CommeUnJeu · L2 MP

Countability

⌚ ~43 min ▢ 5 blocks ✓ 13 exercises ➣ Prerequisites : Maps, Counting

I Countable sets

When $f : E \to \mathbb{C}$ ($E \subset \mathbb{R}$), limit and continuity transfer to the real and imaginary parts. The bridge from the previous sections is purely formal: every result for real-valued functions extends to $\mathbb{C}$-valued ones with no additional theory, except those statements that are intrinsically about the order of $\mathbb{R}$ (TVI, bornes atteintes, monotone limit) which do not extend.

Definition — Limit of a complex-valued function

Let $f : E \to \mathbb{C}$, $a \in \overline{\mathbb{R}}$ adherent to $E$, and $\ell \in \mathbb{C}$. We say $f$ tends to $\ell$ at $a$, written $\textcolor{colordef}{f \to \ell}$ at $a$, if any of the following equivalent conditions holds:

$|f - \ell| \to 0$ at $a$ in $\mathbb{R}$;
$\mathrm{Re}(f) \to \mathrm{Re}(\ell)$ and $\mathrm{Im}(f) \to \mathrm{Im}(\ell)$ at $a$.

The equivalence comes from $\max(|\mathrm{Re}\,z|, |\mathrm{Im}\,z|) \le |z| \le |\mathrm{Re}\,z| + |\mathrm{Im}\,z|$.

Definition — Continuity of a complex-valued function

Let $f : E \to \mathbb{C}$ and $a \in E$. We say $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$. Equivalently, $\mathrm{Re}(f)$ and $\mathrm{Im}(f)$ are both continuous at $a$. The set of continuous functions on $I \subset \mathbb{R}$ to $\mathbb{C}$ is denoted $\textcolor{colordef}{C^0(I, \mathbb{C})}$.

Proposition — Operations on complex limits / continuity

Let $f, g : E \to \mathbb{C}$ admit limits $\ell, m \in \mathbb{C}$ at $a$ (resp. continuous at $a$). Then for any $\lambda, \mu \in \mathbb{C}$, $\textcolor{colorprop}{\lambda f + \mu g}$, $\textcolor{colorprop}{f g}$, and $\textcolor{colorprop}{f / g}$ (when $m \ne 0$, resp. $g(a) \ne 0$) admit the corresponding limits in $\mathbb{C}$ (resp. are continuous at $a$). Same statements as P4.1 and P5.3 with $\mathbb{C}$ in place of $\mathbb{R}$. The proofs reduce to the real and imaginary parts via D8.1.

Example

Compute $\lim_{x \to 0} (\mathrm{e}^{i x} - 1) / x$ in $\mathbb{C}$.

Answer

Decompose: $$ \frac{\mathrm{e}^{i x} - 1}{x} = \frac{\cos x - 1}{x} + i \, \frac{\sin x}{x}. $$ Imaginary part. By Ex4.1, $\sin x / x \to 1$ at $0$.
\noindent Real part. Use the half-angle identity $\cos x - 1 = -2 \sin^2(x/2)$: $$ \frac{\cos x - 1}{x} = \frac{-2 \sin^2(x/2)}{x} = -\sin(x/2) \cdot \frac{\sin(x/2)}{x/2}. $$ As $x \to 0$, $\sin(x/2) \to 0$ and $\sin(x/2)/(x/2) \to 1$ (Ex4.1 again, with $u = x/2$). Hence the real part tends to $0 \cdot 1 = 0$.
\noindent Conclusion. By D8.1, $$ \lim_{x \to 0} \frac{\mathrm{e}^{i x} - 1}{x} = 0 + i \cdot 1 = i. $$

