7Complete Spaces: Baire, Ascoli, Stone–Weierstrass

Proof. For the two statements: a Cauchy sequence of the closed $F$ converges in $X$, and its limit, adherent to $F$, lies in $F$; a convergent-in-$X$ sequence of a complete $A$ is Cauchy, so converges in $A$, and limits are unique. ∎

Proof. Uniqueness: two continuous extensions agree on the dense $D$, hence everywhere (the agreement set $\{g = h\}$ is closed: preimage of the closed diagonal under $x\mapsto(g(x), h(x))$). Existence: for $x \in X$ pick $d_n \to x$, $d_n \in D$. The sequence $(f(d_n))$ is Cauchy: given $\varepsilon$, uniform continuity provides $\delta$ with $d(u,v) < \delta \Rightarrow d(f(u), f(v)) < \varepsilon$, and $(d_n)$ is Cauchy. Define $\bar f(x) = \lim f(d_n)$; the limit does not depend on the chosen sequence (interlace two of them). $\bar f$ extends $f$ (constant sequences) and inherits the modulus of continuity: if $d(x, x') < \delta$, approximating both by points of $D$ at distance $< \frac{\delta - d(x,x')}2$ gives $d(\bar f(x), \bar f(x')) \leq \varepsilon$ in the limit — $\bar f$ is uniformly continuous. ∎

Proof. Existence. Let $\mathcal X$ be the set of Cauchy sequences of $X$, with the pseudo-distance

the limit existing because $\abs{d(x_n, y_n) - d(x_m, y_m)} \leq d(x_n, x_m) + d(y_n, y_m)$ makes the real sequence Cauchy. Let $\hat X = \mathcal X/{\sim}$, identifying sequences at $D$-distance $0$; $D$ descends to a distance. Embed $X$ by constant sequences: an isometry, with dense image (a Cauchy sequence is $D$-approximated by the constants built on its own terms: $D\bigl((x_n), (x_k)_{\rm const}\bigr) = \lim_n d(x_n, x_k) \to 0$ as $k \to \infty$ by Cauchyness). Completeness of $\hat X$: let $(\xi^k)$ be Cauchy in $\hat X$; by density pick $x_k \in X$ with $D(\xi^k, x_k) \leq 2^{-k}$; then $(x_k)$ is Cauchy in $X$ (triangle inequality through the $\xi$’s), defines a point $\xi \in \hat X$, and $D(\xi^k, \xi) \leq 2^{-k} + D(x_k, \xi) \to 0$ (the distance from the constant $x_k$ to the class of $(x_j)_j$ is $\lim_j d(x_k, x_j)$, small for large $k$).

Uniqueness: two completions $\hat X_1, \hat X_2$ contain $X$ densely; the identity of $X$, an isometry, is uniformly continuous, so extends to $\hat X_1 \to \hat X_2$ (Theorem 7.2), still an isometry on a dense set hence everywhere; symmetrically in the other direction, and the composites fix the dense $X$: they are the identities. ∎

Proof. (Year 2 proved this; we re-record the two-line argument for self-containedness.) The orbit $x_{n+1} = f(x_n)$ has $d(x_{n+1}, x_n) \leq k^nd(x_1, x_0)$, hence is Cauchy (geometric series); its limit $x^*$ is fixed (continuity of $f$), unique since two fixed points satisfy $d \leq k\,d$. The rate: sum the geometric tail. ∎

Proof. Let $(U_n)$ be dense opens and $B_0 = B(x_0, r_0)$ any open ball; we find a point of $\bigcap U_n$ in $B_0$. Inductively: $U_{n}$ being dense and open, it meets the open ball $B_{n-1}$ in an open set, which contains a closed ball $\bar B(x_n, r_n)$ with $0 < r_n \leq r_{n-1}/2$ and $\bar B(x_n, r_n) \subseteq B_{n-1}\cap U_n$. The centers form a Cauchy sequence ($x_m \in B_n$ for $m \geq n$, radii $\to 0$); the limit $x$ lies in every $\bar B(x_n, r_n)$ (closedness), hence in every $U_n$ and in $B_0$. For the second form: if no $F_n$ has interior, the $U_n = X \setminus F_n$ are open and dense, and a point of $\bigcap U_n$ escapes $\bigcup F_n = X$: absurd. ∎

