Definition: A set \(U \subseteq \mathbb R\) is open when
\[\begin{align*} \forall a \in U, \exists r \gt 0, (a - r, a + r) \subseteq U \end{align*}\]Proposition:
Proof:
Example: The infinite intersection of open sets is not open. Consider \(U_n = (-1/n, 1/n)\) for \(n \in \mathbb Z_+\). Then \(U_n\) is open, \(\forall n \geq 1\), but \(\bigcap_{n = 1}^\infty U_n = \lbrace 0 \rbrace\), which is not open.
Proposition: Any open interval in \(\mathbb R\) is an open set.
Proof: If \(I = (-\infty, \infty)\), then \(I = \mathbb R\), which is open. So suppose otherwise. Let \(a, b \in \mathbb R\) and WLOG, suppose \(a \lt b\). Let \(c \in (a, b)\). Choose \(r = \min \lbrace c - a, b - c \rbrace\). Let \(x \in (c - r, c + r)\). Then \(a \leq c - r \lt x \lt c + r \leq b\), so \(x \in (a, b)\). So \((c - r, c + r) \subseteq (a, b)\), so \((a, b)\) is open. $\blacksquare$
Proposition: Any closed interval is a closed set.
Proof: For any closed interval $[a, b]$, one has $\mathbb R \setminus [a, b] = (-\infty, a) \cup (b, \infty)$, which is an open set. $\blacksquare
If \((a, \infty)\), choose \(r = c - a\), and if \((-\infty, b)\), choose \(r = b - c\).
Definition: A set \(F \subseteq \mathbb R\) is closed, when \(\mathbb R \setminus F\) is open.
Proposition:
Proof: TODO.
Definition: A point \(a \in \mathbb R\) is a limit point of a set \(A \subseteq \mathbb R\) when \(\exists (x_n)_{n = 1}^\infty \subseteq A \setminus \lbrace a \rbrace\) such that \(\lim_{n \to \infty} x_n = a\).
Theorem 1: A set \(X\) is closed if and only if
\[[\forall x_0 \in \mathbb R, \exists (x_n)_{n = 1}^\infty, \lim_{n \to \infty} x_n = x_0] \implies x_0 \in X\]That is, a set is closed if and only if it contains all its limit points.
Proof: (\(\rightarrow\)) Assume \(X \subseteq \mathbb R\) is closed, and let \(x_0 \in \mathbb R\) and \((x_n)_{n = 1}^\infty \subseteq X\) be such that \(\lim_{n \to \infty} x_n = x_0\). Suppose \(x_0 \not\in X\). Then \(x_0 \in \mathbb R \setminus X\), which is open. Choose \(\epsilon \gt 0\) such that \((x_0 - \epsilon, x_0 + \epsilon) \subseteq \mathbb R \setminus X\). Then, \(\forall n \in \mathbb Z_+\), \(\vert x_n - x_0 \vert \geq \epsilon\), contradicting the assumption that \(\lim x_n = x_0\). $\blacksquare$
(\(\leftarrow\)) Assume the right side and suppose for the sake of contradiction that $X$ is not closed. Then $\mathbb R \setminus X$ is not open. Choose $x_0 \in \mathbb R \setminus X$ such that a neighborhood $(x_0 - 1/n, x_0 + 1/n) \not\subseteq \mathbb R \setminus X$. Then, $\exists x_n \in X$ such that $\vert x_n - x_0 \vert t 1/n$. So the sequence $(x_n)$ converges to $x_0$, and hence $x_0 \in X$, by assumption; a contradiction. $\blacksquare$
Definition: A set \(K \subseteq \mathbb R\) is compact, when \(\forall (x_n)_n \subseteq K\), \(\exists (x_{n_k})_k \subseteq (x_n)_n\) convergent to an element of \(K\).
Proposition: Every closed interval is a compact set.
Proof: Let \(I = [a, b]\), with \(a \lt b\) since if \(a = b\) then \(I\) is a singleton and there is nothing to show. Let \((x_n)_n \subseteq [a, b]\). By Bolzano-Weistress, we can choose a subsequence \((x_{n_k})_k \subseteq (x_n)_n\) such that \(\lim_{k \to \infty} x_{n_k} = x_0 \in \mathbb R\). By the above theorem and since \(I = \bar I\), it follows that \(x_0 \in I\). \(\blacksquare\)
Theorem (Heine-Borel): A set \(K \subseteq \mathbb R\) is compact iff \(K\) is closed and bounded.
