These practice questions accompany the Proof and Real Analysis lecture. Each asks you to prove a result stated in the lecture. A few of the harder proofs are marked (Challenging) — attempt the others first.
Question 1
Prove De Morgan’s laws for sets: for \(A, B \subseteq X\), \[(A \cup B)^c = A^c \cap B^c \qquad\text{and}\qquad (A \cap B)^c = A^c \cup B^c.\]
We prove each equality by a chain of “if and only if” statements on membership.
First law. For any \(x \in X\), \[
x \in (A\cup B)^c \iff x \notin A \cup B \iff \big(x\notin A \text{ and } x \notin B\big) \iff x \in A^c \cap B^c.
\] The middle step is the negation of “\(x\in A\) or \(x\in B\)”, which is “\(x\notin A\) and \(x\notin B\)”. Since \(x\) was arbitrary, \((A\cup B)^c = A^c \cap B^c\).
Second law. Similarly, \[
x \in (A\cap B)^c \iff \neg\big(x\in A \text{ and } x\in B\big) \iff \big(x\notin A \text{ or } x\notin B\big) \iff x \in A^c \cup B^c.
\] Hence \((A\cap B)^c = A^c \cup B^c\). (Alternatively, this follows from the first law applied to \(A^c, B^c\) together with \((E^c)^c = E\).)
Question 2
Prove the following stability properties of open and closed sets in \({\mathbb{R}}^n\).
- The union of an arbitrary family of open sets is open.
- The intersection of finitely many open sets is open.
- Using De Morgan’s laws, deduce that an arbitrary intersection of closed sets is closed, and a finite union of closed sets is closed.
Let \(\{U_i\}_{i\in I}\) be open and \(U = \bigcup_{i\in I} U_i\). Take \(x \in U\); then \(x \in U_{i_0}\) for some \(i_0 \in I\). Since \(U_{i_0}\) is open, there is \(\varepsilon>0\) with \(B(x,\varepsilon)\subseteq U_{i_0} \subseteq U\). As \(x\in U\) was arbitrary, \(U\) is open. (No finiteness is needed here.)
Let \(U_1,\dots,U_n\) be open and \(U=\bigcap_{k=1}^n U_k\). Take \(x\in U\), so \(x\in U_k\) for every \(k\). For each \(k\) choose \(\varepsilon_k>0\) with \(B(x,\varepsilon_k)\subseteq U_k\), and set \(\varepsilon=\min\{\varepsilon_1,\dots,\varepsilon_n\}\). The minimum of finitely many positive numbers is positive, so \(\varepsilon>0\), and \(B(x,\varepsilon)\subseteq B(x,\varepsilon_k)\subseteq U_k\) for every \(k\); hence \(B(x,\varepsilon)\subseteq U\). So \(U\) is open. (Finiteness is essential: it is what guarantees \(\varepsilon>0\).)
Let \(\{F_i\}_{i\in I}\) be closed, so each \(F_i^c\) is open. By De Morgan, \(\big(\bigcap_{i} F_i\big)^c = \bigcup_{i} F_i^c\), an arbitrary union of open sets, which is open by (a); hence \(\bigcap_i F_i\) is closed. Likewise, for finitely many closed sets \(F_1,\dots,F_n\), \(\big(\bigcup_{k=1}^n F_k\big)^c = \bigcap_{k=1}^n F_k^c\) is a finite intersection of open sets, open by (b); hence \(\bigcup_{k=1}^n F_k\) is closed.
Question 3
Let \(A\subseteq{\mathbb{R}}\) be non-empty and let \(s\in{\mathbb{R}}\) be an upper bound of \(A\). Prove that \(s = \sup A\) if and only if \[\forall\, \varepsilon>0,\ \exists\, a\in A \text{ such that } a > s-\varepsilon.\]
Recall that \(\sup A\) is the least upper bound of \(A\): an upper bound that is \(\le\) every other upper bound. Throughout, only this definition is used — no completeness axiom.
(\(\Rightarrow\)) Suppose \(s=\sup A\) and let \(\varepsilon>0\). Since \(s-\varepsilon < s\) and \(s\) is the least upper bound, \(s-\varepsilon\) cannot be an upper bound of \(A\). So some \(a\in A\) satisfies \(a > s-\varepsilon\).
