Real Analysis

For finite intersections, take the smallest of finitely many radii.

$\varnothing$ and $\R$ are open. If $p\in\bigcup G_\alpha$, then $p\in G_\beta$ for some $\beta$, and a ball around $p$ inside $G_\beta$ lies in the union. If $p\in G_1\cap\dots\cap G_n$, take balls of radii $r_i$ and let $r=\min r_i>0$. But $\bigcap_n(-1/n,1/n)=\{0\}$, which contains no ball: with infinitely many radii the minimum can be 0.

Take $x_n=1/(2\pi n+\pi/2)$ and $y_n=1/(2\pi n-\pi/2)$. Both tend to 0, but $f(x_n)=1$ and $f(y_n)=-1$. If $f$ were continuous at 0, both would converge to $c$, so $c=1=-1$. In the game: for $\varepsilon<1$ no $\delta$ works.

Unions of cofinite sets are cofinite (or empty); finite intersections of cofinite sets have finite complements, since a finite union of finite sets is finite. Two nonempty open sets miss only finitely many points each, so they share infinitely many points. In a metric space two distinct points have disjoint balls of radius $d(p,q)/2$ around them, so no metric gives these open sets. This is the Zariski topology on the affine line: its closed sets are the zero sets of polynomials.

The condition compares values at $x+h$ and $x-h$ but never looks at $f(x)$ itself. Try changing $f$ at a single point.

No. Let $f(0)=1$ and $f(x)=0$ for $x\ne0$. If $x\ne0$, then for $0<|h|<|x|$ both $x\pm h$ are nonzero; if $x=0$, the points $\pm h$ are nonzero for every $h\ne0$. Either way $f(x+h)-f(x-h)=0$ for all small $h\ne0$, so the limit is 0 everywhere. But $f$ is not continuous at 0.

The condition only says $f$ has the same behaviour on both sides of each point. It is satisfied by any function that agrees with a continuous one except on a set with no limit points. Even $\mathbf 1_{\mathbb Q}$ fails it, though: take $x$ irrational and $h$ with $x+h\in\mathbb Q$, $x-h\notin\mathbb Q$.

A point of $\bar E$ is a limit of a sequence in $E$, and continuous maps send convergent sequences to convergent sequences.

Let $p\in\bar E$. Choose $p_n\in E$ with $p_n\to p$ (if $p\in E$, take $p_n=p$). By Rudin 4.2, $f(p_n)\to f(p)$, and every $f(p_n)\in f(E)$, so $f(p)\in\overline{f(E)}$.

An open-set proof: $f^{-1}\big(\overline{f(E)}\big)$ is closed (preimage of a closed set) and contains $E$, so it contains $\bar E$.

Proper inclusion: $X=\R$, $f(x)=1/(1+x^2)$, $E=\R$. Then $\bar E=\R$ and $f(\bar E)=(0,1]$, but $\overline{f(E)}=[0,1]$. Or $f:\mathbb Z\to\R$, $f(n)=1/n$ for $n\ge1$: $E=\mathbb Z_{>0}$ is closed, but $0\in\overline{f(E)}\setminus f(\bar E)$.

$Z(f)$ is a preimage.

$Z(f)=f^{-1}(\{0\})$, and $\{0\}$ is closed in $\R$. Preimages of closed sets under continuous maps are closed (take complements in Rudin 4.8). Sequentially: if $p_n\in Z(f)$ and $p_n\to p$, then $f(p)=\lim f(p_n)=0$.

This is the bridge to algebraic geometry: if only zero sets of polynomials are declared closed, you get the Zariski topology, in which polynomial maps are continuous.

Use Exercise 4.2 for the first part. For the second, look at the set where $f$ and $g$ agree, or use sequences.

Density. By Exercise 4.2, $f(X)=f(\bar E)\subseteq\overline{f(E)}$. So every point of $f(X)$ is in the closure of $f(E)$.

Uniqueness. For $p\in X$, pick $p_n\in E$ with $p_n\to p$. Then $f(p)=\lim f(p_n)=\lim g(p_n)=g(p)$, using continuity of both and uniqueness of limits in the metric space $Y$.

The uniqueness step needs limits in $Y$ to be unique, i.e. $Y$ Hausdorff. For a general topological target it can fail: into the indiscrete space every map is continuous.

If $p\notin\bar E$, some ball $N_r(p)$ misses $E$ (else $p$ would be in $E$ or a limit point). That ball also misses $E'$: any $q\in N_r(p)$ has a smaller ball around it inside $N_r(p)$, which misses $E$. So $\bar E^c$ is open. If $F\supseteq E$ is closed, it contains the limit points of $E$ (they are limit points of $F$), so $F\supseteq\bar E$.