Proof: (\(\leftarrow\)) Assume \(K\) is closed and bounded. Let \(S = (x_n)_n \subseteq K\) be an arbitrary sequence. Since \(K\) is bounded, \(S\) is bounded, and by Bolzano-Weistress, we can choose a convergent subsequence \(S' = (x_{n_k})_k \subseteq S\). Let \(x_0 = \lim_{k \to \infty} x_{n_k}\). By Theorem 1, \(x_0 \in K\).
(\(\rightarrow\)) Assume for the sake of contradiction that \(K\) is not bounded above. Then, \(\forall n \in \mathbb Z_+\), \(\exists x_n \in K\) such that \(x_n \geq n\). We claim that \((x_n)_n\) contains no convergent subsequence. Suppose otherwise that there is \((x_{n_k})_k \subseteq (x_n)_n\) convergent to \(x_0 \in \mathbb R\). Choose \(N_0 \in \mathbb N\) such that \(N_0 \geq x_0 + 1\). Then, \(\forall n \geq N_0\), \(\vert x_n - x_0 \vert \geq 1\). In particular, if \(K_0\) is such that \(n_{k_0} \geq N_0\), then \(\forall k \geq K_0, \vert x_{n_k} - x_0 \vert \geq 1\). So there is no subsequence convergent to an $x_0 \in K$, contradicting the compactness of $K$.
It remains to show that \(K\) is closed. Let \(x_0 \in \mathbb R, (x_n)_n \subseteq K\) be such that \(\lim_{n \to \infty} x_n = x_0\). By assumption of the compcatness of \(K\), we can choose a subsequence \((x_{n_k})_k\) convergent to \(z_0 \in K\). Since \(\lim x_n\) exists, then \(x_0 = z_0 \in K\). So \(K\) is closed by Theorem 1. \(\blacksquare\)
Example: The ternary Cantor set is a compact set.
Definition: A set \(U \subseteq \mathbb R\) is called an (open) neighborhood of a point \(a \in \mathbb R\) when \(U\) is open and \(a \in U\).
Proposition: $a \in \mathbb R$ is a limit point of $A \subseteq \mathbb R$ when $\forall U$ open neighborhoods of $a$, $\exists x \in U \cap A \setminus \lbrace a \rbrace$.
Definition: Given a function \(f: A \to \mathbb R, A \subseteq \mathbb R\), and a limit point of \(A\), \(a \in \mathbb R\), we say that the limit of \(f\) at \(a\) exists and equals \(L\), and we write \(\lim_{x \to a} f(x) = L\) when
\[\begin{align*} \forall \epsilon \gt 0, \exists \delta \gt 0, \forall x \in A, 0 \lt \vert x - a \vert \lt \delta \implies \vert f(x) - L \vert \lt \epsilon \end{align*}\]Example: Show that \(\lim_{x \to 1} \frac{1}{x} = 1\).
Note: Let \(a = 1, f(x) = \frac{1}{x}, A = \mathbb R \setminus \lbrace 0 \rbrace\). We have
\[\begin{align*} \vert 1/x - 1 \vert = \vert \frac{1 - x}{x} \vert = \frac{1}{\vert x \vert} \cdot \vert x - 1 \vert \end{align*}\]Since we can make \(\vert x - 1 \vert\) arbitrarily small, it remains to show that \(\frac{1}{\vert x \vert}\) is bounded by some \(m\) for all \(0 \lt \vert x - 1 \vert \lt \delta\). Assuming that \(\delta \leq \frac{1}{2}\), we have \((1 - 1/2 \lt x \lt 1 + 1/2)\). Then, \(x \gt \frac{1}{2} \implies \frac{1}{x} \lt 2 \implies \frac{1}{\vert x \vert} \lt 2\), since \(x \gt 0\).
