Definitions / Statements for Exam II

Let's generate a list of definitions and statements that we need to know for Exam II. I'll start with the first couple of definitions that you'd be responsible for:

• Definition 3.2.1 (Open set): A set $U\subset\mathbb R$ is open if for all points $x\in U$ there is an $\varepsilon>0$ such that $B_{\varepsilon}(x) \subset U$.
• Definition 3.2.4 (Limit point): A point $x\in\mathbb R$ is a limit point of a set $A$ if every $\varepsilon$-neighborhood
  $B_{\varepsilon}(x)$ intersects the set $A$ in some point other than $x$.
  

Note that I've identified the definition under consideration with both the book's definition number and the concept name. I think this helps place it in context. Also, the wording is very close to our text's wording, though I've used my preferred notation. I think you can use the textbook's exact notation or a slight variant, as long as it's clear.

This list will be "official" after I say so!

• Theorem 4.4.2 (Preservation of Compact Sets) Let $f: A\rightarrow \mathbb{R}$ be contiuous on $A$. If $K\subseteq A$ is compact, then $f(K)$ is compact as well.

• Definition 3.2.7 (Closed Set): A set $F \subseteq \mathbb{R}$ is closed if it contains all its limit points.
• Definition 3.3.1.(Compact): A set $K \subseteq \mathbb{R}$ is compact if every sequence in $K$ has a subsequence that converges to a limit that is also in $K$.
• Theorem 3.3.4.(Heine-Borel Theorem): A set $K \subseteq \mathbb{R}$ is compact iff it is closed and bounded.

• Definition 4.3.1 (Continuous at a point): A function $f: A \rightarrow \mathbb{R}$ is continuous at a point $c \in A$ if $\forall \,\,\varepsilon > 0\,\, \exists\,\, \delta > 0\, \ni$ whenever $|x - c| < \delta$ (and $x \in A$) it follows that $|f(x) - f(c)| < \varepsilon$.
  If $f$ is continuous at every point in the domain $A$ then we say that $f$ is continuous on A.
  

• Definition 4.4.5 (Uniformly continuous): A function $f: A \rightarrow \mathbb{R}$ is uniformly continuous on A if $\forall \, \,\varepsilon > 0\,\, \exists\,\, \delta > 0 \ni |x-y| < \delta$ implies $|f(x) - f(y)| < \varepsilon$.

• Definition 4.2.1 (Functional Limits): Let $f$ be defined on an open interval $I$, and let $a \in I$. We can say $lim_{x \rightarrow a}f(x)=L$ if for every $\epsilon >0$ there exists a $\delta>0$ such that $|f(x)-L| < \epsilon$ whenever $0<|x-a|<\delta$.

• Theorem 4.2.3:(Sequential criterion for Functional Limits) Let $f \mapsto \mathbb{R}$ and a limit point $c$ of $A$ then the following are equivalent:

i. $\lim_{x\to c} f(x) = L $

ii. For all sequences $(x_n) \leq A$ satisfying $x_n \neq c$ and $x_n \mapsto c $ if follows that $f(x_n) \mapsto L. $

• Definition 3.4.1 (Perfect): A set $P \subseteq \mathbb{R}$ is perfect if it is closed and contains no isolated points.
• Theorem 3.4.7 (Connected): A set $E \subseteq \mathbb{R}$ is connected iff whenever $a < c < b$ with $a,b \in E$, it folows that $c \in E$ as well.
• Theorem 4.5.1 (Intermediate Value Theorem): If $f: [a,b] \rightarrow \mathbb{R}$ is continuous, and if L is a real number satisfying $f(a) < L < f(b)$ or $f(a) > L > f(b)$, then there exists a point $c \in (a,b)$ where $f(c) = L$.

A little bit simpler version for IVT that mark gave in class for those of us who are bad a memorizing: Suppose that $f: [a,b] \mapsto \mathbb{R}$ is continuous on $[a,b]$ and that

$f(a) < 0 < f(b). $ Then $\exists$ $c \in (a,b) \ni f(c) = 0.$

definition 5.2.1 (derivative) Let $f: A \rightarrow \mathbb{R}$. Given $a \in A$ the derivative of $f$ at $a$, denoted $f'(a)$, is: $$f'(a) = \lim_{x \to\ a} \frac{f(x) - f(a)}{x - a},$$
provided that this limit exists. Note, if $\exists$ $f'$ $\forall$ $a \in A$, we say that $f$ is differentiable on $A$.

Definition 3.2.11 (closure) Given a set $A\subseteq\mathbb{R}$, let $L$ be the set of all limit points of $A$. The closure of $A$ is defined to be $\bar{A}=A\cup L$.

Theorem 5.2.3. If $g: A \rightarrow \mathbb{R}$ is differentiable at a point $c \in A$, then $g$ is continuous at $c$ as well.

Theorem 5.2.3. Proof
W.T.S $\lim_{x\to c} g(x) = g(c) $ or equivalently, $\lim_{x\to c} g(x) - g(c) = 0 $.

$$\lim_{x\to c} g(x) - g(c) = 0 $$
$$=\lim_{x\to c} \frac{g(x) - g(c)}{x-c}(x-c) $$
$$= \lim_{x\to c} \frac{g(x) - g(c)}{x-c} * \lim_{x\to c} (x-c)$$
$$= g'(c) * 0 = 0 $$

It follows that $\lim_{x\to c} g(x) = g(c) $.

This is all looking very good! I don't think anybody mentioned isolated point or the algebraic properties of limits or continuity, but it's still very good. Let's not worry about those, but a proof of an algebraic property (like $f+g$ is continuous at $c$, whenever $f$ and $g$ are) might be a logical proofy thing to expect.

I don't know if it helps you all much at this point, but I had compiled the theorems and definitions from this discussion into an organized printable document in case it helps any of you studying today.

This is great! Thanks so much for sharing!!!

Proof Let $\varepsilon >0$. We know $f$ and $g$ are continuous at $c$. Therefore, we define $ \delta_1 >0$ such that $\left| x-c \right| <\delta_1 \implies \left| f(x)-f(c)\right| < \frac{\varepsilon}{2}$ and $ \delta_2 >0$ such that $\left| x-c \right| <\delta_2 \implies \left| g(x)-g(c)\right| < \frac{\varepsilon}{2}$. Choose $\delta = \min{\{\delta_1, \delta_2\}}$. Note when $\left| x-c \right| <\delta$ we have $$\left| (f(x)+g(x))-(f(c)+g(c)) \right| \le \left| f(x) -f(c) \right| + \left| g(x)-g(c)\right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} =\varepsilon. $$

I like this one, clean and simple.

@notjeremy
I agree, but saying $f(a)>L>f(b)$ is unnecessary given the interval notation $[a,b]$. It's definitely a cleaner way to look at it.

I agree with this!!!!!!

Extreme value theorem: If a real-valued function f is continuous in the closed and bounded interval [a,b], then f must attain a maximum and a minimum, each at least once.