Science:Math Exam Resources/Courses/MATH220/April 2011/Question 02

MATH220 April 2011

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Question 02

For each of the following subsets of $\mathbb {R}$ write its supremum and infimum if they exist. If they do not exist write ``None``. You do not need to prove your answers. $\left\{x\in \mathbb {R} {\text{ s.t. }}-1<x<5\right\},$ $\left\{x\in \mathbb {Q} {\text{ s.t. }}3\leq x^{2}\leq 7\right\},$ $\bigcap _{n=1}^{\infty }\left[2+{\frac {1}{n}},6-{\frac {2}{n}}\right],$ $\bigcup _{n=1}^{\infty }\left[{\frac {1}{n}},n\right].$

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The supremum of a set of real numbers is the smallest real number that is larger or equal than any number of the set. Conversely, the infimum of a set is the largest real number that is smaller or equal than any number of the set.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Solution to 1. The supremum of this set is 5, since it is clearly larger than any number in (-1,5) and no number smaller than 5 could have that property. (If you're not convinced of this, imagine any number smaller than 5, and call it N then there must exist a number between N and 5 so that N cannot be the supremum). And for a similar reason, the infimum of this set has to be -1. Solution to 2. Here the set that we are looking at is all the rational numbers that are between the square root of 3 and the square root of 7. The supremum and infimum of a set do not have to be members of the set, so our intuition isn't bad: the supremum has to be the square root of 7 and the infimum the square root of 3. Solution to 3. This set is the intersection of closed intervals that grow. At $n=1$ it is the set [3,4] and then [2.5, 5], so we have Failed to parse (syntax error): {\displaystyle \displaystyle [3,4] \cap [2.5,5] \cap [2+\frac{1}{3},6-\frac{2}{3}] \cap [2+\frac{1}{4}, 6-\frac{2}{4}] \cap \ldots = [3,4] } since because we're taking an intersection of larger and larger interval, only the smaller remains. It is now clear that the supremum of that set is 4 and the infimum 3. Solution to 4. Here the sets also are growing in size, but this time we are taking a union. Since 1/n converges to 0, the left side of the interval converges to 0 and clearly the right side of the interval converges to infinity. This means that $\displaystyle \bigcup _{n=1}^{\infty }[{\frac {1}{n}},n]=(0,\infty )$ And so the infimum of this set is 0, while there isn't a supremum since the set does not have an upper bound.

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