Proof: Let \(\epsilon \gt 0\) be arbitrary. Choose \(\delta = \min\lbrace \frac{1}{2}, \frac{\epsilon}{2} \rbrace\). Then, for all \(x \in \mathbb R\) such that \(0 \lt \vert x - 1 \vert \lt \delta\), we have \(\frac{1}{\vert x \vert} \lt 2\) and so \(\left\vert \frac{1}{x} - 1 \right\vert = \frac{1}{\vert x \vert} \cdot \vert x - 1 \vert \lt 2 \delta \leq \epsilon\).
Theorem: Let $f: A \to \mathbb R$ and let $a \in \mathbb R$ be a limit point of $A$. FCAE:
Proof: ($1 \to 2$) Let $(x_n)$ be such a subsequence convergent to $a$. Then let $\epsilon \gt 0$ be arbitrary, and let $\delta \gt 0$ such that the limit definition is satisfied. Then choose $N_0 \in \mathbb N$ such that $\forall n \geq N_0$, $\vert x_n - a \vert t \delta$. Since $\delta, \epsilon$ satisfy the limit definition, $\vert f(x_n) - L \vert t \epsilon$ for all $n \geq N_0$, as required. $\blacksquare$
($2 \to 1$) Suppose otherwise. Then we can choose $\epsilon_0 \gt 0$ such that $\forall n \in \mathbb Z_+, \exists x_n \in A$ such that $0 t \vert x_n - a \vert t 1/n$ and $\vert f(x_n) - L \vert \gt \epsilon_0$. But then, $(x_n)$ converges to $a$ so by assumption, $\lim_{n \to \infty} f(x_n) = L$; a contradiction. $\blacksquare$
Theorem (Algebraic Limit Theorem): Let $f, g: A \to \mathbb R$ and let $a \in \mathbb R$ be a limit point of $A$. Suppose $\lim_{x \to a} f(x) = s$, $\lim_{x \to a} g(x) = t$, and $c \in \mathbb R$ is a constant. Then:
Proof: By the above theorem, it suffices to show that for every sequence $(x_n) \subseteq A \setminus \lbrace a \rbrace$ convergent to $a$, the sequences $(f(x_n) \pm g(x_n)), \dots$ converge to their respective limits. This follows from the Algebriaic Limit THeorem for sequences. $\blacksquare$.
Definition: A point $a \in \mathbb R$ is said to be an isolated point of a set $A \subseteq \mathbb R$ when $a \in A$ and $\exists \delta \gt 0$ such that $(a - \delta, a + \delta) \cap A = \lbrace a \rbrace$.
Definition: We say that a function $f: A \to \mathbb R$ is continuous at a point $a \in A$ when $A$ is an isolated point of $A$, or else $\lim_{x \to a} f(x)$ exists and equals $f(a)$.
Definition: We say that $f: A \to \mathbb R$ is continuous when $\forall a \in A$, $f$ is continuous at $a$.
Theorem: Let $f: A \to \mathbb R$ and $a \in A$. The following conditions are equivalent:
Proof:
(1) TODO.
(2) TODO.
(3) TODO.
(4) (\(\rightarrow\)) Assume \(f\) is continuous at \(a\). Let \(V\) be any neighborhood of \(f(a)\). By the openeness of \(V\), there is some \(\epsilon_0 \gt 0\) such that \((f(a) - \epsilon_0, f(a) + \epsilon_0) \subseteq V\). By continuity of \(f\) at \(a\), choose \(\delta_0 \gt 0\) such that \(\forall x \in A, \vert x - a \vert \lt \delta_0 \implies \vert f(x) - f(a) \vert \lt \epsilon_0\). Define \(U = (a - \delta_0, a + \delta_0)\). Then, \(\forall x \in A \cap U\), so \(\vert f(x) - f(a) \vert \lt \epsilon_0\), or \(f(x) \in (f(a) - \epsilon_0, f(a) + \epsilon_0) \subseteq V\). So \(f(U) \subseteq V\).
