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--- math104-s21:s:genevievebrooks [2021/05/11 21:21]
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+++ math104-s21:s:genevievebrooks [2026/02/21 14:41] (current)
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 .20 Theorem\\
 If $p$ is a limit point of E then every neighborhood of $p$ contains infinitely many points in E.\\
+.23 Theorem\\
+A set is open iff its complement is closed.\\
+.24 Theorem\\
+. arbitrary union of open sets is open\\
+. arbitrary intersection of closed sets is closed\\
+. finite intersection of open sets is open\\
+. finite union of closed sets is closed\\
+Proof uses the relation: $$(\bigcap_{\alpha}F_{\alpha})^c = \bigcup_{\alpha}F_{\alpha}^c$$
-**Continuity**
+Next we introduced notion of a closure of a set which is equal to the set but with the addition of its limit points.\\
+**Compact Sets**\\
+**open cover** of E in metric space X is a collection{$G_{\alpha}$} of open subsets of X such that $E \subset \bigcup_{\alpha} G_{\alpha}$\\
+K in X is compact if every open cover of K contains a finite subcover. i.e. There exist finite indices such that: $$K \subset G_1 \cup G_2 \cup \cdots \cup G_m$$
+**Heine-Borel Theorem**: If E in $\mathbb{R}^k then E compact \Leftrightarrow E closed and bounded.\\
+Example: $(0, 1) \epsilon \mathbb{R}^k$ is not compact because it is not closed and so we have $$\bigcup (\frac{1}{n}, 1 - \frac{1}{n})$$ which is an open cover of (0, 1) that does not have a finite subcover.\\
+Compactness brings us many interesting theorems such as **Theorem 2.35** closed subsets of compact sets are compact\\
+Theorem 2.41: If E is in $\mathbb{R}^k$ then TFAE\\
+(a) E is closed and bounded\\
+(b) E is compact\\
+c) Every infinite subset of E has a limit point in E
+**Continuity**\\
+Definitions:
+(Credit to Kaylene Stocking because I really liked the way she described the definitions)\\
+The limit perspective:  Let $(x_n)$ be a sequence of points in the domain of $f$ that converges to some $x_0 \epsilon S$. The sequence $f(xn)$ must converge to $f(x_0)$
+The bounded rate of change perspective:  Pick any point $x_0 \epsilon S$ and any $\varepsilon > 0$. There must exist some $\delta > 0$ so that moving less than $\delta$ away from $x_0$ results in a change of less than $\varepsilon$ in the value of $f(x)$.\\
+**formal definition**: $f$ is continuous at point $p$ if $\forall \varepsilon > 0 \exists \delta > 0$ such that $$d_y(f(x), f(p)) < \varepsilon$$ $$\forall x \epsilon E$$ for which $$d_x(x, p) < \delta$$ Then if f is continuous at every p in its domain, f is continuous overall.
+The topological perspective:  Let E be an open subset of the range of f. The set of points in the domain of f that f maps into E must also be open.
@@ Line 132: / Line 165: @@
 Example: Let $f_n(x) = nx^n$ for $x \epsilon [0, 1)$. We show that convergence is not uniform. If it were there would exist $N \epsilon \mathbb{N}$ such that $$|nx^n - 0| < 1 \forall x \epsilon [0, 1), n > N$$ In particular we would have $(N + 1)x^{N + 1} < 1 \forall x \epsilon [0, 1)$. But this fails for $x$ sufficiently close to 1: consider $\frac{1}{(N + 1)^\frac{1}{N + 1}}$.
+**Differentiation**
+two definitions: $$ f'(x) = \lim_{t \rightarrow x} \frac{f(t) - f(x)}{t-x}, t \neq x$$ $$ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$\\
+.2 Theorem: If $f$ is defferentiable at $x$ then $f$ is continuous at $x$.
+**Mean Value Theorems**
+These theorems let us extrapolate data about arbitrary points on a function in a range [a, b] for which f is defined. If a point on f is a local min or local max then its derivative at that point is 0.\\
+Generalized MVT: If $f, g$ are continuous on [a, b] and differentiable on (a, b) then $\exists x \epsilon (a, b)$ such that $$[f(b) - f(a)]g'(x) = [g(b) - g(a)]f'(x)$$
+MVT: $\frac{f(b) - f(a)}{b - a} = f'(x)$
+Rolle's Theorem (special case of MVT): If f is continuous and differentiable on (a, b) then $f(a) = f(b)$ then there exists at least one x in (a, b) such that $f'(x) = 0$.
+Continuity of Derivatives Theorem: Suppose $f$ is a real differentiable function on [a, b] and suppose $f'(a) < \lambda < f'(b)$ then there is a point x in (a, b) such that $f'(x) = \lambda$
+-useful corollary: If f is differentiable on (a, b) then f can not have any simple discontinuities on (a, b).
+**L'Hopital's Theorem**\\
+Let $s$ signify $a$, $a^+$, $a^-$, $\infty$, or $-\infty$ where $a \epsilon \mathbb{R}$ and suppose $f$ and $g$ are differentiable functions for which the following limit exists: $$\lim_{x \rightarrow s} \frac{f'(x)}{g'(x)} = L$$ If $$\lim_{x \rightarrow s} f(x) = \lim_{x \rightarrow s} g(x) = 0$$ or if $$ \lim_{x \rightarrow s} |g(x)| = \infty$$ then $$\lim_{x \rightarrow s} \frac{f(x)}{g(x)} = L$$ In other words, we can take the derivative of the top and bottom when f over g is in some indeterminate form and this will give us the same limit and may be easier to calculate the limit from.
+**Taylor's Theorem**\\
+If $f$ is a smooth function (is infinitely differentiable) then $$\sum_{k = 0}^{\infty} \frac{f^k(c)}{k!}(x - c)^k$$ is the Taylor series of f about c. We can stop k at a certain index to get a partial representation of $f$ with some error. The error formula is: $$R_n(x) = f(x) - \sum_{k = 0}{n - 1} \frac{f^k(c)}{k!}(x - c)^k$$ The remainder depends on f and c. Furthermore we may have that $$f(x) = \sum_{k = 0}^{\infty} \frac{f^k(c)}{k!}(x - c)^k$$ iff $$\lim_{n \rightarrow \infty} R_n(x) = 0$$ But the remainder need not always tend to zero. Hence we can have $f$ not given exactly by its Taylor series.\\
+**The actual Taylor's Theorem** \\
+Let $f$ be defined on (a, b) where a < c < b. Suppose the nth derivative of f(x) exists on (a, b). Then for each $x \neq c$ in (a, b) there is some $y$ between c and x such that $$R_n(x) = \frac{f^n(y)}{n!}(x - c)^n$$
@@ Line 189: / Line 249: @@
 . Find a subset K ⊂ Q, such that K is compact and K is an infinite set.
-. If A is open, then f(A) is open. FALSE why?
+. If A is open, then f(A) is open. FALSE why? **Answer**: consider the continuous mapping from $(0, 1) \epsilon \mathbb{R}$ to $\{0\}$. The range is closed but the domain is open.
+. If A is closed, then f(A) is closed. FALSE, why? **Answer**:
-. If A is closed, then f(A) is closed. FALSE, why?
+. Why is $\{f^{-1}(U_0) : U_0 \epsilon U\}$ a cover of E, where E is the domain set of $f$ and $U$ is the open cover of $f(E)$? **Answer**: Let $x \epsilon E$. Then as $U$ is a cover of $f(E)$, $f(x) \epsilon U_0$ for some $U_0 \epsilon U$. Thus $x \epsilon f^{-1}(U).$

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