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Sets: Finding elements or defining rules

Yes, it is that time of the semester when all of my classes are having exams.  It seemed that the last posts going over the practice exam problems for Calculus were helpful, so I will continue to make such posts around exam time.  For this series of posts, I will be working through the practice exam for the exam my introduction to proofs class will take later this week.

In this post, we will be looking at examples of determining the elements of a set if the rule for building the set is given.  Then, we will be looking at find the rule for building a set if the elements are given.

Finding elements in a set

Our first goal will be to take a given rule for sets and write out the elements.

Let \(A=\left\{x \in \mathbb{R} :x^{2}-4=0\right\}\), determine the elements of the \(A\).

When determining if a given element is in this set or not, we have to determine if it satisfies the given conditions.

For example, suppose that we wanted to know if \(1 \in A\).  We then check that, yes, \(1\) is a real number but, no, \(1^{1}-4=-3 \neq 0\).  Therefore, \(1 \notin A\).  As another example, suppose that we check \(\star\) is an element of \(A\).  Here, a \(\star\) is a big ball of burning gas, and is therefore not real number.  Therefore, \(\star \notin A\).

If we were to check every possible thing, we should be able to determine exactly what is and what is not in the set.  Since we are only interested in what is in the set, it is better to instead use the definition to see what would satisfy the given conditions.  In order to do this, we would solve \(x^{2}-4=0\) within the set of real numbers.  Therefore, \((x-2)(x+2)=0\).  Since the real numbers have no \(0\)-divisors, this would imply \(x= \pm 2\).  Both \(2\) and \(-2\) would be in the set, so we can write \[A=\left\{-2,2\right\}.\]

Determining rules for a set

Instead of being given rules for sets, we are often given a list of elements and asked to determine a defining rule for a set.  Let’s look at the example \(A=\left\{1,3,5,7,9,\ldots \right\}\).

There are indeed multiple ways to define a rule that will give you this set. Let’s look at 2 such ways.

The odd numbers

As we look at this, we may be tempted to say \(A=\left\{x: x \text{ is odd}\right\}\).  However, we need to be careful with this definition in that there is plenty of ambiguity.  That is, we should state what odd is.  So, let’s try to do that.

By \(x\) is odd, we mean that \(x\) is not even.  If \(x\) is even, we would think of this as \(x\) is divisible by 2.  When we say divisible by 2, we would mean that \(x\) is the product of 2 and some integer.  That is \(x=2k\) for some integer.  Therefore, we would say that \(A=\left\{x: x \neq 2k \text{ for any} k \in \mathbb{Z}\right\}\).

We should actually check this definition to make it gives the above set.  Note that \(1 \neq 2k \) for any integer \(k\), so we would get that \(1\) in our defined set.  As are \(3,5,7,9\).  However, since the dots go to the right, we would think that \(-1\) would be in the original set.  In the set we defined; however, we do get that \(-1 \neq 2k \) for any integer \(k\), so \(-1\) is in our defined set.  Furthermore, \(\star \neq 2k \) for any integer, so \(\star\) would also be in our defined set.  We, therefore, have to refine our definition of \(A\).

If we take a moment to think about the problem, we note that, what we want is to only have natural number in our set.  Therefore, we can fix our definition by stating \(A=\left\{x \in \mathbb{N}: x \neq 2k \text{ for any } k \in \mathbb{Z}\right\}\).  If we now check this definition, we note that we will only get the positive odd numbers.

Using an equality

Instead of seeing that we have only odd numbers here, we could also just start writing out the terms in order to determine an equational relation.  In doing so, we would have
\begin{align*}a_{1}&=1, \\
a_{2}&=3, \\
a_{3}&=5 \end{align*}
and so on.

Now, how do we get from \(1 \) to \(3\).  We do indeed have a few options.  We could add 2, multiply by 3, square and add 2, take \(3^{1}\) or other number of options.  As we make a list here, we want to check the pattern to see if will continue.  For this, we note that \(3+2=5 =a_{3} \), \(3*3=9 \neq 5\), \(3^{2}+2 =11 \neq 5\), and \(3^{3}=27 \neq 5\).  If you came up with other options, we would again check these against the second term.

Since the only one of our options that worked was adding 2, we want to check this against the fourth term.  Now \(5+2=7\) and \(7+2=9\), so it appears that the pattern continues to work.  Therefore, we would say that \(a_{1}=1\), \(a_{2}=1+2\), \(a_{3}=1+2+2\) and in general \[a_{n}=1+2+2 + \ldots +2,\] where there are \(n-1\) twos being added to 1.  Now, since \(2+2 +\ldots +2\) with \(n-1\) twos is just another way of writing \(2(n-1)\), we could instead write that \(a_{n}=2(n-1)+1\) or \(a_{n}=2n-1\) if we want to simplify.  We can now say that \[A=\left\{2n-1\right\}.\]

Or can we?  Well, if we read this as written, this would say that \(2n-1\in A\).  However, since we don’t say anything about \(n\), this would mean precisely that “2n-1” is the only element of \(A\).  Therefore, \(1 \notin A\), \(3 \notin A\) only “2n-1.”  We therefore have to let \(n\) be variable by stating what values \(n\) can take.  Since we want \(n\) to be all the positive whole numbers, we would write\[A=\left\{2n-1: n \in \mathbb{N}.\right\}\]

Final Note

As a final note, we generally assume that the dots imply that the pattern will continue on to the right.  Therefore, we can work with the formula given above as our general definition.  On the other hand, there may be some formula that does not agree with the formula we found.  If it agreed with the given values, there is no way for us to determine which one is accurate.

This is main difference between inductive and deductive reasoning, and we are limited by the information shown when we are not given a rule for determining what is in the set.  As an example of this, suppose that we were given \(A=\left\{1,3,5,7, \ldots\right\}\).  Then, the above definition would still work, but so would \(A=\left\{2n-1:n \in \mathbb{Z}_{12}\right\}\), that is the integers mod 12. (For definition of mod see Modular Arithmetic: 2+2=1!).


I hope this post helped you spend more time thinking about sets and definition thereof.  Even though some of this work may seem easy at first glance, it is important to be precise about the definitions you give.  Forgetting even a small part of a definition can lead to drastically different looking sets.

If you found the post helpful or entertaining, please like it below and share it with anyone else that may be looking at set theory or logic.  You can also find the rest of the solutions to this practice exam here.



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