(\(\leftarrow\)) Assume next that \(\forall V\) open neighborhood of \(f(a)\), \(\exists U\) open neighborhood of \(a\) such that \(f(U) \subseteq V\). Let \(\epsilon \gt 0\). Consider \(V := (f(a) - \epsilon, f(a) + \epsilon)\). By assumption, we can choose \(U\) and open neighborhood of \(a\) such that \(f(x) \in V\) for all \(x \in U \cap A\). By openness of \(U\), we can choose \(\delta \gt 0\) such that \((a - \delta, a + \delta) \subseteq U\). Then, \(\forall x \in A, \vert x - a \vert \lt \delta \implies x \in U \cap A \implies f(x) \in V\), so \(\vert f(x) - f(a) \vert \lt \epsilon\). \(\blacksquare\)
Corollary: The following conditions are equivalent:
Theorem: If \(f, g: A \to \mathbb R\) are continuous at \(a \in A, c \in \mathbb R\), then \(f + g, f - g, c \cdot f, f \cdot g\) are continuous at \(a\) and \(f/g\) are continuous at \(a\) if \(g(a) \neq 0\).
Proof: TODO.
Corollary: Every polynomial function is continuous on \(\mathbb R\), and every rational function \(P(x)/Q(x)\) is continuous on \(\mathbb R \setminus Q^{-1}(0)\).
Proof: Suffices to show that for any \(c \in \mathbb R\), \(\lbrace x \in \mathbb R \mapsto c \in \mathbb R \rbrace\) is continuous and \(\lbrace x \in \mathbb R \mapsto x \in \mathbb R \rbrace\) is continuous. TODO. \(\blacksquare\).
Theorem: Given \(f: A \to \mathbb R, g: B \to \mathbb R\), \(a \in A\), \(f(a) \in B\), \(f(A) \subseteq B\), if \(f\) is continuous at \(a\) and \(g\) continuous at \(f(a)\), then \(g \circ f\) is continuous at \(a\).
Proof: By assumption, \(\forall (a_n)_n \subseteq A\), \(\lim_{n \to \infty} a_n = a \implies \lim_{n \to \infty} f(a_n) = f(a)\) and \(\forall (b_n)_n \subseteq B\), \(\lim_{n \to \infty} b_n = f(a) \implies \lim_{n \to \infty} g(b_n) = g(f(a))\). We want to show that \(\forall (a_n)_n \subseteq A\), \(\lim_{n \to \infty} a_n = a \implies \lim_{n \to \infty} (g \circ f) (a_n) = (g \circ f)(a)\).
Let \((a_n)_n \subseteq A\) be any such sequence such that \(\lim a_n = a\). Then \((f(a_n))_n\) converges to \(f(a)\). By assumption, \(\lim_{n \to \infty} g(f(a_n)) = g(f(a))\). \(\blacksquare\)
Theorem: Suppose \(f: A \to \mathbb R\) is a continuous function, \(K \subseteq A\) is compact. Then \(f(K)\) is also compact.
Proof: Let \((y_n)_n\) be any sequence in \(f(K)\). Choose, for \(n \geq 1\), \(x_n \in K\) such that \(f(x_n) = y_n\). Then, \((x_n)_n\) is a sequence in \(K\). By compactness, we can choose a subsequence \((x_{n_k})_k \in (x_n)_n\) such that \(\lim_{k \to \infty} x_{n_k} = x_0 \in K\). Consider \((y_{n_k})_k\) where \(y_{n_k} = f(x_{n_k})\). By continuity of \(f\) at \(x_0\), \(\lim_{k \to \infty} f(x_{n_k}) = f(x_0) \in f(K)\). \(\blacksquare\)
Corollary (Extreme Value Theorem): If \(f: K \to \mathbb R\) is continuous and \(K \subseteq \mathbb R\) is non-empty and compact, then \(f\) admits its maximum and minimum value on \(K\).
Proof: By the theorem above, \(f(K) \subseteq \mathbb R\) is a compact subset, so by Heine-Borel, it is closed and bounded. Then, \(\inf f(K), \sup f(K) \in \mathbb R\). We have \(\inf f(K), \sup f(K)\) are limit points of \(f(K)\), and \(f(K)\) being closed, it follows that \(\inf f(K), \sup f(K) \in f(K)\). \(\blacksquare\)
Definition: We say \(f: A \to \mathbb R\) is uniformly continuous when
\[\begin{align*} \forall \epsilon \gt 0, \exists \gt 0, \forall x_1, x_2 \in A, \vert x_1 - x_2 \vert \lt \delta \implies \vert f(x_1) - f(x_2) \vert \lt \epsilon \end{align*}\]Theorem: If \(f: K \to \mathbb R\) is continuous and \(K\) is compact, then \(f\) is uniformly continuous.
