Second hour

1 files changed, 49 insertions, 1 deletions

diff --git a/lecture_3.typ b/lecture_3.typ
index ee3bec6..72d0c8b 100644
--- a/lecture_3.typ
+++ b/lecture_3.typ
@@ -117,7 +117,7 @@ $
                             & --> 0 + 0
 $
 
-= Sets
+== 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) >=
@@ -162,3 +162,51 @@ $
   <==> 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 ${x^n}$.
+
+