Collection of Maths Problems

Set operations

Task number: 2825

• Variant 1

  \(A\cup B\)

• Resolution

  Any upper bound of \(A\cup B\) must be an upper bound of both \(A\) and \(B\). Therefore we can find the upper bounds of \(A\cup B\) by intersecting the upper bounds of \(A\) with the upper bounds of \(B\). Because these two sets are the intervals \([\sup A,\infty)\) and \([\sup B,\infty)\), their intersection is the interval \([\max\{\sup A, \sup B\},\infty)\).

  We can find the lower bounds analogously.

• Result

  \(\sup(A\cup B)=\max\{\sup A, \sup B\}\),\ \(\inf(A\cup B)=\min\{\inf A, \inf B\}\).

• Variant 2

  \(A\cap B\), under the assumption that this intersection is non-empty.

• Resolution

  Every upper bound of \(A\) or \(B\) is also an upper bound of \(A\cap B\). We can find the upper bounds of \(A\cap B\) by taking the union of the upper bounds of \(A\) with the upper bounds of \(B\). Since these two sets are the intervals \([\sup A,\infty)\) a \([\sup B,\infty)\), their union is the interval \([\min\{\sup A, \sup B\},\infty)\).

  The argument for lower bounds is analogous.

• Result

  \(\sup(A\cap B)=\min\{\sup A, \sup B\}\),\ \(\inf(A\cap B)=\max\{\inf A, \inf B\}\).

• Variant 3

  \(A\setminus B\), under the assumption that this difference is non-empty.

• Resolution

  The only relationship we can deduce is that the upper bounds of \(A\) are also upper bounds of subsets of \(A\), and therefore of \(A\setminus B\).

  By an appropriate choice of \(B\) we can make the supremum be substantially different from \(\sup A\), as long as the extent of \(A\) is sufficiently large. Conversely, if we add elements outside \(A\) to \(B\), we can influence the value of \(\sup B\) and \(\inf B\), without having any effect on the supremum of the set difference.

  The argument works analogously for lower bounds.

• Result

  \(\sup (A\setminus B) \le \sup A\), \(\inf (A\setminus B) \ge \inf A\).

• Variant 4

  \(\ominus A\) where the operation \(\ominus\) is defined as \(\ominus A=\{-a: a\in A\}\)

• Resolution

  \(a\) is an upper bound of \(A\) exactly when \(-a\) is a lower bound of \(\ominus A\). From here the conclusion is straightforward.

  The argument for lower bounds is analogous.

• Result

  \(\sup(\ominus A)=-\inf A\),\ \(\inf(\ominus A)=-\sup A\).

• Variant 5

  \(A\oplus B\) where the operation \(\oplus\) is defined as \(A\oplus B=\{a+b: a\in A,\ b\in B\}\)

• Resolution

  For every \(m\in A\oplus B\) there exist \(a\in A\), \(b\in B\), such that \(m=a+b\). Furthermore \(a\le \sup A\), \(b\le \sup B\), so \(\sup A+\sup B\) is an upper bound of \(A\oplus B\).

  For \(s<\sup A+\sup B\) we compute \(\varepsilon=\frac12(\sup A+\sup B-s)\). Now there must exist \(a\in A\cap(\sup A-\varepsilon,\sup A]\) and \(b\in B\cap(\sup B-\varepsilon,\sup B]\) for which \(a+b>s\), and so \(s\) cannot be an upper bound of \(A\oplus B\).

  The argument for lower bounds is analogous.

• Result

  \(\sup(A\oplus B)=\sup A + \sup B\),\ \(\inf(A\oplus B)=\inf A + \inf B\).