math 104 midterm

$x \le \beta$ for every $x \in E$. NOT an upper bound: some $x \in E$ has $x > \beta$.

(i) $\alpha$ is an upper bound of $E$. (ii) If $\gamma < \alpha$, then $\gamma$ is not an upper bound: some $x \in E$ has $x > \gamma$. The sup may or may not be in $E$.

(i) $\alpha \le x$ for every $x \in E$ (lower bound). (ii) If $\gamma > \alpha$, some $x \in E$ has $x < \gamma$.

EVERY nonempty $E \subseteq S$ that is bounded above has $\sup E$ in $S$. To show it fails: name ONE such $E$ with no sup in $S$. $\mathbb{R}$ has it, $\mathbb{Q}$ doesn't.

(i) $y < z \Rightarrow x + y < x + z$. (ii) $x > 0$ and $y > 0 \Rightarrow xy > 0$.

If $x > 0$ and $y \in \mathbb{R}$, some $n \in \mathbb{N}$ has $nx > y$. Most-used form: choose $n > 1/x$, then $1/n < x$.

If $x < y$ are real, some $p \in \mathbb{Q}$ has $x < p < y$.

Countable: a bijection $J \to A$, where $J = \{1, 2, 3, \dots\}$ (so countable sets are INFINITE). At most countable: finite or countable. Uncountable: neither.

(a) $d(p,q) > 0$ if $p \ne q$, and $d(p,p) = 0$. (b) $d(p,q) = d(q,p)$. (c) $d(p,q) \le d(p,r) + d(r,q)$.

$N_r(p) = \{q \in X : d(p,q) < r\}$. In $\mathbb{R}$: the open interval $(p - r, p + r)$.

For EVERY $r > 0$, $N_r(p)$ contains a point $q \in E$ with $q \ne p$. ($p$ does not have to be in $E$.)

There is ONE $r > 0$ such that $N_r(p)$ contains no point of $E$ other than (maybe) $p$.

$p \in E$ and $p$ is not a limit point of $E$.

Every limit point of $E$ is in $E$ ($E' \subseteq E$). Same thing: $E^c$ is open (Rudin 2.23).

There is SOME $r > 0$ with $N_r(p) \subseteq E$.

For EVERY $r > 0$, $N_r(p)$ contains a point outside $E$.

Every point of $E$ is an interior point of $E$.

$\overline{E} = E \cup E'$. It is closed, and it is the smallest closed set containing $E$ (Rudin 2.27).

Every point of $X$ is in $E$ or is a limit point of $E$. Not dense: some ball in $X$ misses $E$.

There are $M$ and $q \in X$ with $d(p,q) < M$ for every $p \in E$. In $\mathbb{R}$: $|x| \le M$ for all $x \in E$.

A collection $\{G_\alpha\}$ of OPEN sets with $E \subseteq \bigcup_\alpha G_\alpha$. A subcover is some of the $G_\alpha$ that still cover $E$.

EVERY open cover of $K$ has a FINITE subcover.

There is ONE open cover of $K$ with NO finite subcover. You get to choose that cover.

$\overline{A} \cap B = \varnothing$ AND $A \cap \overline{B} = \varnothing$. Disjoint is not enough.

$E$ is NOT a union of two nonempty separated sets.

$\forall \varepsilon > 0\ \exists N\ \forall n \ge N:\ d(p_n, p) < \varepsilon$. Order: ε first (not yours), then $N$ (yours, may depend on ε), then EVERY $n \ge N$.

$\exists \varepsilon > 0\ \forall N\ \exists n \ge N:\ d(p_n, p) \ge \varepsilon$.

$p_{n_1}, p_{n_2}, p_{n_3}, \dots$ with $n_1 < n_2 < n_3 < \cdots$.

Increasing: $s_n \le s_{n+1}$ for all $n$. Decreasing: $s_n \ge s_{n+1}$ for all $n$.

Some $M$ has $|s_n| \le M$ for all $n$.

$s = r - d(p,q) > 0$. If $d(q,x) < s$: $d(p,x) \le d(p,q) + d(q,x) < r$. So $N_s(q) \subseteq N_r(p)$.

