complete, but not ordered? If so, give an example and explain why the field is complete.

>>8433390
Problems:
1. Prove that if X is a set, then [math](\mathcal P(X), \subseteq)[/math] is an ordered set. Prove that every nonempty subset has an infimum and a supremum (it is said to be a complete lattice).

2. Prove that [math](\mathbb N, |)[/math] is an ordered set (where [math]a | b \leftrightarrow \exists k \in \mathbb N, b = ka[/math]). Prove that it is a complete lattice.

3. Prove that [math]E = \{x \in \mathbb Q, x^2 < 2\}[/math] is a nonempty bounded above subset of [math]\mathbb Q[/math] that does not have a least upper-bound. Comment.

4. An ordered set is said to be well-ordered if every non-empty set has a least element. Prove that the statement that [math](\mathbb N, \le)[/math] is well-ordered is equivalent to the induction principle. Is [math](\mathbb N, |)[/math] well-ordered ?

5.a) Let [math](L, \le)[/math] be a complete lattice and [math]f: L \to L[/math] be an order-preserving function. Prove that f has a fixed point (Hint: Consider the set [math]\{x \in L, x \le f(x)\}[/math]).
b) (Schroeder-Bernstein) Let X and Y be two sets and [math]f: X \to Y[/math] and [math]g: Y \to X[/math] two injections.
Prove that [math]\Phi: S \mapsto X \setminus g(Y \setminus f(S))[/math] is an order-preserving map from [math]\mathcal P(X)[/math] to itself.
Let S be a fixed point of [math]\Phi[/math]. Prove that the map [math]h[/math] defined on X by [math]h(x) = f(x)[/math] if [math]x \in S[/math] and [math] f(x) = g^{-1}(x)[/math] otherwise is well-defined and is bijection from X to Y.

>>8433804
The poset in 2. doesn't have infinite joins

>>8433824
I should have said that I consider 0 to be in [math]\mathbb N[/math]