Ex 4

Skills to practice

Exhibiting an explicit bijection
Enumerating a subset of $\mathbb{N}$

II Characterising countability

Ex 5

Differentiability formalises the secondary-school slope of a tangent line as the limit of a difference quotient. From this single notion springs a whole calculus: algebraic rules, the chain rule, the derivative of an inverse map, then three pillars of analysis --- Fermat's theorem, Rolle's theorem, and the mean value theorem. These produce the mean value inequality, the link sign-of-$f'$ $\leftrightarrow$ monotonicity, and the theorem of the limit of the derivative for rigorous $C^1$ extensions. We close with iterated derivatives, the class $C^k$, the Leibniz formula for $(fg)^{(n)}$, and a brief extension to complex-valued functions.
Endpoint convention used throughout. When $a \in I$ is an interior point of $I$, « differentiable at $a$ » is the bilateral notion (limit from both sides). When $a$ is an endpoint of $I$, it is the one-sided notion (limit from inside $I$). Most global theorems use the hypothesis « $f$ continuous on $I$, differentiable on $\mathring{I}$ »: differentiability is then only required at interior points, while endpoint behavior is handled by continuity.
Forward references to Standard functions. This chapter uses polynomials, rational functions, $\sqrt{\cdot}$, $|\cdot|$, $\sqrt[3]{\cdot}$ as a self-contained example toolbox. Wherever an example invokes $\exp$, $\sin$, $\cos$, the derivative is admitted from Real functions: lycée recap; derivations of $\ln$, $\arcsin$, $\arctan$, hyperbolic inverses are deferred to Standard functions, where they will use P2.5 of this chapter.

Definition — Derivative at a point

Let $f : I \to \mathbb{R}$ and $a \in I$. For every $h \in \mathbb{R}^*$ such that $a + h \in I$, the difference quotient of $f$ at $a$ with increment $h$ is $$ \tau_a(h) = \frac{f(a + h) - f(a)}{h}. $$ We say that $f$ is differentiable at $a$ if $\tau_a(h)$ admits a finite limit as $h \to 0$ with $a + h \in I$. This limit is called the derivative of $f$ at $a$ and is denoted $$ \textcolor{colordef}{f'(a)} \qquad \text{or} \qquad \textcolor{colordef}{\frac{df}{dx}(a)}. $$ Equivalently, by setting $x = a + h$, this amounts to requiring the finite limit of $$ \frac{f(x) - f(a)}{x - a} \quad \text{as } x \to a. $$

Definition — Tangent line

Suppose $f$ is differentiable at $a$. The tangent line to the graph of $f$ at $a$ is the line of equation $$ \textcolor{colordef}{y = f(a) + f'(a)(x - a)}. $$ If $a$ is an endpoint of $I$, we speak of a half-tangent line on the side that lies inside $I$. If $\tau_a(h)$ admits no finite limit but tends to $\pm \infty$ as $h \to 0$, we speak of a vertical tangent of equation $x = a$ (a vertical half-tangent at an endpoint).

Example

Compute $f'(a)$ for $f(x) = x^2$, for every $a \in \mathbb{R}$.

Answer

For $h \ne 0$, $$ \tau_a(h) = \frac{f(a + h) - f(a)}{h} = \frac{(a + h)^2 - a^2}{h} = \frac{2 a h + h^2}{h} = 2 a + h. $$ As $h \to 0$, $\tau_a(h) \to 2 a$. Hence $f'(a) = 2 a$.

Definition — Left and right derivatives

Let $a \in I$. We say $f$ is right-differentiable at $a$ if $\tau_a(h)$ admits a finite limit as $h \to 0^+$ with $a + h \in I$. This limit is denoted $\textcolor{colordef}{f'_d(a)}$.
We say $f$ is left-differentiable at $a$ if $\tau_a(h)$ admits a finite limit as $h \to 0^-$ with $a + h \in I$. This limit is denoted $\textcolor{colordef}{f'_g(a)}$.
If $a$ is an interior point of $I$, then $f$ is differentiable at $a$ if and only if $f'_g(a)$ and $f'_d(a)$ exist and are equal; the common value is $f'(a)$. At an endpoint, only the derivative from inside $I$ is considered.

Example

Show that $f(x) = |x|$ is not differentiable at $0$.

Answer

For $h > 0$, $\tau_0(h) = |h|/h = 1$, so $f'_d(0) = 1$. For $h < 0$, $\tau_0(h) = |h|/h = -1$, so $f'_g(0) = -1$. Since $f'_g(0) \ne f'_d(0)$, $f$ is not differentiable at $0$.

Skills to practice

Applying the injection and surjection criteria

III Operations on countable sets

Proposition — Differentiable implies continuous

If $f$ is differentiable at $a$, then $\textcolor{colorprop}{f}$ is continuous at $a$.

Proof

For $h \ne 0$ with $a + h \in I$, $$ f(a + h) - f(a) = h \, \tau_a(h). $$ As $h \to 0$, $h \to 0$ and $\tau_a(h) \to f'(a) \in \mathbb{R}$, so $h \, \tau_a(h) \to 0$ by the product rule for function limits (P3.1 of Limits and continuity). Hence $f(a + h) \to f(a)$, which is continuity at $a$.