If $q\in N_r(p)$, let $h=r-d(p,q)>0$. For $s\in N_h(q)$, $d(p,s)\le d(p,q)+d(q,s)<d(p,q)+h=r$. So $N_h(q)\subseteq N_r(p)$.

(a) Balls are open. (d) $p\notin E^\circ$ iff every ball around $p$ meets $E^c$. For (e) and (f), try $\mathbb Q\subseteq\R$.

(a) If $p\in E^\circ$, some $N_r(p)\subseteq E$. Each $q\in N_r(p)$ has a ball $N_h(q)\subseteq N_r(p)\subseteq E$ (balls are open), so $q\in E^\circ$. Thus $N_r(p)\subseteq E^\circ$.

(b) If $E$ is open, every point is interior. Conversely $E^\circ$ is open by (a).

(c) Each $p\in G$ has a ball inside $G\subseteq E$.

(d) $p\notin E^\circ$ iff no ball around $p$ lies in $E$, iff every ball around $p$ meets $E^c$, iff $p\in\overline{E^c}$.

(e) No: $\mathbb Q$ has empty interior, but $\bar{\mathbb Q}=\R$ has interior $\R$.

(f) No: $\mathbb Q^\circ=\varnothing$ has closure $\varnothing$, but $\bar{\mathbb Q}=\R$.

So $E^\circ$ is the largest open set inside $E$, dual to $\bar E$ being the smallest closed set containing $E$.

Check $d(x,x)=0$, positivity, symmetry and the triangle inequality. Concave increasing functions $\varphi$ with $\varphi(0)=0$ are subadditive.

• $d_1$: not a metric. $d_1(0,2)=4>d_1(0,1)+d_1(1,2)=2$.
• $d_2$: a metric. $\sqrt{a+b}\le\sqrt a+\sqrt b$, so $\sqrt{|x-z|}\le\sqrt{|x-y|+|y-z|}\le\sqrt{|x-y|}+\sqrt{|y-z|}$.
• $d_3$: not a metric. $d_3(1,-1)=0$ but $1\ne-1$.
• $d_4$: not a metric. $d_4(1,1)=1\ne0$, and it isn't symmetric.
• $d_5$: a metric. $\varphi(t)=t/(1+t)$ is increasing and $\varphi(a+b)\le\varphi(a)+\varphi(b)$, which with $|x-z|\le|x-y|+|y-z|$ gives the triangle inequality.

Both $d_2$ and $d_5$ give the usual open sets on $\R$, though $d_5$ is bounded: the same topology from very different metrics.

Use the points with rational coordinates.

$\mathbb Q^k$ is countable (a finite product of countable sets). Given $x\in\R^k$ and $r>0$, choose rationals $q_i$ with $|q_i-x_i|<r/\sqrt k$ (density of $\mathbb Q$ in $\R$, Rudin 1.20). Then $|q-x|<r$, so every ball meets $\mathbb Q^k$, i.e. $\mathbb Q^k$ is dense.

Group the points of $G$ by the largest interval in $G$ containing them. Each such interval contains a rational.

For $x\in G$, let $I_x$ be the union of all open intervals containing $x$ and contained in $G$. It is an open interval (possibly infinite) inside $G$, and it is the largest one containing $x$. If $I_x\cap I_y\ne\varnothing$, then $I_x\cup I_y$ is an interval in $G$ containing $x$, so $I_y\subseteq I_x$ by maximality, and symmetrically, so $I_x=I_y$. Hence the distinct $I_x$ are disjoint and their union is $G$.

Each $I_x$ contains a rational number, and distinct ones contain distinct rationals because they are disjoint. Since $\mathbb Q$ is countable, there are at most countably many intervals.

Both ratios tend to 1, so the test is inconclusive, yet one series diverges and the other converges. The ratio and root tests only detect geometric decay.

If $e=m/n$, then $n!e$ and $n!\sum_{k\le n}1/k!$ are integers, so their difference $x$ is too. But $x=\sum_{k>n}n!/k!<\frac1{n+1}+\frac1{(n+1)^2}+\dots=\frac1n\le1$, and $x>0$. No integer lies strictly between 0 and 1.

The reverse triangle inequality.

$\big||s_n|-|s|\big|\le|s_n-s|$, so $|s_n|\to|s|$. In the language of the story: $x\mapsto|x|$ is continuous, and continuous maps preserve limits.

The converse is false: $s_n=(-1)^n$ diverges, while $|s_n|=1$ converges. It is true when the limit is 0: $|s_n|\to0$ iff $s_n\to0$.

Show by induction that the sequence is increasing and bounded by 2, then use monotone convergence.

Bounded: $s_1<2$, and if $s_n<2$ then $s_{n+1}<\sqrt{2+\sqrt2}<2$.