$r = \min\{r_A, r_B\} > 0$. Then $N_r(x) \subseteq A \cap B$.

$\delta = d(p,c)/2$ (or $/3$ for extra room). Then $N_\delta(p)$ and $N_\delta(c)$ don't overlap.

Def 3.1 with $\varepsilon = r$: some $N$ has $s_n \in N_r(s)$ for all $n \ge N$. Only finitely many terms are outside the ball.

$r = \min\{\delta, d(p,q_1), \dots, d(p,q_m)\} > 0$. Then $N_r(p)$ has no point of $E$ except $p$. Only works for FINITELY many points.

Choose $n > 1/r$ (Archimedean). Then $0 < 1/n < r$.

Full distance for $p = 1$ gives the ball $(0, 2)$, which reaches $0$: infinitely many $1/n$ get in, and you can't shrink past infinitely many. Half leaves a buffer, so only finitely many get in.

Circle the key word and write its definition with quantifiers. Line 1 comes from the hypothesis or a definition, never from the claim. One reason per line.

Let $x$ be arbitrary. (You don't get to choose it.)

Let $x = $ (your choice). Then check it works.

Let $x \in A$ ... so $x \in B$. For $A = B$: prove both inclusions.

Name the messy thing: $b = \sqrt{a}$, so $b > 0$, $b^2 = a$. Multiply only by things shown positive. Suppose $b \le 1$: then $b^2 \le b$, so $a \le 1$, contradiction. So $b > 1$, multiply by $b > 0$: $b < b^2 = a$.

Two parts. (i) Let $x \in E$, show $x \le \alpha$. (ii) Let $\gamma < \alpha$, find $x \in E$ with $x > \gamma$ (often by Archimedean). Or (ii'): let $\beta$ be any upper bound, show $\alpha \le \beta$.

(1) Bijection from $J$ (or $\mathbb{Z}$) AND write its inverse. (2) Countable union of countable sets (2.12). (3) Infinite subset of a countable set (2.8). (4) Tuples from a countable set (2.13).

(1) It contains an uncountable set (an interval). (2) Its complement in $\mathbb{R}$ is countable, so if it were countable, $\mathbb{R}$ would be too (irrationals, HW2 P4). (3) Cantor diagonal: list it, change the $n$-th digit of the $n$-th entry.

Two inclusions. ($\subseteq$) The ball inside $A \cap B$ is inside $A$ and inside $B$. ($\supseteq$) Balls of radius $r_A$ in $A$ and $r_B$ in $B$, $r = \min\{r_A, r_B\}$ gives a ball inside $A \cap B$.

Let $r > 0$ (arbitrary). Build $q \in E$ with $q \ne p$ and $d(p,q) < r$. Usually: choose $n > 1/r$, take $q = p + 1/n$.

(1) Every limit point of $E$ is in $E$. (2) $E^c$ is open: for $p \notin E$ find $r > 0$ with $N_r(p) \cap E = \varnothing$. (3) Build it from closed sets: finite unions, any intersections, finite sets are closed.

$U_n = (1/n, 1)$, $n \ge 2$. Covers: for $x \in (0,1)$ pick $n$ with $1/n < x$. Any finite pick has union $U_N$ ($N$ = largest index, nested), which misses $1/(2N)$.

$U_n = (-n, n)$. Any finite pick sits inside $U_N$, which misses the point $N$.

Let $\{G_\alpha\}$ be ANY open cover. $0 \in G_{\alpha_0}$, which is open, so $(-\delta, \delta) \subseteq G_{\alpha_0}$. Choose $N > 1/\delta$: every $1/n$ with $n \ge N$ is in $G_{\alpha_0}$. Add one set for each of $1, 1/2, \dots, 1/(N-1)$. Finite subcover.

Only in $\mathbb{R}^k$. Show bounded (e.g. convergent ⇒ bounded, 3.2c) AND closed (complement open, ball moves). Cite Heine-Borel (2.41).

