Maxima, minima, suprema and infima

Exercises · Numbers and logic · with worked solutions · PDF

Exercise 1

Determine whether the set

\[ A=[0,\sqrt{2}]\cap \Q \]

has a maximum, a minimum, a supremum, an infimum and, if so, determine these elements.

Solution

The minimum of \(A\) is \(0\). \(A\) has no maximum; its supremum is \(\sqrt{2}\).

Exercise 2

Determine whether the set

\[ A=[\sqrt{2},\sqrt{3}]\cap \Q \]

has a maximum, a minimum, a supremum, an infimum and, if so, determine these elements.

Solution

\(A\) has neither a minimum nor a maximum. The infimum is \(\sqrt{2}\), the supremum is \(\sqrt{3}\).

Exercise 3

Determine whether the set

\[ A=[0,\sqrt{2}]\cap(\R- \Q) \]

has a maximum, a minimum, a supremum, an infimum and, if so, determine these elements.

Solution

\(A\) has no minimum; the infimum is \(0\). The maximum of \(A\) is \(\sqrt{2}\).

Exercise 4

In each of the following cases, say whether the set

\[ A=(-\infty,\sqrt{2}]\cap \Q \]

has a maximum, a minimum, a supremum, an infimum and, if so, determine these elements.

Solution

\(A\) is not bounded below, hence it has no infimum in \(\R\). \(A\) has no maximum; the supremum is \(\sqrt{2}\).

Exercise 5

Say whether the following set \(A \subset\R\) has a maximum, a minimum, a supremum, an infimum, and whether it is bounded:

\[ A=\left\{\frac{2n+1}{n}: \ n\in \N \setminus \{0\} \right\} \]

Solution

From \((2n+1)/n=2+(1/n)\) we have that the maximum of \(A\) is attained for \(n=1\) and is equal to \(3\). \(A\) has no minimum; the infimum is \(2\). In particular, \(A\) is bounded.

Exercise 6

Say whether the following set \(A \subset\R\) has a maximum, a minimum, a supremum, an infimum, and whether it is bounded:

\[ A=\left\{\frac{n-1}{n}: \ n\in \N \setminus \{0\} \right\} \]

Solution

From \((n-1)/n=1-(1/n)\) we have that the minimum of \(A\) is attained for \(n=1\) and is equal to \(0\). \(A\) has no maximum; the supremum is \(1\). In particular, \(A\) is bounded.

Exercise 7

Say whether the following set \(A \subset\R\) has a maximum, a minimum, a supremum, an infimum, and whether it is bounded:

\[ A=\left\{\frac{3n^{2}+1}{n^{2}}: \ n\in \N \setminus \{0\} \right\} \]

Solution

From \((3n^{2}+1)/n^{2}=3+(1/n^{2})\) we have that the maximum of \(A\) is attained for \(n=1\) and is equal to \(4\). \(A\) has no minimum; the infimum is \(3\). In particular, \(A\) is bounded.

Exercise 8

Say whether the following set \(A \subset\R\) has a maximum, a minimum, a supremum, an infimum, and whether it is bounded:

\[ A=\left\{\frac{1}{n^{2}+1}: \ n\in \N \setminus \{0\} \right\} \]

Solution

The maximum value of \(A\) is attained for \(n=1\) and is equal to \(1/2\). \(A\) has no minimum; the infimum is \(0\). In particular, \(A\) is bounded.

Exercise 9

Let

\[ A=\left\{\left|\frac{3x}{x+1}\right| :\ -\frac{1}{2}<x\leq2\right\}. \]

Choose the correct statements among the following:

\[ (a)~~ \sup A\notin A ~~~~~~~~~~~~~~(b)~~ \inf A=\min A=2 \]

\[ (c)~~ \max A=3 ~~~~~~~~~~~~~~(d)~~ {\rm ~~none~of~the~other~answers~is~correct} \]