Proof: Suppose for the sake of contradiction that there is \(\epsilon_0 \gt 0\) such that \(\forall \delta \gt 0\), \(\exists x_1, x_2 \in K\) such that \(\vert x_1 - x_2 \vert \lt \delta \land \vert f(x_1) - f(x_2) \vert \geq \epsilon_0\). In particular, \(\forall n \in \mathbb Z_+\), \(\exists x_n^1, x_n^2 \in K\) such that \(\vert x_n^1 - x_n^2 \vert \lt 1/n \land \vert f(x_n^1) - f(x_n^2) \vert \geq \epsilon_0\). Hence, we get two sequences \((x_n^1)_n, (x_n^2)_n \subseteq K\). Pick a subsequence \((x^1_{n_k})_k \subseteq (x^1_n)_n\) such that \(\lim_{k \to \infty} x^1_{n_k} = x_0 \in K\). Consider \((x^2_{n_k})_k \subseteq (x^2_n)_n\). We have \(\forall k, \vert x^1_{n_k} - x^2_{n_k} \vert \lt 1/n_k \to_{n \to \infty} 0\), so \(\lim_{k \to \infty} x^2_{n_k} = \lim_{k \to \infty} x^1_{n_k} = x_0\). By continuity of \(f\) at \(x_0\), \(\lim f(x^1_{n_k}) = f(x)_0 = \lim f(x^2_{n_k})\). So \(\exists K_0 \in \mathbb N\) such that \(\forall k \geq K_0\), \(\vert f(x^1_{n_k}) - f(x_0) \vert \lt \epsilon_0/2 \land \vert f(x^2_{n_k}) - f(x_0) \vert \leq \epsilon_0/2\). When adding these two inequalities, we get a contradiction, since we already determined that \(\vert f(x^1_n) - f(x^2_n) \vert \geq \epsilon_0\). \(\blacksquare\)
Examples (Warning): The theorem fails if \(K\) is not closed or not bounded.
Proof: Suppose otherwise. Consider \(\epsilon = 1\). Let \(\delta_0 \gt 0\) such that \(\forall 0 \leq x_1 \lt x_2\), \(x_2 - x_1 \lt \delta_0\), then \(\vert x_2^2 - x_1^2 \vert \lt 1\). Choose \(x_1 = 2/\delta_0\) and \(x_2 = 2/\delta_0 + \delta_0/2\). Then, \(x_2 - x_1 = \delta_0/2 \lt \delta_0\), and \(\vert x_2^2 - x_1^2 \vert = (x_2 - x_1)(x_2 + x_1) = \delta_0/2 \cdot (2/\delta_0 + 2/\delta_0 + \delta_0/2) \gt \delta_0/2 \cdot 4/\delta_0 = 2 \gt \epsilon = 1\). \(\blacksquare\)
Proof: Suppose otherwise. Consider \(\epsilon = 1\). Choose \(0 \lt \delta_0 \lt 1\) such that \(\forall 0 \lt x_1 \lt x_2 \leq 1, x_2 - x_1 \lt \delta_0 \implies \vert 1/x_2 - 1/x_1 \vert \lt 1\). Let \(x_1 = \delta_0/4\), \(x_2 = \delta_0/2\). Then \(x_2 - x_1 = \delta_0/2 - \delta_0/4 = \delta_0/4 \lt \delta_0\), and \(1/x_1 - 1/x_2 = \frac{x_2 - x_1}{x_1x_2} = \delta_0/4 \cdot 8/\delta_0^2 = 2/\delta_0 \gt 2 \gt 1\); a contradiction. \(\blacksquare\)
Theorem (Intermediate Value Theorem): Let \(f: I \to \mathbb R\) be continuous, where \(I\) is any interval. Then, \(f(I) \subseteq \mathbb R\) is an interval.