Let $p \notin K$, $\delta = |p - c|/3$. Tail: $|c_n - c| < \delta$ for $n \ge N$, so $|p - c_n| > 2\delta$. Early terms: finitely many, all $\ne p$. $r = \min\{\delta, \frac{1}{2}|p - c_1|, \dots, \frac{1}{2}|p - c_{N-1}|\}$. Then $N_r(p)$ misses $K$.

Cut at a point $t \notin E$ with points of $E$ on both sides: $A = E \cap (-\infty, t)$, $B = E \cap (t, \infty)$. Both nonempty and separated.

Exactly the intervals: if $x, y \in E$ and $x < z < y$, then $z \in E$ (Rudin 2.47).

Suppose $E = A \cup B$ with $A$, $B$ separated. Show one of them must be empty.

(1) Scratch: bound $|s_n - s| \le C/n$ (top bigger, bottom smaller). (2) Let $\varepsilon > 0$. (3) Choose $N > C/\varepsilon$ (Archimedean). (4) For $n \ge N$: $|s_n - s| \le C/n \le C/N < \varepsilon$. (5) So $s_n \to s$.

Top: $2n + 3 \le 5n$. Bottom: $8n^3 + n^2 \ge 8n^3$. So it's $\le \frac{5}{8n^2} \le \frac{1}{n}$. Choose $N > 1/\varepsilon$.

Keep half: $n^2 - 4 \ge n^2/2$ once $n \ge 3$. Then take $N = \max\{3, \dots\}$.

Multiply by the conjugate: $\sqrt{n^2+n} - n = \frac{n}{\sqrt{n^2+n} + n}$. (HW5 P4: the limit is $1/2$.)

Let $\varepsilon > 0$. Use the hypothesis with that $\varepsilon$ to get $N$. Show the new quantity is $< \varepsilon$ for all $n \ge N$ (usually the triangle inequality).

Suppose $s_n \to s$. Take $\varepsilon = 1$ (half the gap). For $n \ge N$: $2 = |s_{n+1} - s_n| \le |s_{n+1} - s| + |s - s_n| < 1 + 1 = 2$. Contradiction.

Bounded: induction. Monotone: induction. Then monotone + bounded ⇒ converges (3.14). Increasing ⇒ the limit is the sup.

Check (a) positive off the diagonal and $0$ on it, (b) symmetric, (c) triangle inequality, usually by cases.

$N_{1/2}(p) = \{p\}$, so every set is open, so every set is also closed. Every point is isolated.

$|a + b| \le |a| + |b|$. Reverse: $\big||a| - |b|\big| \le |a - b|$.

It's a calculus rule, not Def 3.1. The definition means: let $\varepsilon > 0$, choose $N$, check every $n \ge N$.

Rudin 1.21: positive square roots exist. Turns the goal into $b < b^2$.

Multiply by $b > 0$: a positive multiplier keeps the direction (1.18). Then $a = b^2 \le b \le 1$ contradicts $a > 1$.

$b > 1$ (from the contradiction), multiplied by $b > 0$ (1.18).

$\ell \le x \Rightarrow -x \le -\ell$: multiplying by $-1$ flips the inequality.

LUB property of $\mathbb{R}$: $-A$ is nonempty and bounded above.

For $x \in A$: $-x \le s$ ($s$ is an upper bound of $-A$), so $-s \le x$.

$-c$ is an upper bound of $-A$, and $s$ is the LEAST one, so $s \le -c$.

Its inverse is $r \mapsto rq$. Writing the inverse IS the proof.

$S_n$ is a finite union of the countable sets $\{p/q : p \in \mathbb{Z}\}$, $q = 1, \dots, n$ (2.12).

Membership rule of $S_n$ with $n = q$ ($1 \le q \le q$). That's the $\mathbb{Q} \subseteq \bigcup S_n$ direction.

$\mathbb{Q} = \bigcup_n S_n$, a countable union of countable sets (2.12).

$A \cap B$ sits inside $A$ and inside $B$: the same ball works for both. So $x \in A^\circ \cap B^\circ$.

Move 2: the smaller ball fits inside $N_{r_A}(x) \subseteq A$ and $N_{r_B}(x) \subseteq B$, so inside $A \cap B$.