Solution

\(A\) is the set of values of \(|f(x)|\) on the interval \((-1/2,2]\), with \(f(x)=\frac{3x}{x+1}\). From

\[ \frac{3x}{x+1}=3-\frac{3}{x+1} \]

we have that the function \(f\) is strictly increasing on the given interval \((-1/2,2]\), negative on \((-1/2,0)\), zero for \(x=0\), and positive on \((0,2]\). It follows that

\[ |f(x)|=\left|\frac{3x}{x+1}\right| \]

is strictly decreasing on \((-1/2,0]\), where it takes all the values in \([0,3)\), and strictly increasing on \([0,2]\), where it takes all the values in \([0,2]\). The minimum value is therefore attained for \(x=0\) and is equal to \(0\). The supremum is \(3\); there is no maximum value. The correct answer is (a).

Exercise 10

For \(I=[1/3,+\infty)\) consider the function

\[ f:I\rightarrow\R,\ f(x)=\exp \left(\left|\frac{2x-1}{x}\right|\right). \]

Determine, among the following intervals, the set \(J=f(I)\) of the values taken by \(f\).

\[ (a)~~ [e,e^{2}) ~~~~~~~~~~~~~~(b)~~ (e^{-2},e] ~~~~~~~~~~~~~~(c)~~ [1,e^{2}) \]

\[ ~~~~~~~~~~~~~~(d)~~ (e,+\infty) ~~~~~~~~~~~~~~(e)~~ (0,e^{2}) ~~~~~~~~~~~~~~(f)~~ {\rm ~~another~interval} \]

Solution

The function \(f\) has the same monotonicity behavior as

\[ |g(x)|=\left|\frac{2x-1}{x}\right| \]

with

\[ g(x)=\frac{2x-1}{x}=2-\frac{1}{x}. \]

Solution

The function \(g\) is strictly increasing on the interval \(I=[1/3,+\infty)\), negative on \([1/3,1/2)\), zero for \(x=1/2\), and positive on \((1/2,+\infty)\). It follows that \(|g(x)|\) is strictly decreasing on \([1/3,1/2]\), where it takes all the values in \([0,1]\), and strictly increasing on \([1/2,+\infty)\), where it takes all the values in \([0,2)\). The set of values \(|g|(I)\) is \([0,2)\), hence for \(f(x)=e^{|g(x)|}\) we have

\[ f(I)=[1,e^{2}). \]

The correct answer is (c).

More precisely, \(f\) is strictly decreasing on \([1/3,1/2]\), where it takes all the values in \([1,e]\), and strictly increasing on \([1/2,+\infty)\), where it takes all the values in \([1,e^{2})\).

Exercise 11

Let \(A=A_{+}\cup A_{-}\) with

\[ A_{+}=\left\{x+\frac{2}{x}:x>0\right\},~~~\ A_{-}=\left\{x+\frac{2}{x}:x<0\right\}. \]

Choose the correct statements among the following:

\[ (a)~~ \sup A=+\infty ~~~~~~~~~~~~~~(b)~~ \inf A_{+}=2\sqrt{2} ~~~~~~~~~~~~~~(c)~~ A_{+} {\rm~has~no~minimum} \]

\[ ~~~~~~~~~~~~~~(d)~~ A_{-} {\rm~has~no~maximum} ~~~~~~~~~~~~~~(e)~~ {\rm none~of~the~other~answers~is~correct} \]

Solution

Clearly, \(A_{+}\) is not bounded above, hence statement (a) is correct (and statement (e) is false). For \(x\neq0\), we consider the equation

\[ x+\frac{2}{x}=y, \]

\[ x^{2}-yx+2=0. \]

There are solutions \(x\in\R\) if and only if \(y\in(-\infty,-2\sqrt{2}]\cup[2\sqrt{2},+\infty)\). Moreover, for every \(y\in(-\infty,-2\sqrt{2}]\) the solutions \(x\) are negative, while for every \(y\in[2\sqrt{2},+\infty)\) the solutions \(x\) are positive. This means

\[ A_{+}=[2\sqrt{2},+\infty),\ A_{-}=(-\infty,-2\sqrt{2}]. \]

Therefore (b) is true, (c) is false, (d) is false.