(\(\Leftarrow\)) Suppose \(s\) is an upper bound and for every \(\varepsilon>0\) there is \(a\in A\) with \(a>s-\varepsilon\). Let \(t\) be any upper bound of \(A\); we show \(t\ge s\). If instead \(t<s\), put \(\varepsilon = s-t>0\). By hypothesis there is \(a\in A\) with \(a> s-\varepsilon = t\), contradicting that \(t\) is an upper bound. Hence every upper bound \(t\) satisfies \(t\ge s\), so \(s\) is the least upper bound: \(s=\sup A\).
Question 4
Using the definition compact \(=\) closed and bounded (in \({\mathbb{R}}^n\)), justify each claim.
- \([0,1]^n\) is compact.
- \((0,1)\) is not compact.
- \({\mathbb{R}}\) is not compact.
- The budget set \(\{x\in{\mathbb{R}}^n : x\ge 0,\ p\cdot x \le w\}\) is compact, where \(p\gg 0\) (all prices strictly positive) and \(w>0\).
Bounded: for \(x\in[0,1]^n\) each coordinate lies in \([0,1]\), so \(\|x\|\le \sqrt{n}\). Closed: if \((x_k)\subseteq[0,1]^n\) and \(x_k\to x\), each coordinate satisfies \(0\le x_{k,i}\le 1\), so the order limit theorem gives \(0\le x_i\le 1\); thus \(x\in[0,1]^n\). By the sequential characterisation, \([0,1]^n\) is closed. Closed and bounded, hence compact.
\((0,1)\) is bounded but not closed: \(x_k = 1/k \in (0,1)\) and \(x_k\to 0\notin(0,1)\), so it does not contain the limit point \(0\). Not closed \(\Rightarrow\) not compact.
\({\mathbb{R}}\) is closed but not bounded: for any \(M>0\), the point \(M+1\) satisfies \(|M+1|>M\), so no single \(M\) bounds every element. Not bounded \(\Rightarrow\) not compact.
Write \(K=\{x : x\ge 0,\ p\cdot x\le w\}\). Bounded: let \(\underline p=\min_i p_i>0\). For \(x\in K\) each \(x_i\ge 0\), so \(\underline p\sum_i x_i \le \sum_i p_i x_i = p\cdot x\le w\), giving \(0\le x_i \le w/\underline p\) for every \(i\); hence \(\|x\|\le \sqrt n\, w/\underline p\). Closed: each set \(\{x : x_i\ge 0\}\) and \(\{x : p\cdot x\le w\}\) is closed (if \(x_k\to x\) with \(x_{k,i}\ge 0\) then \(x_i\ge0\), and if \(p\cdot x_k\le w\) then \(p\cdot x\le w\), both by the order limit theorem). \(K\) is a finite intersection of these closed sets, hence closed by Question 2(c). Closed and bounded, hence compact.
Question 5
Prove that if a sequence \((x_k)\) in \({\mathbb{R}}^n\) converges, then its limit is unique.
Suppose \(x_k\to L\) and \(x_k\to L'\); we show \(L=L'\). Let \(\varepsilon>0\). By convergence there are \(K_1,K_2\) with \(\|x_k-L\|<\varepsilon/2\) for \(k\ge K_1\) and \(\|x_k-L'\|<\varepsilon/2\) for \(k\ge K_2\). For any \(k\ge\max\{K_1,K_2\}\), the triangle inequality gives \[
\|L-L'\| \le \|L-x_k\| + \|x_k-L'\| < \tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2} = \varepsilon.
\] Thus \(\|L-L'\|<\varepsilon\) for every \(\varepsilon>0\), which forces \(\|L-L'\|=0\), i.e. \(L=L'\).
Question 6
Let \(x_k\to x\) and \(y_k\to y\) in \({\mathbb{R}}\). Prove:
- (Algebraic Limit Theorem) \(x_k+y_k\to x+y\); \(\ c\,x_k\to cx\); \(\ x_k y_k\to xy\); and, if \(y\ne0\) and \(y_k\ne0\), \(\ x_k/y_k\to x/y\).
- (Order Limit Theorem) if \(x_k\le y_k\) for all \(k\), then \(x\le y\).
- (Squeeze Theorem) if \(x_k\le z_k\le y_k\) for all \(k\) and \(x_k,y_k\to L\), then \(z_k\to L\).