Proof. (a) $\R = \bigcup_{x}\{x\}$ over a countable set would make some singleton have interior. (b) If $\Q = \bigcap_n V_n$ with $V_n$ open (necessarily dense, as $\supseteq \Q$), then the sets $V_n$ and the complements $\R\setminus\{q\}$, $q \in \Q$, form a countable family of dense opens with empty intersection — contradicting Baire. If $X$ is complete without isolated points and countable, $X = \bigcup_{x \in X}\{x\}$ exhibits it as a countable union of closed sets with empty interior (no isolated points): Baire again. ∎

Proof. For $n \geq 1$ let

If $f$ is differentiable at $x$, then $f \in F_n$ for some $n$: the quotient $\abs{f(x+h)-f(x)}/\abs h$ is bounded for $\abs h \leq \delta$ (differentiability: it tends to $\abs{f'(x)}$) and bounded by $2\norm f_\infty/\delta$ for $\abs h \geq \delta$. So $\bigcup F_n$ contains all somewhere-differentiable functions, and it suffices to show each $F_n$ is closed with empty interior.

Closed: let $f_k \to f$ uniformly, $f_k \in F_n$ with witnesses $x_k \to x$ (compactness, after extraction). For $h$ with $x + h \in \intcc01$: choose $h_k \to h$ with $x_k + h_k \in \intcc01$ (e.g. $h_k = h + x - x_k$ clipped); then $\abs{f(x + h) - f(x)} = \lim \abs{f_k(x_k + h_k) - f_k(x_k)} \leq \lim n\abs{h_k} = n\abs h$, using uniform convergence and continuity of $f$ at the relevant points: $f \in F_n$.

Empty interior: given $f \in F_n$ and $\varepsilon > 0$, we find $g$ with $\norm{g - f}_\infty \leq \varepsilon$ and $g \notin F_n$. First approximate $f$ within $\varepsilon/2$ by a piecewise affine $\varphi$ (uniform continuity: interpolate on a fine grid), of slopes bounded by some $M$. Add a small sawtooth: $g = \varphi + \frac\varepsilon2\,s_N$, where $s_N(x)$ is the $\frac1N$-periodic zigzag of amplitude $1$ and slope $\pm 2N$. At every $x$, on one side there is $h$ arbitrarily small with the sawtooth contributing slope $\pm 2N\cdot\frac\varepsilon2 = \pm\varepsilon N$ over $[x, x+h]$: the difference quotient of $g$ exceeds $\varepsilon N - M > n$ for $N$ large. So $g \notin F_n$, at any uniform distance $\leq \varepsilon$ from $f$. Conclusion: $\bigcup F_n$ is meagre; by Baire its complement — made of nowhere differentiable functions — is dense in $\mathcal C(\intcc01)$: such functions exist in abundance. ∎

Proof. ($\Leftarrow$) Let $(f_k)$ be a sequence of $\mathcal F$. $K$ compact metric is separable: for each $n$, finitely many balls of radius $\frac1n$ cover $K$ (Theorem 6.16); their centers form a countable dense set $D = \{x_1, x_2, \dots\}$. By pointwise boundedness and Bolzano–Weierstrass, extract successively subsequences converging at $x_1$, then also at $x_2$, etc., and take the diagonal subsequence $(g_j)$: it converges at every point of $D$. Equicontinuity upgrades this to uniform Cauchy: given $\varepsilon$, take $\delta$ as in the definition, cover $K$ by finitely many balls $B(x_i, \delta)$ with $x_i \in D$ ($i \leq m$), and pick $J$ so large that $\norm{g_j(x_i) - g_{j'}(x_i)} < \varepsilon$ for $j, j' \geq J$, $i \leq m$. For arbitrary $x \in B(x_i, \delta)$:

So $(g_j)$ is uniformly Cauchy, and converges in the complete $\mathcal C(K)$. Thus every sequence of $\mathcal F$ has a convergent subsequence: $\bar{\mathcal F}$ is (sequentially, hence by Theorem 6.16) compact.

