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#set document(
title: [Lecture 3],
author: "Gidon Rosalki",
date: auto,
)
#set heading(numbering: "1.")
#set text(font: "New Computer Modern")
#set page(margin: (
bottom: 0.5cm,
right: 1.5cm,
left: 1.5cm,
top: 1.5cm,
))
#set list(indent: 10pt)
#set enum(indent: 10pt)
#show title: set align(center)
#show link: set text(rgb("005eff"))
#show link: underline
#show math.equation.where(block: true): set block(breakable: true)
#title()
#align(center)[
Gidon Rosalki \
2026-11-03
]
#let colourmaths(x, color) = text(fill: color)[$#x$]
// author: eric1102 - https://sitandr.github.io/typst-examples-book/book/typstonomicon/math_display.html
// may slightly interfere with maths blocks
#show math.equation: it => {
if it.body.fields().at("size", default: none) != "display" {
return math.display(it)
}
it
}
#let qquad = h(0.5em)
= Introduction
We will be speaking about Infi today! Hooray! Not Linear Algebra, but Infi! Praise be, and gird thy loins. It appears
that this will be a recap of stuff we discussed in the bachelors degree courses, but I'm writing this paragraph at the
start of the lecture, so that may yet change...
= Limits, Sets, Continuity
*Definition* $RR^d$: This is the $d$-dimensional euclidean vector space over $RR$.
$
RR^d = {vec(x_1, x_2, dots.v, x_d) : x_1, dots, x_d in RR}
$
We define the standard basis, which has elements
$
e_1 = vec(1, 0, dots.v, 0), quad e_2 = vec(0, 1, dots.v, 0), dots, quad e_d = vec(0, 0, dots.v, 1)
$
*Definition* Euclidean norm:
$
norm(x) = sqrt(sum_(i=1)^(d)x_i^2)
$
Properties of $norm(dot.c)$:
- Positive definite: $forall x qquad norm(x) >= 0, norm(x) = 0 <=> x = 0$
- Triangle inequality: $forall x, y qquad norm(x + y) <= norm(x) + norm(y)$
- Homogeneity: $forall alpha in RR, x in RR^d, qquad norm(alpha x) = abs(alpha) norm(x)$
*Definition* (distance): The _distance_ between $x, y in RR^d$ is $d(x, y) = norm(x - y)$ \
Properties of $d(dot.c, dot.c)$:
- Positive definite: $d(x, y) >= 0, d(x, y) = 0 <=> x = y$
- Triangle inequality: $forall x, y, z, qquad d(x, z) <= d(x, y) + d(y, z)$
- Symmetry :$forall x, y qquad d(x, y) = d(y, x)$
*Definition* (Open ball): The _open ball_ of radius $r > 0$, centred at $x in RR^d$ is the set
$
B_r(x) = {y in RR^d : d(x, y) < r}
$
*Definition* (Closed ball): The _closed ball_ of radius $r > 0$, centred at $x in RR^d$ is the set
$
overline(B_r)(x) = {y in RR^d : d(x, y) <= r}
$
== Limits
*Definition* A sequence ${x_n}_(n = 1)^(infinity) subset.eq RR^d$ _converges_ to $x in RR^d$ if $forall i in [d],
qquad lim_(n -> infinity) x_i^n = x_i$
#pagebreak()
*Proposition*: Let ${x^n}_(n = 1)^infinity subset.eq RR^d$. It converges to $x in RR^d$ *if and only if* $lim_(n ->
infinity) norm(x^n - x) = 0$. We will note that this is arguing an alternate definition to the one above. \
_Proof_: Let us assume $lim_(n_> infinity) x^n = x$:
$
forall n, qquad norm(x^n - x) & = sqrt(sum_(i = 1)^(d) (x_i^n - x_i)^2) \
& underbrace(-->, n -> infinity) 0
$
Which follows since all of squared, sum, and square root are continuous functions, and by definition $x_i^n - x_i$ tends
to 0. \
Let us assume that $norm(x^n -> x)$:
$
sqrt((x_i^n - x_i)^2) & = abs(x_i^n - x_i) \
& <= sqrt(sum_(i =1)^(d) (x_j^n - x_j)^2) \
7 & underbrace(-->, n -> infinity) 0
$
*Definition* (Subsequence): Let ${x^n}_(n = 1)^(infinity) subset.eq RR^d$. A _subsequence_ is any sequence ${x^(n_k)}_(k =
1)^(infinity)$, where ${n_k}_(k = 1)^infinity$ is a strictly increasing sequence in $NN$. \
In short, a subsequence is an infinitely large subset of the original sequence, selected in increasing order of indices.