The quotient limit in (a) is the most delicate part.
Sum. Given \(\varepsilon>0\), choose \(K\) with \(|x_k-x|<\varepsilon/2\) and \(|y_k-y|<\varepsilon/2\) for \(k\ge K\). Then \(|(x_k+y_k)-(x+y)|\le|x_k-x|+|y_k-y|<\varepsilon\).
Scalar. If \(c=0\) this is trivial. For \(c\ne0\), given \(\varepsilon>0\) choose \(K\) with \(|x_k-x|<\varepsilon/|c|\); then \(|cx_k-cx|=|c|\,|x_k-x|<\varepsilon\).
Product. A convergent sequence is bounded, so there is \(M>0\) with \(|x_k|\le M\) for all \(k\). Then \[
|x_ky_k-xy| = |x_k y_k - x_k y + x_k y - xy| \le |x_k|\,|y_k-y| + |y|\,|x_k-x| \le M|y_k-y| + |y|\,|x_k-x|.
\] Given \(\varepsilon>0\), choose \(K\) so that \(|y_k-y|<\dfrac{\varepsilon}{2M}\) and \(|x_k-x|<\dfrac{\varepsilon}{2(|y|+1)}\) for \(k\ge K\); then \(|x_ky_k-xy|<\varepsilon\).
Quotient. It suffices to show \(1/y_k\to 1/y\) and then apply the product rule. Since \(y\ne0\), taking \(\varepsilon_0=|y|/2\) gives \(K_0\) with \(|y_k-y|<|y|/2\) for \(k\ge K_0\), so \(|y_k|\ge|y|-|y_k-y|>|y|/2\) for such \(k\). Then for \(k\ge K_0\), \[
\left|\frac1{y_k}-\frac1y\right| = \frac{|y-y_k|}{|y_k|\,|y|} \le \frac{|y-y_k|}{(|y|/2)|y|} = \frac{2}{|y|^2}\,|y_k-y|.
\] Given \(\varepsilon>0\), also require \(|y_k-y|<\tfrac{|y|^2}{2}\varepsilon\); then \(|1/y_k-1/y|<\varepsilon\). Hence \(1/y_k\to1/y\), and \(x_k/y_k = x_k\cdot(1/y_k)\to x\cdot(1/y)=x/y\) by the product rule.
Suppose for contradiction \(x>y\), and set \(\varepsilon=(x-y)/2>0\). For large \(k\), \(|x_k-x|<\varepsilon\) and \(|y_k-y|<\varepsilon\), so \(x_k> x-\varepsilon = \tfrac{x+y}{2}\) and \(y_k< y+\varepsilon=\tfrac{x+y}{2}\), giving \(y_k< x_k\) — contradicting \(x_k\le y_k\). Hence \(x\le y\).
Given \(\varepsilon>0\), choose \(K\) with \(|x_k-L|<\varepsilon\) and \(|y_k-L|<\varepsilon\) for \(k\ge K\), i.e. \(L-\varepsilon<x_k\) and \(y_k<L+\varepsilon\). Then \(L-\varepsilon<x_k\le z_k\le y_k<L+\varepsilon\), so \(|z_k-L|<\varepsilon\). Hence \(z_k\to L\).
Question 7
- Prove the Monotone Convergence Theorem: every bounded, monotone sequence in \({\mathbb{R}}\) converges. (Prove the non-decreasing case; the non-increasing case is symmetric.)
- Show directly from the \(\varepsilon\)-\(K\) definition — without invoking the MCT — that \(x_k = 1-\tfrac1k\) converges to \(1\).
Let \((x_k)\) be non-decreasing and bounded above. The set \(A=\{x_k : k\in{\mathbb{N}}\}\) is non-empty and bounded above, so by the completeness axiom \(s=\sup A\) exists. We claim \(x_k\to s\). Let \(\varepsilon>0\). By the \(\varepsilon\)-characterisation of the supremum (Question 3), there is an index \(K\) with \(x_K > s-\varepsilon\). Since \((x_k)\) is non-decreasing, \(x_k\ge x_K> s-\varepsilon\) for all \(k\ge K\); and \(x_k\le s\) because \(s\) is an upper bound. Hence \(s-\varepsilon< x_k\le s< s+\varepsilon\), i.e. \(|x_k-s|<\varepsilon\) for \(k\ge K\). Therefore \(x_k\to s=\sup_k x_k\). (For a non-increasing sequence bounded below, apply this to \((-x_k)\), giving convergence to \(\inf_k x_k\).)
