Science:Math Exam Resources/Courses/MATH220/December 2009/Question 09 (a)

MATH220 December 2009

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Question 09 (a)

Find the supremum, maximum, infimum, and minimum of the following subsets of $\mathbb {R}$ (if they exist). If they do not exist write "none". You do not need to justify your answers. set supremum maximum infimum minimum $\{x\in \mathbb {R} \quad {\textrm {s.t.}}\quad {}7<x\leq 10\}$ $\{x\in \mathbb {Q} \quad {\textrm {s.t.}}\quad {}x^{2}<3\}$ $\{x\in \mathbb {Z} \quad {\textrm {s.t.}}\quad {}x^{2}<3\}$ $\bigcap _{n=1}^{\infty }\left(1-{\frac {1}{n}},4+{\frac {1}{\sqrt {n}}}\right)$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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
Remember that max and mins (if they exist) are contained inside the set whereas supremums and infimums are always real numbers (in this course) and may or may not be inside the set.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. $\{x\in \mathbb {R} \quad {\textrm {s.t.}}\quad {}7<x\leq 10\}$ Supremum: 10. All elements in this set are less than 10 and this is the least upper bound of this set as anything slightly smaller than (or equal to) 10 is inside the set. Maximum: 10. The supremum is in the set and thus it is the maximum. Infimum: 7. All elements in this set are at least as large as (or equal to) 7 and this is the greatest lower bound as anything less than 7 is strictly not in the set. Minimum: None. The infimum is not in the set and thus there is no minimum. $\{x\in \mathbb {Q} \quad {\textrm {s.t.}}\quad {}x^{2}<3\}$ Supremum: $\displaystyle {\sqrt {3}}$. Any positive real number smaller than this value satisfies $x^{2}<3$ and any real number larger than or equal to this element has a square larger than 3. Maximum: None. Since the supremum is irrational, it is not contained in this set and so this set has no maximum. Infimum: $\displaystyle -{\sqrt {3}}$. Any negative real number larger than this element satisfies $x^{2}<3$ and any real number smaller than or equal to this element has a square larger than 3. Minimum: None. Since the infimum is irrational, it is not contained in this set and so this set has no minimum. $\{x\in \mathbb {Z} \quad {\textrm {s.t.}}\quad {}x^{2}<3\}$ Supremum: $\displaystyle {\sqrt {3}}$. Any positive real number smaller than this element satisfies $x^{2}<3$ and any real number larger than or equal to this element has a square larger than 3. Maximum: None. Since the supremum is irrational, it is not contained in this set and so this set has no maximum. Infimum: $\displaystyle -{\sqrt {3}}$. Any negative real number larger than this element satisfies $x^{2}<3$ and any real number smaller than or equal to this element has a square larger than 3. Minimum: None. Since the infimum is irrational, it is not contained in this set and so this set has no minimum. $\bigcap _{n=1}^{\infty }\left(1-{\frac {1}{n}},4+{\frac {1}{\sqrt {n}}}\right)$ Supremum: 4. This element is inside every interval. If there were an element c larger than 4 inside the intersection, then take n so large such that $\displaystyle 4+{\frac {1}{\sqrt {n}}}<c$. This would mean that c is not in the interval $\left(1-{\frac {1}{n}},4+{\frac {1}{\sqrt {n}}}\right)$ and hence not in the intersection of all the given intervals. Maximum: 4. Even though the interval is open, the element 4 is inside every interval and hence in the intersection. Infimum: 1. This element is inside every interval. If there were an element c smaller than 1 inside the set, then there take n so large such that $\displaystyle 1-{\frac {1}{n}}>c$. This would mean that c is not in the interval $\left(1-{\frac {1}{n}},4+{\frac {1}{\sqrt {n}}}\right)$ and hence not in the intersection of all the given intervals. Minimum: 1. Even though the interval is open, the element 1 is inside every interval and hence in the intersection.

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