Example

The converse of P1.2 is false: $f(x) = |x|$ is continuous at $0$ (Ex1.3 above) but not differentiable at $0$.

Method — Establishing differentiability at a point

Three classical approaches:

Direct via the difference quotient. Compute $$ \tau_a(h) = \frac{f(a + h) - f(a)}{h} $$ and check whether it has a finite limit as $h \to 0$.
Via $DL_1$. Find $\ell \in \mathbb{R}$ and $\varepsilon$ with $\varepsilon(h) \to 0$ such that $f(a+h) = f(a) + \ell h + h \varepsilon(h)$. Then $f'(a) = \ell$.
Via the theorem of the limit of the derivative (T5.2 below). If $f$ is continuous at $a$ and differentiable on $I \setminus \{a\}$, compute $\lim_a f'$.

Ex 6 Ex 7 Ex 8 Ex 9 Ex 10

Skills to practice

Proving countability through operations

IV Uncountable sets

Ex 11

Linearity, product, quotient, chain rule, derivative of the inverse: the same toolbox as the lycée, but with rigorous proofs. The derivative of the inverse map (P2.5) is the genuinely new atom at this level; it underlies the rigorous derivatives of $\arcsin$, $\arctan$, $\ln$ to be proved in Standard functions.

Proposition — Product rule

Let $f, g : I \to \mathbb{R}$ be differentiable at $a \in I$. Then $f g$ is differentiable at $a$ and $$ \textcolor{colorprop}{(f g)'(a) = f'(a) g(a) + f(a) g'(a)}. $$

Proof

For $h \ne 0$ with $a + h \in I$, $$ f(a + h) g(a + h) - f(a) g(a) = (f(a + h) - f(a)) g(a + h) + f(a) (g(a + h) - g(a)). $$ Dividing by $h$ gives $$ \tau_a^{f g}(h) = \tau_a^f(h) \cdot g(a + h) + f(a) \cdot \tau_a^g(h). $$ As $h \to 0$: $\tau_a^f(h) \to f'(a)$; $g(a + h) \to g(a)$ by continuity of $g$ at $a$ (P1.2); $\tau_a^g(h) \to g'(a)$. Hence $\tau_a^{f g}(h) \to f'(a) g(a) + f(a) g'(a)$.

Proposition — Quotient rule

Let $f, g : I \to \mathbb{R}$ be differentiable at $a \in I$, with $g(a) \ne 0$. Then $f/g$ is defined on a neighborhood of $a$ in $I$, differentiable at $a$, and $$ \textcolor{colorprop}{(f/g)'(a) = \frac{f'(a) g(a) - f(a) g'(a)}{g(a)^2}}. $$

Proof

First, the derivative of $1/g$ at $a$. By continuity of $g$ at $a$ and $g(a) \ne 0$, $g$ does not vanish on a neighborhood of $a$. For small $h$ with $a + h$ in this neighborhood and $h \ne 0$, $$ \frac{1}{g(a + h)} - \frac{1}{g(a)} = \frac{g(a) - g(a + h)}{g(a + h) \, g(a)}. $$ Dividing by $h$: $$ \tau_a^{1/g}(h) = -\frac{\tau_a^g(h)}{g(a + h) \, g(a)} \to -\frac{g'(a)}{g(a)^2}. $$ Hence $1/g$ is differentiable at $a$ with $(1/g)'(a) = -g'(a)/g(a)^2$. The case $f/g = f \cdot (1/g)$ follows from the product rule (P2.2).

Example

By induction on $n \in \mathbb{N}$, show that $(x^n)' = n x^{n - 1}$ on $\mathbb{R}$.

Answer

Case $n = 0$: $f(x) = x^0 = 1$ is constant, so $f'(x) = 0$. (The formula $(x^n)' = n x^{n-1}$ is interpreted at $n = 0$ as $0$; the expression $x^{-1}$ does not need to be defined.) Case $n \ge 1$, by induction. Base $n = 1$: $f(x) = x$, $\tau_a(h) = ((a + h) - a)/h = 1 \to 1$, so $f'(a) = 1 = 1 \cdot x^0$. Heredity: assume $(x^n)' = n x^{n-1}$ for some $n \ge 1$. Then by the product rule (P2.2), $(x^{n+1})' = (x \cdot x^n)' = 1 \cdot x^n + x \cdot n x^{n-1} = x^n + n x^n = (n+1) x^n$.

Skills to practice

Running a diagonal argument