If we consider $NN$ to be our sequence, then sub-sequences could be the even numbers, the odd numbers, the prime
numbers, randomly selected numbers in increasing order, and so on.
*Proposition*: Fix sequences ${x^n}_(n = 1)^(infinity), {y^n}_(n = 1)^(infinity) subset.eq RR^d$ and
${alpha^n}_(n = 1)^(infinity) subset.eq RR$, with respective limits $x, y in RR^d, qquad alpha in RR$. Then:
+ $lim_(n -> infinity) (x^n + y^n) = x + y$
+ $lim_(n -> infinity) alpha_n x^n = alpha x$
+ Every subsequence of ${x^n}$ converges to $x$
_Proof_:
+ For every $n$:
$
norm(x^n + y^n - (x + y)) & overbrace(<=, triangle) norm(x^n - x) + norm(y^n - y) \
& --> 0 + 0
$
== Sets
*Definition* (closed sets): A set $A subset.eq RR^d$ is _closed_ if for every convergent sequence ${x^n}_(n = 1)^infinity subset.eq A$, their limit is also in $A$ (i.e. $lim_(n -> infinity) x^n in A)$. \
*Example* (half spaces): Let $w in RR^d, qquad a in RR$, and define $H = {x in RR^d : lr(chevron.l x, w chevron.r) >=
a}$ where \ $lr(chevron.l x, w chevron.r) = sum_(i = 1)^(d) x_i w_i$. Suppose ${x^n}_(n = 1)^infinity subset.eq H$
converges to $x$. We must prove that $x in F$, i.e. $lr(chevron.l x, w chevron.r) >= a$:
$
lr(chevron.l x, w chevron.r) & = sum_(i = 1)^(d)x_i w_i \
& = sum_(i = 1)^(d) w_i lim_(n -> infinity)x_i^n \
& = lim_(n -> infinity) underbrace(
sum_(i = 1)^(d)w_i x_i^n, lr(
chevron.l x^n, w
chevron.r
) >= a
) \
& >= a
$
#pagebreak()
*Example* (sphere): $S^(d - 1) = {x in RR^d : norm(x) = 1}$. Suppose ${x^n}_(n = 1)^(infinity) subset.eq S^(d - 1)$
that converges to $x$.
$
norm(x) & = sqrt(sum_(i in [d])^() x_i^2) \
& = sqrt(sum_(i in [d])^()(lim_(n -> infinity)x_i^n)^2) \
& = lim_(n -> infinity) sqrt(sum_(i in [d])^() (x_i^n)^2) \
& = lim_(n -> infinity) norm(x^n) \
& = 1
$
Thus the sphere is a closed set.
*Definition* (open set): A set $U$ is _open_ if $forall x in U qquad exists epsilon > 0 : B_epsilon (x) subset.eq U$.
*Example* (open half spaces): Let $w in RR^d, a in RR$, and define $H = {x in RR^d : lr(chevron.l w, x chevron.r) > a}$.
