Collection of Maths Problems

Suprema and infima of concrete sets

Task number: 2824

• Variant 1

  \(\displaystyle M=\left\{\frac1n,n\in\mathbb N\right\}\)

• Solution

  For every natural number \(n\) it is true that \(\frac1n\le 1\).
  Conversely, for any \(a<1\), if \(n=1\) then \(\frac1n=1>a\), so \(a\) cannot be an upper bound.
  The set of upper bounds of \(M\) is therefore the interval \([1,\infty)\), from which we have \(\sup(M)=1\).

  Similarly for every natural number \(n\) it is true that \(\frac1n> 0\).
  Conversely, for any \(a>0\), if \(n=\lceil \frac1a\rceil+1\) then \(\frac1n<a\), so \(a\) cannot be a lower bound.
  The set of lower bounds of \(M\) is the interval \((-\infty,0]\), so \(\inf(M)=0\).

• Answer

  \(\sup(M)=1\), \(\inf(M)=0\).

• Variant 2

  \(\displaystyle M=\left\{-\frac1n,n\in\mathbb N\right\}\)

• Solution

  Similarly for every natural number \(n\) it is true that \(-\frac1n< 0\).
  Conversely, if we take any \(a<0\), then for \(n=\lceil -\frac1a\rceil+1\) we get: \(-\frac1n<a\), so \(a\) cannot be an upper bound.
  The set of upper bounds of \(M\) is the interval \([0,\infty)\), and so \(\sup(M)=0\).

  For every natural number \(n\) it is true that \(-\frac1n\ge -1\).
  Conversely, if we take any \(a>-1\), then for \(n=-1\) we get: \( \frac1n=-1>a\), so \(a\) cannot be a lower bound.
  The set of lower bounds of \(M\) is therefore the interval \((-\infty,-1,]\), and so \(\inf(M)=-1\).

• Answer

  \(\sup(M)=0\), \(\inf(M)=-1\).

• Variant 3

  \( M=\{0.3; 0.33; 0.333; 0.3333; …\}\)

• Solution

  \(0.3<0.33<0.333…\), and so \(0.3\) is a lower bound, which is also a member of \(M\), and so \(i\) is an infimum.

  The number \(\frac13\) is an upper bound. For any \(a<\frac13\) we can choose a number \(0.3…3\) with more decimal places than \(\log \left(\frac13-a\right)\), yielding a number greater than \(a\). So any such \(a\) cannot be an upper bound.

• Answer

  \(\sup(M)=\frac13\), \(\inf(M)=\frac3{10}\).

• Variant 4

  \(\displaystyle M=\left\{q: q<\sqrt{3},\ q\in\mathbb Q\right\}\)

• Solution

  If we take \(a<\sqrt{3}\), we can find a rational number \(\frac{p}{q}\) between \(a\) a \(\sqrt{3}\) , for example \(q=\left\lceil\frac{1}{\sqrt{3}-a}\right\rceil+1\), \(p=\left\lfloor\frac{a}{q}\right\rfloor+1\). And so \(a\) cannot be an upper bound.
  The set of upper bounds \(M\) is the interval \([\sqrt{3},\infty)\), and so \(\sup(M)=\sqrt{3}\).

  The set \(M\) is unbounded from below, therefore it has no lower bound and therefore no real infimum. In this circumstance is is sometimes customary to define the infimum as \(-\infty\).

• Answer

  \(\sup(M)=\sqrt{3}\); a real infimum does not exist.

• Variant 5

  \( M=\{\sin x: x\in [ 0, 2\pi)\}\)

• Solution

  The set \(M\) is the range of values of the \(\sin\) function, namely the interval \([-1{,}1]\). Both bounds are attained, and are therefore a supremum and infimum of the set \(M\).

• Answer

  \(\sup(M)=1\), \(\inf(M)=-1\).

• Variant 6

  \( M=\{\sin x: x\in (0, 2\pi)\}\)

• Answer

  \(\sup(M)=1\), \(\inf(M)=-1\).

• Variant 7

  \( M=\{\sin x: x\in (0, \pi)\}\)

• Answer

  \(\sup(M)=1\), \(\inf(M)=0\).