Let \(\varepsilon>0\). We want \(|x_k-1| = \big|(1-\tfrac1k)-1\big| = \tfrac1k<\varepsilon\). By the Archimedean property, choose \(K\in{\mathbb{N}}\) with \(K>1/\varepsilon\). Then for all \(k\ge K\), \(\tfrac1k\le\tfrac1K<\varepsilon\), so \(|x_k-1|<\varepsilon\). Hence \(x_k\to1\).
Question 8
- Prove the sequential characterisation of closed sets: \(F\subseteq{\mathbb{R}}^n\) is closed if and only if every convergent sequence \((x_k)\subseteq F\) has its limit in \(F\).
- Use it to show \([0,1]\) is closed and \((0,1)\) is not.
Recall \(F\) is closed iff its complement \(F^c\) is open.
(\(\Rightarrow\)) Suppose \(F\) is closed. Let \((x_k)\subseteq F\) with \(x_k\to x\); we show \(x\in F\). If instead \(x\in F^c\), then since \(F^c\) is open there is \(\varepsilon>0\) with \(B(x,\varepsilon)\subseteq F^c\). But \(x_k\to x\) yields some \(k\) with \(\|x_k-x\|<\varepsilon\), i.e. \(x_k\in B(x,\varepsilon)\subseteq F^c\), contradicting \(x_k\in F\). Hence \(x\in F\).
(\(\Leftarrow\)) Suppose every convergent sequence in \(F\) has its limit in \(F\). We show \(F^c\) is open. If not, some \(x\in F^c\) has no ball contained in \(F^c\): for every \(\varepsilon>0\), \(B(x,\varepsilon)\) meets \(F\). Taking \(\varepsilon=1/k\), pick \(x_k\in F\cap B(x,1/k)\). Then \(\|x_k-x\|<1/k\to0\), so \(x_k\to x\) with \((x_k)\subseteq F\); by hypothesis \(x\in F\), contradicting \(x\in F^c\). Hence \(F^c\) is open and \(F\) is closed.
\([0,1]\) closed: if \((x_k)\subseteq[0,1]\) and \(x_k\to x\), then \(0\le x_k\le1\) gives \(0\le x\le1\) by the order limit theorem, so \(x\in[0,1]\); by (a), \([0,1]\) is closed. \((0,1)\) not closed: \(x_k=1/k\in(0,1)\) and \(x_k\to0\notin(0,1)\), so (a) fails and \((0,1)\) is not closed.
Question 9
- Prove that if \(x_k\to L\), then every subsequence \((x_{k_j})\) converges to \(L\).
- (Challenging.) Prove the Bolzano–Weierstrass theorem: every bounded sequence in \({\mathbb{R}}\) has a convergent subsequence. Hint: first show every real sequence has a monotone subsequence, then apply the Monotone Convergence Theorem.
Let \(\varepsilon>0\). Since \(x_k\to L\), there is \(K\) with \(|x_k-L|<\varepsilon\) for all \(k\ge K\). The subsequence indices satisfy \(k_1<k_2<\cdots\), so \(k_j\ge j\); thus \(j\ge K \Rightarrow k_j\ge K \Rightarrow |x_{k_j}-L|<\varepsilon\). Hence \(x_{k_j}\to L\).
Step 1 — every sequence has a monotone subsequence. Call an index \(m\) a peak if \(x_m\ge x_k\) for all \(k>m\).
- If there are infinitely many peaks \(m_1<m_2<\cdots\), then by definition of peak \(x_{m_1}\ge x_{m_2}\ge\cdots\), a non-increasing subsequence.
- If there are only finitely many peaks, choose \(k_1\) larger than every peak. Since \(k_1\) is not a peak, there is \(k_2>k_1\) with \(x_{k_2}>x_{k_1}\). Since \(k_2\) is also past every peak, it is not a peak, so there is \(k_3>k_2\) with \(x_{k_3}>x_{k_2}\). Continuing, we obtain a strictly increasing subsequence \(x_{k_1}<x_{k_2}<\cdots\).
Either way we get a monotone subsequence.
