\3. Proof 1
-S\coloneqq\{-x|x\in S\} \text{Let } d=sup(S)\Leftrightarrow d&\geq x,\forall x\in S\\ d&=min(ub(S))\\ &\text{if } u\gt x,\forall x\in S \text{ then } u\geq d\Leftrightarrow -u\leq-d (3)\\ d\geq x\leftrightarrow-d\leq-x\\ -d\in lb(-S) (1) u\geq x\Leftrightarrow-u\leq-x\implies-u\in lb(-S) (2)\\ (1),(2),(3)\implies -d=inf(-S) \end{align} $$Proof 2 (less rigorous)\begin{align} ub(S)=[sup(S),\infty)\ lb(-s)=[-\infty, -sup(S)]\ \implies inf(-s)=-sup(S) \end{align} 4. Let $f : D → \mathbb{R}$ and $g : D → \mathbb{R}$ be two functions defined on a nonempty set $D$. Prove that \underset{x\in D}{inf} (f(x)+g(x)\geq \underset{x\in D}{inf}f(x)+\underset{x\in D}{inf}g(x)sup_{x\in D}(f(x)+g(x)\leq sup_{x\in D}f(x)+sup_{x\in D}g(x)$$Give examples where the above inequalities are strict. Solution:
f(x)\leq \underset{x\in D}{sup\ f(x)}=F, \forall x\in D\\ g(x)\leq \underset{x\in D}{sup\ g(x)}=G, \forall x\in D\\ f(x)+g(x)\leq F+G\forall x\in D\implies sup(f(x)+g(x))\leq F+G\\ \\ D=[0,1], x\in D\\ f(x)=x, sup\ f(x)=1\\ g(x)=-x, sup\ g(x)=0 f(x)+g(x)=0 \end{align}Definition
is a neighborhood of if
- Which of the following sets are neighborhoods of 0? a. \varepsilon= \frac{1}{2},\ (-\frac{1}{2},\frac{1}{2})\subseteq[-1,1]\cup\{2\}\\ \end{align} b.
\nexists\varepsilon\text{ s.t. } (-\varepsilon,\varepsilon)\subseteq(-1,1)\cup\mathbb{Q}
\end{align}
$$
c. $\bigcap_{n=1}^\infty[-\frac{1}{n},\frac{1}{n}]$
$$\begin{align}
\bigcap_{n=1}^\infty[-\frac{1}{n},\frac{1}{n}]=0\\
0\in[-\frac{1}{n},\frac{1}{n}], \forall n\geq1
\end{align}$$
7. Let $x \in \mathbb{R}$ and $U, V \in \mathcal{V}(x)$. Prove that $U \cup V \in \mathcal{V}(x)$.
$$\begin{align}
U \cup V \in \mathcal{V}(x),&\exists\varepsilon_1,\text{ s.t. } (x-\varepsilon_1, x+\varepsilon_1)\subset\mathcal{V}\\
&\exists\varepsilon_2,\text{ s.t. } (x-\varepsilon_2, x+\varepsilon_2)\subset\mathcal{V}\\
Let\ \varepsilon=min(\varepsilon_1,\varepsilon_2)\\
(x-\varepsilon,x+\varepsilon)\subset U,V\implies\\
\implies (x-\varepsilon, x+\varepsilon)\subset U\cup V\implies U\cup V\in\mathcal{V}(X)
\end{align}
$$$
- 📅 2024-10-14 ⏫ exercises 5,8,10
Interiors & Closures
If you take any number in the interior you can find a neighborhood inside the set A So an element is in the closure if any neighborhood will intersect the set A
- Find the interior and the closure for each of the following sets: a. b. c. d. Since for any interval that we take there will always be neighborhoods that aren’t subsets of (because we have only whole numbers)