We claim that $H$ is open: Let $x in H$. Fix $epsilon > 0$, and suppose that $y in B_epsilon (x)$:
$
lr(chevron.l y, w chevron.r) & = lr(chevron.l y - x + x, w chevron.r) \
& = lr(chevron.l y - x, w chevron.r) + underbrace(lr(chevron.l x, w chevron.r), > a) \
abs(lr(chevron.l y - x, w chevron.r)) & underbrace(<=, "Cauchy-Schwarz") norm(y - x) norm(w) \
& < epsilon norm(w) \
==> lr(chevron.l y - x, w chevron.r) + lr(chevron.l x, w chevron.r) & > lr(chevron.l x, w chevron.r) - epsilon norm(w) \
lr(chevron.l x, w chevron.r) - epsilon norm(w) & > a \
<==> epsilon norm(w) & < lr(chevron.l x, w chevron.r) - a \
<==> epsilon & < (lr(chevron.l x, w chevron.r) - a) / norm(w)
$
*Definition* (boundary): The _boundary_ of a set $A subset.eq RR^d$ denoted $partial A$ is the set of all $x in RR^d$
s.t. \ $ forall epsilon > 0 qquad B_epsilon (x) inter A != emptyset and B_epsilon (x) inter A^c != emptyset $
*Example*: Let $B^d = {x in RR^d : norm(x) <= 1}$ (the unit ball). Then $partial B^d = S^(d - 1)$, i.e., the unit sphere:
- $S^(d - 1) subset.eq partial B^d$: Let $x in S^(d - 1), qquad epsilon > 0$. Then $x in B_epsilon (x)$. Also, $(1 +
epsilon / 2) x in B_epsilon (x) inter (B^d)^(c)$
- $partial B^d subset.eq S^(d - 1)$: Let $x in partial B^d, qquad epsilon > 0$. By definition $exists x_epsilon^- in B^d
inter B_epsilon (x) and x_epsilon^+ in (B^d)^c inter B_epsilon (x)$.
$
norm(x) & overbrace(<=, triangle) norm(x_epsilon^- - x) + norm(x_epsilon^-) \
& <= 1 + epsilon \
norm(x) & >= norm(x_epsilon^+) - norm(x - x_epsilon^+) \
& >= 1 - epsilon
$
Therefore $1- epsilon <= norm(x) <= 1 + epsilon qquad forall epsilon$, so $norm(x) = 1 => x in S^(d - 1)$.
We have shown both directions of containment, and so the sets are equivalent.
*Definition* (bounded sets): A set $A in RR^d$ is _bounded_ if $exists R > 0 : A subset.eq B_R^d := {x in RR^d :
norm(x) < R}$. I.e., a set is bounded if there exists a ball that contains the entire set.
#pagebreak()
*Definition* (compact sets): A set $A subset.eq RR^d$ is _compact_ if every sequence ${x_n}_(n = 1) ^ (infinity)
subset.eq A$ has a convergent subsequence whose limit is in $A$.
*Examples*
- The closed $d$ dimensional ball $overline(B)^d$
- $[0, 1]$
- $S^(d - 1)$
*Non-Examples*:
- $B^d$ (open)
- $(0, 1)$ (consider $1 / n -> 0 in.not (0, 1)$)
*Theorem*: A set $K subset.eq RR^d$ is compact *if and only if* it is closed, and bounded. \
_Proof_:
- $==>$: we have assumed that $K$ is compact, let us prove that it is closed. Let ${x_n}_(n = 1) ^ (infinity)
subset.eq K$ converging to $x in RR^d$. We need to prove that $x in K$. Since $K$ is compact, $exists {x_(n_k)}_(k=1) ^
(infinity)$ subsequence of ${x_n}$ that converges to $y in k$. Since ${x_n}$ is convergent, so every subsequence
converges to $ lim_(n -> infinity) x_n & = x \ => y & = x \ => x & in K $ \
If $K$ were _not_ bounded, then $forall n in NN qquad exists x_n in K : norm(x_n) >= n$. We claim that ${x_n}_(n = 1) ^
(infinity)$ has no convergent subsequence. Indeed, if $lim_(k -> infinity) x_(n_k) = x$, then $lim_(k -> infinity)
norm(x_n_k) = norm(x)$. However; $lim_(k -> infinity) norm(x_n_k) >= lim_(k -> infinity) n_k = infinity$. \
- $<==$: Suppose that $K$ is closed and bounded. Let ${x_n} subset.eq K$ be a sequence. Since $K$ is closed, it suffices
to prove that ${x_n}$ has a convergent subsequence. We proceed by induction on $d$: The base case is
Bolzano–Weierstrass. Let us suppose that ${x^(n_k)} : forall i < d qquad {x_i^(n_k)}$ converges. Then ${x_d^n_k}$ is a
bounded sequence in $RR$, and by Bolzano-Weierstrass, has a convergent subsequence ${x_d^n_k_l}$, but then
${x^(n_k_l)}$ is a convergent subsequence of ${x^n}$.
|
