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 look at graphically displaying sets, their union, intersections and complements.

**Definitions**

If we let \(A\) and \(B\) be sets, then we define \[A \cup B=\left\{x: x \in A \text{ or } x \in B\right\}.\] Before we move on from the definition, I want to remind you of exactly what this means. Here, the notation is just a way to write an English sentence using shorthand, so let’s read the sentence. We have, \(A\) union \(B\) is the set consisting of all elements \(x\) such that \(x\) is an element of \(A\) or an element of \(B\).

Now, we do need to be careful with that as well. By x is an element of A <strong>or</strong> an element of B, I mean that x is in <strong>at least</strong> one of the two sets. That is, if \(x\) is in both sets, it will still be in the union. I want to point this out, because or is sometimes used to colloquially to mean exclusively \(A\) or \(B\), so not both.

We similarly define \[A \cap B=\left\{x: x \in A \text{ and } x \in B\right\},\] That is, \(A\) intersect \(B\) is the set of all elements \(x\) such that \(x\) is in both \(A\) and \(B\).

By the \(\overline{A}\), we mean the complement of \(A\). We formally define this as \(\overline{A}=\left\{x: x \notin A\right\}\). A given, the complement can be quite large. For example \(\overline{\left\{2\right\}}\) is the set consisting of everything but the number \(2\). That is, \(1,3, 1.5, \pi,\) Apple \(\pi\), \(\star\), horses or just about anything you can think of.

In most situations, we define a universal set, \(U\) which contains all elements of interests. We then write \(\overline{A}=\left\{x \in U: x \notin A \right\}\) in order to limit the amount of elements we have to consider. In a similar way, we can also denote this as \(U-A\), that is \(U\) not \(A\). When using this notation; however, there is no need to have the first set contain the second. Therefore, we would say \(B-A=\left\{x \in B: x \notin A\right\}.\)

## Venn Diagrams

### Let \(A=\left\{1,2,4,5,7\right\}\), \(B=\left\{2,5,6,7,8\right\}\), \(C=\left\{1,5,7,8\right\}\). and \(U=\left\{x \in \mathbb{N}: x \leq 10\right\}\). Then draw a venn-diagram for \(A,B\) and \(C\) putting each of the elements of \(U\) in the appropriate region of the venn-diagram.

Here we are going to find any combinations of sets \(A,B,C\) as unions, intersections or differences by denoting these graphically. The way we normally do this is by using a Venn diagram. The way we use this, is that each set will be represented by some convex shape (usually a circle, but rectangles may also be used). Therefore, we would denote the set \(A\) as

We are using colors to show that the set \(A\) would consist of anything in the red region. Therefore, if we use the above example, we will have

I do need to note; however, that \(A\) is not an element of itself. Rather, the \(A\) is just a label. The numbers within the red circle are the elements of \(A\).

**2 sets at a time**

We can then generalize this idea by including another set in the picture. Therefore, in order to include \(B\), we will draw another circle, with the inside being blue, so that everything inside that circle is in \(B\). In general, the sets \(A\) and \(B\) may have some overlap, so we draw these circles so that there is a portion that is overlapping and a portion that is not. The overlapping portion will denote our intersection.

Here, we have that all the elements in \(A\) are indeed in the red circle, and all the elements of \(B\) are in the blue circle. However, we want to be a little more precise with this. In particular, the purple region is both red and blue. Therefore, this is where any elements that are in both \(A\) and \(B\) should appear. That is, the purple region is \(A \cap B\). Now, in order to change our picture, we will move any elements that are in both from the outer regions into this region.

Note that, while we can repeat elements and have the same set, we don’t want to leave the \(2,5\) and \(7\) in the outer portions. The reason being is that portion that is red but not purple is the now the set \(A -B\). That is, by looking at the visual we find that \(A-B=\left\{1,4\right\}\), \(B-A=\left\{6,8\right\}\) and \(A \cap B=\left\{2,5,7\right\}\).

**3 sets at a time**

If we add a third arbitrary set, \(C\), we want to do so in a similar manner as we did before. However, there is a little more we need to pay attention to. Instead of just intersecting or not intersection the previous sets, we can have \(C\) intersecting both \(A\) and \(B\), \(A\) but not \(B\), \(B\) but not \(A\) or neither \(A\) nor \(B\). Therefore, we get the following picture.

Now, because we want to know what is in each of the intersections, we will move the numbers so that they appear in the appropriate region. That is, anything that is in all three, (\(5\) and \(7\)), will appear in the white area. This is then \(A \cap B \cap C\). If the element is in \(A\) and \(B\) but not \(C\), (\(2\)), we will put it in the purple region. This is then \(A \cap B \cap \overline{C}\). Continuing this process, we will get the following picture.

**3 sets plus a universal set**

The last thing we need to do is insert \(U\) into the picture. When drawing \(U\); however, we want \(U\) to contain all the other sets since it is the universal set. Therefore, when we draw it, its region should surround the circles for \(A\), \(B\) and \(C\). Here, we denote not being in any set as gray instead of black because it is easier to see the labels this way.

Now, the last thing to fill in would be the rest of \(U\). Since we have already drawn \(A\), \(B\) and \(C\), we now have to find all elements that are in the complement of \(A\), the complement of \(B\) and the complement of \(C\). That is, we find \(\overline{A } \cap \overline{B} \cap \overline{C}=\left\{3,9,10\right\}.\) Graphically, we then have

**Conclusion**

Now that we have are Venn diagram drawn, we can quickly find the sets we may be interested in. If, for example, we want \(A \cup B\), we would find all the elements in a region with the red or the blue property, thus, \(A \cup B=\left\{1,2,4,5,6,7,8\right\}\).

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.