For $x \in (0,1)$, Archimedean gives $n$ with $1/n < x$, so $x \in U_n$.

Nested: $n \le m \Rightarrow U_n \subseteq U_m$, since $1/m \le 1/n$.

$1/(2N) < 1/N$, yet $1/(2N) \in (0,1)$. So no finite subcover.

Compact = EVERY open cover has a finite subcover. 'For all' means arbitrary: you don't choose it.

$0 \in G_{\alpha_0}$ and $G_{\alpha_0}$ is open, so $0$ is an interior point.

Chose $1/N < \delta$ (Archimedean), $n \ge N \Rightarrow 0 < 1/n \le 1/N < \delta$.

$G_{\alpha_0}$ takes $0$ and the tail, one set each for the finitely many points $1, \dots, 1/(N-1)$. Finite subcover.

Top bigger ($3 \le 3n$), bottom smaller (drop $n^2$). Bigger top + smaller bottom = bigger fraction.

Def 3.1 starts with 'for every ε': it's arbitrary. You don't pick it.

Archimedean (1.20a). $N$ is allowed to depend on $\varepsilon$.

The bound $b_n \le 1/n$, $n \ge N \Rightarrow 1/n \le 1/N$, the choice of $N$. That's Def 3.1.

$\{c_n\}$ converges, so it's bounded (3.2c). Adding one point $c$ keeps it bounded.

$p \notin K$, so $p \ne c$. Dividing by 3 leaves room (Move 3).

Reverse triangle: $|p - c_n| \ge |p - c| - |c_n - c| > 3\delta - \delta$.

Min of FINITELY many positive numbers is positive (Move 5). Each $|p - c_k| > 0$ since $p \notin K$.

Closed and bounded in $\mathbb{R}$, so Heine-Borel (2.41).

Triangle inequality: $d(p,x) \le d(p,q) + d(q,x) < d(p,q) + s = r$ (Move 1).

$A$ is open, so $N_r(a) \subseteq A$, which misses $B$ ($A \cap B = \varnothing$). So $a \notin B$ and $a$ isn't a limit point of $B$.

$\overline{A}$ is closed and contains $E$, $\overline{E}$ is the smallest closed set containing $E$ (2.27). Then $B \subseteq \overline{A}$ forces $B = \varnothing$.

Reverse triangle inequality, then the hypothesis $s_n \to s$ with the same $\varepsilon$.

Induction: $s_1 = \sqrt{2} < 2$, if $s_n < 2$, then $s_{n+1} < \sqrt{2 + \sqrt{2}} < \sqrt{4} = 2$.

Increasing (induction) and bounded above by 2, so it converges (3.14).

TRUE. Every nonempty subset has a largest element, and that's its sup.

FALSE. $E = (0,1)$: $\sup E = 1 \notin E$. (A finite set won't work: its sup is its max.)

FALSE. $\{p \in \mathbb{Q} : p^2 < 2\}$ is bounded above but has no least rational upper bound (Rudin 1.1).

TRUE. Density, Rudin 1.20(b).

TRUE. Pick rational $q$ with $x - \sqrt{2} < q < y - \sqrt{2}$, then $q + \sqrt{2}$ is irrational and lies between $x$ and $y$.

TRUE. $\alpha - \varepsilon < \alpha$, so it is not an upper bound (Def 1.8(ii)).

FALSE. $x = 0$, $y = 1$, $z = -1$: $xz = 0 > -1 = yz$. Negatives flip.

TRUE. List it $0, 1, -1, 2, -2, \dots$ (Rudin Ex. 2.5).

TRUE. A countable union of countable sets $S_n$ (PMT 1b), or Rudin 2.13.

TRUE. Rudin 2.8.

FALSE. $\{1\} \subseteq \mathbb{N}$ is finite: at most countable, not countable (Rudin's sense).

TRUE. Rudin 2.12.

FALSE. $\mathbb{N}$ (or $\mathbb{Q}$).

FALSE. $[0,1]$: the interval $(2,3)$ contains none of its points.