• Variant 8

  \(\displaystyle M=\left\{1-\frac1{n^2},n\in\mathbb N\right\}\)

• Solution

  The set of upper bounds of \(M\) is the interval \([1,\infty)\), from which \(\sup(M)=1\).

  The set of lower bounds of \(M\) is the interval \((-\infty,0]\), from which \(\inf(M)=0\).

• Answer

  \(\sup(M)=1\), \(\inf(M)=0\).

• Variant 9

  \(\displaystyle M=\left\{\frac{n-1}n,n\in\mathbb Z\setminus\{0\}\right\}\)

• Solution

  For positive \(n\) the set \(M\) contains the fractions \(0,\frac12,\frac23,\frac34,\frac45,…\).

  For negative \(n\) the set \(M\) contains the fractions \(2,\frac32,\frac43,\frac54,\frac65,…\).

  The set of upper bounds of \(M\) is therefore the interval \([2,\infty)\), from which \(\sup(M)=2\).

  The set of lower bounds of \(M\) is the interval \((-\infty,0]\), from which \(\inf(M)=0\).

• Answer

  \(\sup(M)=1\), \(\inf(M)=0\).

• Variant 10

  \(\displaystyle M=\left\{\frac{p}{p+q},p,q\in\mathbb N\right\}\)

• Solution

  For all natural numbers \(p,q\) we have \(p<p+q\), and so \(\frac{p}{p+q}<1\).
  For any \(a<1\), if we take \(q=1\) and \(p=\lceil \frac1{1-a} \rceil\) then \(\frac{p}{p+q}<a\) and so \(a\) cannot be an upper bound.
  The set of upper bounds of \(M\) is the interval \([1,\infty)\), from which \(\sup(M)=1\).

  Similarly, any ratio of positive numbers must also be positive.
  For any \(a>0\), if we take \(p=1\) and \(q=\lceil \frac1a \rceil\) then \(\frac{p}{p+q}>a\), and so \(a\) cannot be a lower bound.
  The set of lower bounds of \(M\) is the interval \((-\infty,0]\), from which \(\inf(M)=0\).

• Answer

  \(\sup(M)=1\), \(\inf(M)=0\).

• Variant 11

  \(\displaystyle M=\left\{\frac{n+(-1)^n}n,n\in\mathbb N\right\}\)

• Answer

  \(\sup(M)=\frac32\), \(\inf(M)=0\).

• Variant 12

  \(\displaystyle M=\left\{n^{(-1)^n},n\in\mathbb N\right\}\)

• Answer

  \(\sup(M)\) does not exist, \(\inf(M)=0\).

• Variant 13

  \( M=\{n^2-m^2: n,m\in \mathbb N\}\)

• Answer

  Neither \(\sup(M)\) nor \(\inf(M)\) exists.

• Variant 14

  \( M=\{n^2-m^2: n,m\in \mathbb N, n>m\}\)

• Answer

  \(\sup(M)\) does not exist, \(\inf(M)=3\).

• Variant 15

  \( M=\{n^2-m^2: n,m\in \mathbb N, n\le m\}\)

• Answer

  \(\sup(M)=0\), \(\inf(M)\) does not exist.

• Variant 16

  \( M=\{2^{-n}+3^{-n}: n\in \mathbb N\}\)

• Answer

  \(\sup(M)=\frac56\), \(\inf(M)=0\).

• Variant 17

  \( M=\{2^{-n}+3^{-n}: n\in \mathbb Z\}\)

• Answer

  \(\sup(M)\) does not exist, \(\inf(M)=0\).

• Variant 18

  \( M=\{5^{(-1)^j3^k}: j,k\in \mathbb Z\}\)

• Answer

  \(\sup(M)\) does not exist, \(\inf(M)=0\).

• Variant 19

  \(\displaystyle M=\left\{\cos\left(\frac{n+1}n\pi\right),n\in\mathbb N\right\}\)

• Answer

  \(\sup(M)=1\), \(\inf(M)=-1\).

• Variant 20

  \(\displaystyle M=\left\{\cos\left(\frac{n+1}n\pi\right),n\in\mathbb N, n \text{ sudé}\right\}\)

• Variant 21

  \(\displaystyle M=\left\{\cos\left(\frac{n+1}n\pi\right),n\in\mathbb N, n \text{ liché}\right\}\)

• Answer

  \(\sup(M)=1\), \(\inf(M)=-1\).