Science:Math Exam Resources/Courses/MATH220/December 2010/Question 05 (a)

MATH220 December 2010

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Question 05 (a)
Prove that the set $S=\{x\in \mathbb {R} \quad {\textrm {s.t.}}\quad \sin(x)=0\}$ is countable.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
A set is countable if it can be written in bijection with a subset of the natural numbers, or any other countable set. Can you rewrite the given set and set up a bijection with an already known countable set?

Hint 2
Recall that the integers is a countable set. Setting up a bijection between S and the set of integers is your best bet.

Checking a solution serves two purposes: helping you if, after having used all the hints, 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. Firstly, we examine the set $S=\{x\in \mathbb {R} \quad {\textrm {s.t.}}\quad \sin(x)=0\}$ We claim that $\displaystyle S=T$ where $T=\{n\pi :n\in \mathbb {Z} \}$. The inclusion that $\displaystyle T\subseteq S$ is easy from trigonometry, we know that $\displaystyle \sin(n\pi )=0$. Conversely, notice that $\displaystyle \sin(x)$ is a $\displaystyle 2\pi$ periodic function. So solving $\displaystyle \sin(x)=0$ requires only that we solve this between 0 and $\displaystyle 2\pi$ and then use periodicity. We know that on the interval $\displaystyle [0,2\pi )$, we get only two solution at $\displaystyle 0,\pi$ and so taking $\displaystyle 2\pi$ multiples of this, we have that $\displaystyle 0+2\pi n,\pi +2\pi n$ for $n$ an integer give all the solutions. This is precisely the elements of $T$ and thus we have shown that $S$ equals $T$. Now the mapping from $S=T$ to $\mathbb {Z}$ is trivial. We can map ${\begin{aligned}T&\to \mathbb {Z} \\x&\mapsto x/\pi \end{aligned}}$ and in the opposite direction, ${\begin{aligned}\mathbb {Z} &\to T\\n&\mapsto n\pi \end{aligned}}$ These maps are injective, surjective and thus are bijective. Thus, $\displaystyle S=T$ is countable since it is in bijection with a countable set, namely $\displaystyle \mathbb {Z}$. Aside: A quick example showing that $\displaystyle \mathbb {Z}$ injects into $\displaystyle \mathbb {N}$ is given by $f:\mathbb {Z} \to \mathbb {N}$, $f(k)=2k+1$ if $k\geq 0$ and $f(k)=-2k$ if $k<0$. That is, $f$ maps $0,1,2,3,\dots$ to $1,3,5,7,\dots$ and $f$ maps $-1,-2,-3,-4,\dots$ to $2,4,6,8,\dots$.

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