($\Rightarrow$) If $\bar{\mathcal F}$ is compact: pointwise boundedness is clear (evaluation is continuous). For equicontinuity, cover $\mathcal F$ by finitely many balls $B(f_i, \varepsilon)$ of $\mathcal C(K)$; each $f_i$ is uniformly continuous (Heine, Corollary 6.17), giving a common $\delta$ for $i \leq m$; then for $f \in B(f_i, \varepsilon)$ and $d(x,y) < \delta$: $\norm{f(x)-f(y)} \leq 2\varepsilon + \norm{f_i(x)-f_i(y)} < 3\varepsilon$. ∎

Proof. Say $f_n \uparrow f$; let $g_n = f - f_n \downarrow 0$, continuous. Given $\varepsilon$, the open sets $U_n = \{g_n < \varepsilon\}$ increase and cover $K$ (pointwise convergence); extract a finite subcover: $K = U_{n_0}$ for some $n_0$ (increasing family), i.e. $0 \leq g_n < \varepsilon$ everywhere for $n \geq n_0$. ∎

Proof. Define $u_0 = 0$, $u_{n+1}(t) = u_n(t) + \frac12\bigl(t - u_n(t)^2\bigr)$: polynomials. By induction $0 \leq u_n(t) \leq \sqrt t$ on $\intcc01$: granting it for $n$,

since $\sqrt t + u_n \leq 2$; and $u_{n+1} \geq u_n \geq 0$. So $(u_n(t))$ is nondecreasing, bounded by $\sqrt t$: it converges pointwise, and the limit $\ell(t)$ satisfies $\ell = \ell + \frac12(t - \ell^2)$: $\ell(t) = \sqrt t$, continuous. Dini (Lemma 7.13) upgrades to uniform. ∎

Proof. Let $\bar{\mathcal A}$ be the closure, again an algebra (products of uniform limits on bounded sets converge).

Step 1: $\bar{\mathcal A}$ is a lattice, i.e. stable under $\max$ and $\min$. Since $\max(f,g) = \frac{f + g + \abs{f-g}}2$ and $\min$ likewise, it suffices that $f \in \bar{\mathcal A} \Rightarrow \abs f \in \bar{\mathcal A}$: with $M = \norm f_\infty > 0$, $\abs f = M\sqrt{(f/M)^2}$, and Lemma 7.14 gives polynomials $u_n$ with $u_n\bigl((f/M)^2\bigr) \to \abs f/M$ uniformly; polynomials in members of the algebra (with constant term: constants are there) stay in $\bar{\mathcal A}$.

Step 2: two-point interpolation. For $x \neq y$ and $a, b \in \R$, some $g \in \mathcal A$ has $g(x) = a$, $g(y) = b$: take $h$ separating $x, y$ and set $g = a + (b - a)\frac{h - h(x)}{h(y) - h(x)}$.

Step 3. Let $f \in \mathcal C(K)$, $\varepsilon > 0$. For each pair $x, y$ pick $g_{x,y} \in \mathcal A$ with $g_{x,y}(x) = f(x)$, $g_{x,y}(y) = f(y)$ (Step 2; for $x = y$ take the constant function $g_{x,x} = f(x)$). Fix $x$: for each $y$, the open set $V_y = \{g_{x,y} < f + \varepsilon\}$ contains $y$; compactness extracts $y_1, \dots, y_m$ with $K = \bigcup V_{y_j}$, and $h_x = \min_j g_{x, y_j} \in \bar{\mathcal A}$ (Step 1) satisfies $h_x < f + \varepsilon$ everywhere, $h_x(x) = f(x)$. Now vary $x$: $W_x = \{h_x > f - \varepsilon\}$ is open, contains $x$; extract $x_1, \dots, x_l$ covering $K$, and $h = \max_i h_{x_i} \in \bar{\mathcal A}$ satisfies $f - \varepsilon < h < f + \varepsilon$: $\norm{f - h}_\infty \leq \varepsilon$. Hence $f \in \bar{\mathcal A}$. ∎

Proof. (a) Polynomials form an algebra with constants; the coordinate functions separate points of $\R^n$. (b) The real and imaginary parts $\frac{f + \bar f}2$, $\frac{f - \bar f}{2\iu}$ of members of $\mathcal A$ form a real algebra $\mathcal A_\R \subseteq \mathcal C(K, \R)$ with constants; it separates points ($f(x) \ne f(y)$ forces $\operatorname{Re}f$ or $\operatorname{Im}f$ to separate). Apply the real theorem and recombine. (c) View $2\pi$-periodic functions as $\mathcal C(S^1, \C)$ ($S^1 = \R/2\pi\Z$, compact: Exercise 6.5); the algebra generated by $\eu^{\iu t}, \eu^{-\iu t}$ and constants is stable under conjugation and separates points of the circle ($\eu^{\iu t}$ is injective on it). Apply (b). ∎