Proof: We want to show that \(\forall u, v, w \in \mathbb R; u \lt v \lt w; u, w \in f(I) \implies v \in f(I)\). Let \(u, v, w \in \mathbb R\) such that \(u \lt v \lt w\) and \(u, w \in f(I)\). Choose \(x_u, x_w \in I\) be such that \(f(x_u) = u\) and \(f(x_w) = w\). Either \(x_u \lt x_w\) or \(x_u \gt x_w\). WLOG, suppose \(x_u \lt x_w\). Then, \([x_u, x_w] \subseteq I\). Consider \(A = \lbrace x \in [x_u, x_w] : f(x) \lt v \rbrace\), \(B = \lbrace x \in [x_u, x_w] : f(x) \gt v \rbrace\). We have \(x_u \in A\) and \(x_w \in B\). So \(A, B\) are non-empty and bounded, so they have an infimum and supremum. Define \(\alpha := \sup A \in [x_u, x_w]\). By continuity at \(\alpha, f(\alpha) = \lim_{n \to \infty} f(x_n)\) for any \((x_n) \subseteq A\) such that \(\lim x_n = \alpha\). It follows that \(f(\alpha) \leq v\), since \(f(x_n) \lt v\). Now, \(\forall n \in \mathbb Z_+\), \(\alpha + 1/n \not\in A\). It follows that \(\forall n \in \mathbb Z_+, f(\alpha + 1/n) \geq v\). Consider the sequence \((z_n)_n \subseteq I\) where \(z_n = \alpha + 1/n\). So \(\vert z_n - \alpha \vert = 1/n \to_{n \to \infty} 0\), so \(\lim_{n \to \infty} z_n = \alpha\). By continuity, \(\lim f(z_n) = f(\alpha)\). Hence, \(f(\alpha) \geq v\). So \(f(\alpha) = v\). So \(v \in f(I)\). \(\blacksquare\)
Example: Let \(g: [0, 1] \to [0, 1]\) be continuous. Show that \(\exists a \in [0, 1]\) such that \(g(a) + 2a^5 = 3a^7\).
Proof: Consider* \(f(x) := g(x) + 2x^5 - 3x^7\). Then \(f\) is continuous on \([0, 1]\). We want to show that \(\exists a \in [0, 1]\) such that \(f(a) = 0\). Note that \(f(0) = g(0) \geq 0\). Also, \(f(1) = g(1) - 1 \leq 0\). Either \(f(0) = 0\) so \(a = 0\), or \(f(1) = 0\) so \(a = 1\), or else \(f(0) \gt 0\) and \(f(1) \lt 0\). So \(0 \in (f(1), f(0))\). By IVT, \(\exists a \in (0, 1)\) such that \(f(a) = 0\). \(\blacksquare\)
Corollary: If \(f: I \to \mathbb R\) is continuous and \(I\) is a closed interval, then \(f(I)\) is a closed interval.
Proof: By IVT, \(f(I)\) is an interval. By EVT, \(f(I)\) is compact, so it is bounded. So \(\inf f(I), \sup f(I) \in \mathbb R\). Since \(\inf f(I)\) and \(\sup f(I)\) are limit points of \(f(I)\) and \(f(I)\) is closed, \(\inf f(I), \sup f(I) \in f(I)\). So \(f(I) = [\inf f(I), \sup f(I)]\). \(\blacksquare\)
Definition: Let \(f: A \to \mathbb R\) with \(A \neq \emptyset\). We say that \(f\) is:
A function is monotone if it is increasing or decreasing, and strictly montoone if it is strictly increasing or strictly decreasing.
Theorem: Let \(I \subseteq \mathbb R\) be an interval and let \(f: I \to \mathbb R\) be a continuous injection. Then
(1) \(f\) is strictly monotone.
(2) \(f^{-1}: f(I) \to I\) is a strictly monotone continuous injection.