FALSE. Then $\mathbb{R} = \mathbb{Q} \cup$ (irrationals) would be countable (HW2 P4).

FALSE. $E = (0,1)$: $0$ is a limit point, not in $E$. (True exactly when $E$ is closed.)

FALSE. $X = \mathbb{Z}$, $E = \{0\}$: $N_{1/2}(0) = \{0\} \subseteq E$, so $0$ is interior, but no other points are nearby.

TRUE. $(p - r, p + r) \subseteq E$, so every ball around $p$ holds points of $E$ other than $p$.

FALSE. $\bigcap_n (-1/n, 1/n) = \{0\}$. (Finite intersections: TRUE, 2.24.)

FALSE. $\bigcup_n [1/n, 1] = (0, 1]$. (Finite unions: TRUE.)

TRUE. Rudin 2.24(a).

TRUE. Rudin 2.24(b).

TRUE. $\varnothing$ and $X$, every set in the discrete metric.

TRUE. Finite sets have no limit points (2.20).

TRUE. HW3 P2.

FALSE. $E = \{1/n\}$: $E' = \{0\}$, but $\{0\}$ has no limit points (HW3 P2).

FALSE. $A = [0,1]$, $B = [1,2]$: $1 \in (A \cup B)^\circ = (0,2)$, but $1 \notin (0,1) \cup (1,2)$. (The $\cap$ version is TRUE.)

TRUE. PMT 2(a): take the min of the two radii.

FALSE. $A_i = \{1/i\}$: $0$ is in the closure of the union but not in $\bigcup \{1/i\}$ (HW3 P3). For finite unions: TRUE.

TRUE. Pick one cover set per point.

FALSE. $(0,1)$, with the cover $U_n = (1/n, 1)$.

FALSE. $\mathbb{Z}$ or $[0,\infty)$, with the cover $U_n = (-n, n)$.

FALSE. $E = \{p \in \mathbb{Q} : 2 < p^2 < 3\}$ inside $X = \mathbb{Q}$ (HW4 P4). Heine-Borel is for $\mathbb{R}^k$ only.

TRUE. Put the two finite subcovers together (PMT T/F 5).

TRUE. $K_1 \cap K_2$ is closed (compact ⇒ closed, 2.34) and sits inside compact $K_1$ (2.35).

TRUE. Rudin 2.35.

TRUE. Rudin 2.34.

TRUE. Rudin 2.37.

TRUE. HW4 P1 (or closed + bounded, Heine-Borel).

FALSE. $[0,1]$ and $(1,2)$: $1 \in [0,1]$ is a limit point of $(1,2)$ (Rudin 2.46).

TRUE. HW5 P1(b). Disjoint closed sets too (P1(a)).

FALSE. $[0,1] \cup [2,3]$.

FALSE. $\mathbb{Q} \cap (-\infty, \sqrt{2})$ and $\mathbb{Q} \cap (\sqrt{2}, \infty)$ separate it.

TRUE. Rudin 2.47.

TRUE. PMT T/F 6 / HW5 P2.

FALSE. $[-1,0]^2 \cup [0,1]^2$ in $\mathbb{R}^2$: its interior is two separate open squares (HW5 P2).

FALSE. $(-1)^n$.

TRUE. Rudin 3.14.

TRUE. Rudin 3.2(c).

FALSE. $x_n = (-1)^n$: $|x_n| = 1$ converges, $x_n$ doesn't. (The other direction is TRUE, HW5 P3.)

FALSE. $(-1)^n$ diverges, but the subsequence $x_{2n} = 1$ converges.

TRUE. The same $N$ works, since $n_k \ge k$.

FALSE. $1/n \to 0$. (You only get $s \ge 0$.)

FALSE. $s_n = t_n = (-1)^n$: the product is always $1$.

FALSE. $s_n = (-1)^n$.

TRUE. $[0,1]$ is compact (3.6).

FALSE. Limits are unique (3.2(b)).

FALSE. $1/n \in (0,1]$, but the limit $0 \notin (0,1]$. (TRUE if $E$ is closed.)