Increasing: $s_2=\sqrt{2+2^{1/4}}>\sqrt2=s_1$. If $s_{n}>s_{n-1}$, then $\sqrt{s_n}>\sqrt{s_{n-1}}$, so $s_{n+1}>s_n$.

By Rudin 3.14 it converges, to some $L\in(\sqrt2,2]$. Passing to the limit (the map $t\mapsto\sqrt{2+\sqrt t}$ is continuous), $L=\sqrt{2+\sqrt L}$, i.e. $L^2-2=\sqrt L$. Numerically $L\approx1.831$.

Split $d(p_n,p)$ through a term of the subsequence with a large index.

Given $\varepsilon$, choose $N$ with $d(p_m,p_n)<\varepsilon/2$ for $m,n\ge N$, and choose $i$ with $n_i\ge N$ and $d(p_{n_i},p)<\varepsilon/2$. For $n\ge N$, $d(p_n,p)\le d(p_n,p_{n_i})+d(p_{n_i},p)<\varepsilon$.

Combined with Bolzano–Weierstrass this proves completeness of $\R^k$: a Cauchy sequence is bounded, so it has a convergent subsequence, and then the whole sequence converges.

$\{(1/n,1)\}_{n\ge2}$: finitely many of them have a largest $n$, and miss $(0,1/n]$. For $[0,1]$, any cover must have a set containing 0, and that set contains some $[0,r)$; the sets $(1/n,1)$ alone don't cover $[0,1]$.

Take $\delta=\varepsilon/L$. For $\sqrt x$: it is continuous on the compact $[0,1]$, hence uniformly continuous; but $\sqrt x/x=1/\sqrt x$ is unbounded near 0, so no $L$ works.

Start with an open set of the cover that contains 0.

Let $\{G_\alpha\}$ be an open cover. Some $G_{\alpha_0}\ni0$. Being open, it contains $(-r,r)$ for some $r>0$, hence every $1/n$ with $n>1/r$. The finitely many remaining points $1/n$, $n\le1/r$, each lie in some $G_{\alpha_n}$. These finitely many sets cover $K$.

Apply the intermediate value theorem to $g(x)=f(x)-x$.

$g$ is continuous, $g(0)=f(0)\ge0$ and $g(1)=f(1)-1\le0$. Since $[0,1]$ is connected, $g(I)$ is a connected subset of $\R$ (Rudin 4.22), i.e. an interval, containing a number $\ge0$ and a number $\le0$, hence 0.

This is the one-dimensional Brouwer fixed point theorem. It uses only topological facts: continuity and connectedness, no distances. Compare the contraction principle (Rudin 9.23), which needs a metric but gives uniqueness and a way to compute the fixed point.

Numerator and denominator tend to 0 and are differentiable near 0 with $2x\ne0$ for $x\ne0$. The quotient of derivatives $\sin x/(2x)\to\tfrac12$, so the limit is $\tfrac12$.

The degree-4 Taylor polynomial of $\sin$ at 0 is $x-x^3/6$ (the $x^4$ term vanishes), and the remainder is $\sin^{(5)}(\xi)x^5/5!=\cos(\xi)x^5/120$ for some $\xi$ between 0 and $x$. Its absolute value is at most $|x|^5/120$.

Divide by $|x-y|$ and let $y\to x$.

For $y\ne x$, $\left|\dfrac{f(y)-f(x)}{y-x}\right|\le|y-x|\to0$. So $f'(x)=0$ for every $x$, and by the mean value theorem $f(x)-f(y)=f'(c)(x-y)=0$.

The mean value theorem for monotonicity. For $g'$, write the difference quotient of $g$ in terms of $f$, and use that $g$ is continuous.

For $x<y$, $f(y)-f(x)=f'(c)(y-x)>0$. So $f$ is strictly increasing, hence injective, and maps $(a,b)$ onto an interval $J$ (it is continuous, so the image is connected). $g:J\to(a,b)$ is increasing and its image is an interval, so it has no jumps: $g$ is continuous.

Fix $y_0=f(x_0)$. For $y\ne y_0$ in $J$, with $x=g(y)$,

As $y\to y_0$, continuity of $g$ gives $x\to x_0$, so the right side tends to $1/f'(x_0)$.

$f'=1+\varepsilon g'$. When is it positive?

$f'(x)=1+\varepsilon g'(x)\ge1-\varepsilon M$. If $\varepsilon<1/M$, then $f'>0$ everywhere, so $f$ is strictly increasing by Exercise 5.2, hence one-to-one. (If $M=0$, every $\varepsilon$ works.)

Alternatively, $|f(x)-f(y)|\ge|x-y|-\varepsilon M|x-y|>0$ for $x\ne y$. This is a baby version of the inverse function theorem.