Proof:
(1) Observe that for any \(a, b, c \in I\) such that \(a \lt b \lt c\), one cannot have \(f(a) \lt f(b) \gt f(c)\)x or \(f(a) \gt f(b) \lt f(c)\) (by IVT). Therefore, \(\forall x \in I, \forall u, v \in I\) such that \(u \lt x \lt v\), either (a) \(f(u) \lt f(x) \lt f(v)\) or else (b) \(f(u) \gt f(x) \gt f(v)\). Let \(x_1 \in I\). Suppose that (a) is the case. We claim that \(\forall x \in I\), (a) holds at \(x\). Suppose otherwise, that there exists \(x_2 \in I\) such that (b) holds at \(x_2\). If \(x_1 \lt x_2\), then \(f(x_1) \lt f(x_2)\) (by (a) at \(x_1\)) and \(f(x_1) \gt f(x_2)\) (by (b) at \(x_2\)), a contradiction. The argument is symmetric for if \(x_2 \lt x_1\). So, \(f\) is strictly increasing. Case (b) holds symmetrically to show that \(f\) is strictly decreasing. \(\blacksquare\)
(2) We claim that if (and only if) \(f\) is strictly increasing, \(f^{-1}\) is strictly increasing. If \(y_1, y_2 \in f(I), y_1 \lt y_2\), let \(x_1, x_2 \in I\) be such that \(f(x_1) = y_1, f(x_2) = y_2\). Then \(f^{-1}(y_1) = x_1 \lt x_2 = f^{-1}(y_2)\), for otherwise, \(x_2 \lt x_1 \implies f(x_2) \lt f(x_1)\); a contradiction.
Let \(b \in f(I)\) be arbitrary. We claim that \(f^{-1}\) is continuous at \(b\). Let \(\epsilon \gt 0\). Let \(a = f^{-1}(b)\). Consider \(y_1 := f(a - \epsilon), y_2 := f(a + \epsilon)\). Choose \(\delta = \min \lbrace y_2 - b, b - y_1 \rbrace\). Then \(\forall y \in I\),
\[\begin{align*} \vert y - b \vert \lt \delta &\implies y \in (y_1, y_2) \\ &\implies y \in (f(a - \epsilon), f(a + \epsilon)) \\ &\implies \exists x_y \in (a - \epsilon, a + \epsilon), f(x_y) = y \text{ (by IVT)}\\ &\implies f^{-1}(y) \in (a - \epsilon, a + \epsilon) \\ &\implies \vert f^{-1}(y) - f^{-1}(b) \vert \lt \epsilon \end{align*}\]Definition: A function \(f: A \to \mathbb R\) is Lipschitz, when
\[\begin{align*} \exists L \geq 0, \forall x_1, x_2 \in A, \vert f(x_1) - f(x_2) \vert \leq L \cdot \vert x_1 - x_2 \vert \end{align*}\]Any such \(L\) is called a Lipschitz constant for \(f\).
Remark: Any Lipschitz function is uniformly continuous (Lipschitz is a stronger condition than uniform continuity).
Proof: Let \(L\) be a Lipschitz constant for \(f\). Let \(\epsilon \gt 0\) be arbitrary. Set \(\delta := \epsilon/L\). Ten \(\forall x_1, x_2 \in A, \vert x_1 - x_2 \vert \lt \delta \implies \vert f(x_1) - f(x_2) \vert \lt L \cdot \delta = \epsilon\).
Example: Find a function that is uniformly continuous but not Lipschitz.
\[\begin{align*} f(x) = \begin{cases} x\sin(\pi / x^2) & x \in (0, 1] \\ 0 & x = 0 \end{cases} \end{align*}\]Since \(\lim_{x \to 0^+} f(x) = 0 = f(0)\), \(f\) is continuous at \(0\), so \(f\) is continuous on \([0, 1]\), so \(f\) is uniformly continuous. Set \(x_n = \frac{1}{\sqrt{2n}}\) and \(y_n = \frac{1}{\sqrt{2n + \frac{1}{2}}}, n \in \mathbb Z_+\). Then \(f(x_n) = 0\) and \(f(y_n) = y_n\). Then,
\[\begin{align*} \frac{\vert f(x_n) - f(y_n) \vert}{\vert x_n - y_n \vert} = \frac{1}{\sqrt{2n + \frac{1}{2}}} \cdot \frac{1}{\frac{1}{\sqrt{2n}} - \frac{1}{\sqrt{2n + \frac{1}{2}}}} = 2 \cdot \sqrt{2n} (\sqrt{2n + \frac{1}{2}} + \sqrt{2n}) \to_{n \to \infty} \infty \end{align*}\]Definition: A function \(f: A \to \mathbb R\) is called a contraction when \(f\) is Lipschitz with a Lipschitz constant \(L \lt 1\).
