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16:35
Sup
\Bbb R
$\Bbb R$
$\mathbb R$
\mathbb R
$\exists$
$$U_n \rightarrow a\in\mathbb{R} \Leftrightarrow \forall_{\epsilon > 0} \exists_p \forall_{n} n>p \Leftrightarrow U_n \in V_{\epsilon}(a) = |U_n - a| < \epsilon$$
Let $(a_n)_{n\ge 1}\subset \Bbb R$ be a sequence. We say that $a_n\to a$ if for any $\epsilon>0$ there is $N(\epsilon)\in\Bbb N$ such that $n\ge N$ implies $|a_n-a|<\epsilon$.
$N$ depends on $\epsilon$
$$\{a_n : n>p\}\subset V_\epsilon (a)$$
Eg. $a_n = (-1)^n$ is bounded
$|a_n|=1$ for all $n$.
Bounded means there exists some $M\in\Bbb R$ such that $|a_n|\le M$ for all $n$.
16:48
$$\exists _{c>0} \forall_n |a_n|\leq c$$
Let $A\subset \Bbb R$.
$b\in\Bbb R$ is an upper bound of $A$ if $b\ge a$, for all $a\in A$.
We say that $s$ is a supremum of $A$ if $s$ is an upper bound and $s\le b$ for any upper bound $b$ of $A$.
if $a_n$ has a supremum it is unique
$b\in A; b \geq c, \forall c\in A$
Suppose there is a $d \neq b$ s.t. $d$ is also a supremum of $A$
17:04
Since $b$ is a supremum, and $d$ is an upper bound, $b<d$.
Since $d$ is a supremum, and $b$ is an upper bound, $d<b$.
So $b<d<b$, which contradicts trichotomy.
This contradicts $b\ne d$.
$\lim x^n=0$ if $x<1$?
$C^n$
$$U_n = c^n, c\in\mathbb R$$
$a_1,a_2,a_3,...,a_{15},a_{16},...$
$a_1,a_3,a_{15},...$
$1,-1,1,-1,1,-1,\dotsc$
$1,1,1,1,\dotsc$
$(a_n)\subset\Bbb R$ is said to be Cauchy if $(\forall\epsilon>0)(\exists N\in\Bbb N):n,m\ge N\implies |a_n-a_m|<\epsilon$.
$V_\epsilon(x)$
Find $V_{1/2}(0)$ in $\Bbb Z$
17:27
$$u: \mathbb N_1 \rightarrow \mathbb R$$
$$u(n) = u_n \rightarrow\text{ General term of the solution}$$
$$(u_n) \equiv u$$
$$u(\mathbb N_1) = \{u_n : n\in\mathbb N_1\}\subset\mathbb R$$
Eg. $$u_n = \frac{1}{n} \\ \{\frac{1}{n};n\in\mathbb N_1\}$$
$u$ is said to be a real succession
A monotone sequence is a sequence for which $a_n\le a_{n+1}$ for any $n$.
Let $(U_n)$ be a real sequence and $a\in\mathbb R$. It said that $suc(U_n)$ converges to $a$ and it is notated $U_n \rightarrow a$ iff:
$$\forall_{\epsilon>0} \exists_{p\in\mathbb N_1} \forall_{n\in\mathbb N_1}, n>p \Rightarrow U_n \in V_\epsilon(a)$$
$u_n\in V_\epsilon(a)$
Write out what that means.
All the terms of the solution are found in the radius epsilon of a

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