Step 2. Let \((x_k)\) be bounded. By Step 1 it has a monotone subsequence \((x_{k_j})\), which is bounded (being a subsequence of a bounded sequence). By the Monotone Convergence Theorem, \((x_{k_j})\) converges. Hence \((x_k)\) has a convergent subsequence.
(In \({\mathbb{R}}^n\): extract a subsequence convergent in the first coordinate, then a further subsequence convergent in the second, and so on; after \(n\) steps the final subsequence converges in every coordinate, hence in \({\mathbb{R}}^n\).)
Question 10
Let \(f:X\to Y\) and \(A,B\subseteq Y\). Prove the preimage identities: \[
f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B),\quad f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B),\quad f^{-1}(A^c)=\big(f^{-1}(A)\big)^c.
\]
All three follow by unwinding the definition \(f^{-1}(S)=\{x: f(x)\in S\}\) and the matching logical connective.
Union. \(x\in f^{-1}(A\cup B) \iff f(x)\in A\cup B \iff \big(f(x)\in A \text{ or } f(x)\in B\big) \iff x\in f^{-1}(A)\cup f^{-1}(B)\).
Intersection. The same with “or” replaced by “and”: \(x\in f^{-1}(A\cap B)\iff \big(f(x)\in A\text{ and }f(x)\in B\big)\iff x\in f^{-1}(A)\cap f^{-1}(B)\).
Complement. \(x\in f^{-1}(A^c)\iff f(x)\in A^c\iff f(x)\notin A\iff x\notin f^{-1}(A)\iff x\in\big(f^{-1}(A)\big)^c\). (Here \(A^c\) is taken in \(Y\) and \(\big(f^{-1}(A)\big)^c\) in \(X\).)
Question 11
Prove that the \(\varepsilon\)-\(\delta\) and sequential definitions of continuity at a point are equivalent: for \(f:X\to{\mathbb{R}}^m\) and \(x_0\in X\), \[
\big(\forall\varepsilon>0\ \exists\delta>0:\ \|x-x_0\|<\delta\Rightarrow\|f(x)-f(x_0)\|<\varepsilon\big) \iff \big(x_k\to x_0\Rightarrow f(x_k)\to f(x_0)\big).
\]
(\(\Rightarrow\)) Assume the \(\varepsilon\)-\(\delta\) condition and let \(x_k\to x_0\). Fix \(\varepsilon>0\) and take the corresponding \(\delta>0\). Since \(x_k\to x_0\), there is \(K\) with \(\|x_k-x_0\|<\delta\) for \(k\ge K\); then \(\|f(x_k)-f(x_0)\|<\varepsilon\) for \(k\ge K\). Hence \(f(x_k)\to f(x_0)\).
(\(\Leftarrow\)) We prove the contrapositive: if \(\varepsilon\)-\(\delta\) fails, so does the sequential condition. If \(\varepsilon\)-\(\delta\) fails, there is some \(\varepsilon_0>0\) such that for every \(\delta>0\) some point within \(\delta\) of \(x_0\) has image at distance \(\ge\varepsilon_0\) from \(f(x_0)\). Taking \(\delta=1/k\), choose \(x_k\in X\) with \(\|x_k-x_0\|<1/k\) but \(\|f(x_k)-f(x_0)\|\ge\varepsilon_0\). Then \(x_k\to x_0\) while \(f(x_k)\not\to f(x_0)\), so the sequential condition fails. By contraposition, the sequential condition implies \(\varepsilon\)-\(\delta\).
Question 12
Let \(f,g:X\to{\mathbb{R}}\) be continuous at \(x_0\), and let \(c\in{\mathbb{R}}\).
- Using the sequential characterisation of continuity, prove that \(f+g\), \(cf\), and \(fg\) are continuous at \(x_0\), and that \(f/g\) is continuous at \(x_0\) provided \(g(x_0)\ne0\).
- Prove that a composition of continuous functions is continuous: if \(f:X\to Y\) is continuous at \(x_0\) and \(g:Y\to Z\) is continuous at \(f(x_0)\), then \(g\circ f\) is continuous at \(x_0\).
