Real Analaysis

Real and Complex number systems

Theorem

Archimedean and Density Property of Real Numbers: if $x, y \in R$
(a) then there exists a positive integer n such that

n x > y

(b) then there exists $p \in Q$ such that

x < p < y

Proof. use well ordering principle...
$◻$

Proposition

prove

sup_{y \in Y} (inf_{x \in X} f (x, y)) \leq inf_{x \in X} (sup_{y \in Y} f (x, y))

Proof. By definition of supremum we have

sup_{y \in Y} (inf_{x \in X} f (x, y)) - ϵ < inf_{x \in X} f (x, y^{\*})

But we also know that for some $y^{*} \in Y$

inf_{x \in X} f (x, y^{\*}) \leq inf_{x \in X} (sup_{y \in Y} f (x, y)),

because $f (x, y^{*}) \leq sup_{y \in Y} f (x, y)$ for every $x$ .

Combining the above 2 relations gives

sup_{y \in Y} (inf_{x \in X} f (x, y)) - ϵ < inf_{x \in X} (sup_{y \in Y} f (x, y)) .

and because this is true for any $ϵ > 0$ we have

sup_{y \in Y} (inf_{x \in X} f (x, y)) \leq inf_{x \in X} (sup_{y \in Y} f (x, y)) .

$◻$

Basic Toplogy

Definition

A metric space is a set X that is equipped with a metric $d : X \times X \to [0, \infty)$

(Identification) $d (x, y) = 0$ if and only if $x = y$
(Symmetry) $d (x, y) = d (y, x)$ for all $x, y \in X$
(Triangle Inequality) $d (x, y) \leq d (x, z) + d (z, y)$ for all $x, y, z \in X$

Remark

A subset of a metric space X is a metric space. It is clear that if the above properties hold for $x, y, z \in X$ they will hold in the subset of X

Example

$R, R^{k}$ are metric spaces

Limit Point

A point $p$ is a limit point of the set E if every neighborhood of $p$ contains a point $q \neq p$ such that $q \in E$

Example

But that does not mean the neighborhoods of $p$ are contained within/a subset of $E$ ! it only just has to have members in $E$ ! Eg.
attachments/Real Analaysis 2025-06-13 01.49.01.excalidraw.png

Definition

E is closed if every limit point of $E$ is a point of $E$

Definition

A point $p$ is said to be an interior point of $E$ if there is a neighborhood $N$ of $p$ such that $N \subset E$

Note that in the definition for neighborhood there was no requirement for $q \neq p$ to actually be present in $N$ . For all you know it could even be empty. Instead an interior point is basically saying that it is possible to define a radial area around $p$ of a real and non-zero radius $r$ that lies within $E$ . In other words if $p$ lies on the boundary it cannot possibly be an interior point.

Definition

$E$ is open if every point of $E$ is an interior point $E$

Proposition

every neighbourhood is an open set

Proof. Consider $d (p, q) < r$ , a neighbourhood of p with radius r. For any possible point p in this this neighbourhood there must exist some h for which $d (p, q) < h \leq r$ is satisfied. This certainly is possible by Archimedean property of real numbers. Then by definition every possible point in the neighbourhood is contained in a neighbourhood that is a subset of the parent set, that is the original neighborhood

Definition

Set E open subset of metric space X is another way of saying E is open relative to X. That is for all $p \in E$ there must exists $r$ such that for any $q \in X$ within this neighborhood

d (p, q) < r

then it can only possibly be from E.

attachments/Real Analaysis 2025-06-13 02.02.25.excalidraw.png

Once again such a neighborhood may not actually contain points but you must certainly a real nonzero radius where this could be possible. Well say there are points in E that are limit points of $X ∖ E$ then certainly this isn't possible. In other words there is some asshole point in X outside E that is infinitely close to one of the points of E.