Theorem (Fixed Point): Let \(A \neq \emptyset, A \subseteq \mathbb R\), and let \(f: A \to A\) be a contraction. Suppose \(A\) is a closed set. Then \(\exists!p \in A\) such that \(f(p) = p\).
Proof: Pick any \(x_1 \in A\). Construct a sequence \((x_n) \subseteq A\) defined as \(x_{k + 1} = f(x_k), \forall k \in \mathbb Z_+\). Let \(\alpha \in [0, 1)\) be a Lipschitz constant of \(f\).
We claim that \(\forall k \geq 2\), \(\vert x_{k + 1} - x_k \vert \leq \alpha^{k - 1} \cdot \vert x_2 - x_1 \vert\). We proceed by induction. We have \(\vert x_3 - x_2 \vert = \vert f(x_2) - f(x_1) \vert \leq \alpha \vert x_2 - x_1 \vert\) since \(\alpha\) is a Lipschitz constant of \(f\). Now, assuming \(\vert x_{k + 1} - x_k \vert \leq \alpha^{k - 1} \vert x_2 - x_1 \vert\), consider \(\vert x_{k + 2} - x_{k + 1} \vert = \vert f(x_{k + 1}) - f(x_k) \vert \leq \alpha \cdot \vert x_{k + 1} - x_k \vert \leq \alpha^{(k + 1) - 1} \vert x_2 - x_1 \vert\); proving the claim.
Claim 2: \((x_n)\) is Cauchy. Let \(\epsilon \gt 0\). Choose \(N_0 \in \mathbb N\) such that \(\frac{\alpha^{N_0 - 1}}{1 - \alpha} \cdot \vert x_2 - x_1 \vert \lt \epsilon\). Then, \(\forall n \geq m \geq N_0\), and
\[\begin{align*} \vert x_n - x_m \vert &= \vert (x_n - x_{n - 1}) + (x_{n - 1} - x_{n - 2}) + \dots + \vert x_{m + 1} - (x_{m + 1} - x_m) \vert \\ &\leq \vert x_n - x_{n - 1} \vert + \vert x_{n - 1} - x_{n - 2} \vert + \dots + \vert x_{m + 1} - x_m \vert \\ &\leq \alpha^{n - 2} \vert x_2 - x_1 \vert + \alpha^{n - 3} \vert x_2 - x_1 \vert + \dots + \alpha^{m - 1} \vert x_2 - x_1 \vert \\ &= \vert x_2 - x_1 \vert \cdot \alpha^{m - 1} \cdot \alpha^{m - 1} \sum_{k = 0}^{n - m - 1} \alpha^k \\ &\leq \vert x_2 - x_1 \vert \cdot \alpha^{N_0 - 1} \cdot \frac{1}{1 - \alpha}\\ &\lt \epsilon \end{align*}\]Let \(p \in \mathbb R\) be such that \(\lim_{n \to \infty} x_n = p\). Since \(A\) is closed, \(p \in A\). By the continuity of \(f\) at \(p\), \(f(p) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_{n + 1} = \lim_{n \to \infty} x_n = p\) (since \((x_{n + 1}) \subseteq (x_n)\)). So suppose there is some other \(\exists q \in A\) such that \(f(q) = q\). Then \(\vert p - q \vert = \vert f(p) - f(q) \vert \leq \alpha \cdot \vert p - q \vert\), so \(\vert p - q \vert = 0\), so \(p = q\). \(\blacksquare\)
Definition: Given $f: A \to \mathbb R$, $a \in \mathbb R$ is such that $\forall k \in \mathbb Z_+, \exists x_k \in A, x_k \lt a \land \vert x_k - a \vert \lt 1/k$. Then, we say $L = \lim_{x \to a^-} f(x)$, when
\[\begin{align*} \forall \epsilon \gt 0, \exists \delta \gt 0, \forall x \in A, 0 \lt a - x \lt \delta \implies \vert f(x) - L \vert \lt \epsilon \end{align*}\]