Let \(x_k\to x_0\) in \(X\). By continuity (sequential form), \(f(x_k)\to f(x_0)\) and \(g(x_k)\to g(x_0)\). By the Algebraic Limit Theorem (Question 6), \[
f(x_k)+g(x_k)\to f(x_0)+g(x_0),\quad c\,f(x_k)\to c\,f(x_0),\quad f(x_k)g(x_k)\to f(x_0)g(x_0),
\] and, when \(g(x_0)\ne0\), \(f(x_k)/g(x_k)\to f(x_0)/g(x_0)\) (continuity gives \(g(x_k)\to g(x_0)\ne0\), so \(g(x_k)\ne0\) for large \(k\) and the quotient rule applies). Since \(x_k\to x_0\) was arbitrary, each of \(f+g,\,cf,\,fg,\,f/g\) satisfies the sequential criterion, hence is continuous at \(x_0\).
Let \(x_k\to x_0\). Since \(f\) is continuous at \(x_0\), \(f(x_k)\to f(x_0)\). Applying the sequential criterion for \(g\) at \(f(x_0)\) to the sequence \(y_k:=f(x_k)\to f(x_0)\) gives \(g(y_k)\to g(f(x_0))\), i.e. \((g\circ f)(x_k)\to (g\circ f)(x_0)\). Hence \(g\circ f\) is continuous at \(x_0\).
Question 13
(Challenging.)
- Prove the Extreme Value Theorem: if \(K\subseteq{\mathbb{R}}^n\) is compact and \(f:K\to{\mathbb{R}}\) is continuous, then \(f\) attains a maximum and a minimum on \(K\).
- Prove the Intermediate Value Theorem: if \(f:[a,b]\to{\mathbb{R}}\) is continuous and \(y\) lies between \(f(a)\) and \(f(b)\), then \(f(c)=y\) for some \(c\in[a,b]\).
We show \(f\) attains a maximum (for the minimum, apply the result to \(-f\)).
Bounded above. Suppose not: then for each \(k\) there is \(x_k\in K\) with \(f(x_k)>k\). Since \(K\) is bounded, \((x_k)\) is bounded, so by Bolzano–Weierstrass a subsequence \(x_{k_j}\to x^*\); and \(x^*\in K\) because \(K\) is closed. By continuity \(f(x_{k_j})\to f(x^*)\), a finite number — but \(f(x_{k_j})>k_j\to\infty\), a contradiction. So \(f\) is bounded above, and \(M:=\sup_{x\in K}f(x)\) exists by completeness.
Attainment. By the \(\varepsilon\)-characterisation of the supremum (Question 3), for each \(k\) choose \(z_k\in K\) with \(f(z_k)>M-\tfrac1k\); with \(f(z_k)\le M\) this gives \(f(z_k)\to M\). By Bolzano–Weierstrass and closedness, some subsequence \(z_{k_j}\to x^*\in K\), and continuity gives \(f(z_{k_j})\to f(x^*)\). Since also \(f(z_{k_j})\to M\), uniqueness of limits gives \(f(x^*)=M\). Thus \(f\) attains its maximum at \(x^*\).
Assume without loss of generality \(f(a)<y<f(b)\) (if \(f(a)>f(b)\), apply the argument to \(-f\) and \(-y\); if \(y\) equals an endpoint value there is nothing to prove). Let \[S=\{x\in[a,b] : f(x)<y\}.\] \(S\) is non-empty (\(a\in S\)) and bounded above by \(b\), so \(c=\sup S\) exists by completeness, with \(c\in[a,b]\). We rule out both strict inequalities using continuity at \(c\).
- If \(f(c)<y\): by continuity there is \(\delta>0\) with \(f(x)<y\) for all \(x\in[a,b]\) within \(\delta\) of \(c\). Since \(f(b)>y\) we have \(c<b\), so points just to the right of \(c\) belong to \(S\), contradicting \(c=\sup S\).
- If \(f(c)>y\): then \(c\ne a\) (as \(f(a)<y\)), so \(c>a\); by continuity there is \(\delta>0\) with \(f(x)>y\) for \(x\) within \(\delta\) of \(c\). No point of \((c-\delta,c]\) lies in \(S\), so \(c-\delta\) is an upper bound of \(S\) smaller than \(c\), contradicting \(c=\sup S\).
Both cases are impossible, so \(f(c)=y\) (and \(c\in(a,b)\) since \(f(c)=y\) differs from \(f(a)\) and \(f(b)\)).