 So we've already filled up our box with a little bit of information, a little bit of definitions, axioms that we can use in proofs. What I want to talk to you about now is the product set. Product set. Imagine that I have two, I have two sets, a sub 1 and a sub 2. If I take the product set, I put that multiplication sign there, but don't be too concerned about that. We're not really talking about multiplication, because what we're going to do here is create pairs x and y. And this x belongs to A1 and y belongs to A2, because A sub 1 was first and A sub 2 was second. So we're going to create these pairs of elements such that x is an element of A sub 1 and y is an element of A sub 2. Now, does that make good sense? It's probably not. Let's use an example. I'm going to have my A sub 1, and that's going to equal a set 1, 2 and 3, and imagine A sub 2 equals the set 4 and 5. So what is A sub 1 and A sub 2? What is the product set of A sub 1 and A sub 2? Let's have a look at that. A sub 1 was first, so those are always going to be first in the pair, and 4 and 5 should always be second in the pair. So that is going to be the pairs 1 and 4, 1 and 4, and 1 and 5. And then you can see the pattern 2 and 4 and 2 and 5 and 3 and 4 and 3 and 5. And that is going to be my product set. Simple enough. Why would we be concerned about that? The reason why I'm showing you this is that we have to build from this to what we call mappings. And the way that I think about it is to go through an old-fashioned algebra function such as y equals x squared. Can you imagine what we did there? Well, we had one set of real values and we mapped it to another set of real values, or we at least created this product set because think about it. The y-axis was nothing other than a set of all real numbers. The x-axis was a set of all real numbers. So you give me one set of real numbers. That was my a sub 1, and you give me an a sub 2, and I can create these product sets. Because if you gave me 0, that would be 0, because 0 squared is 0. If you gave me 1, that would be 1. If you gave me 2, that would be 4. And there we go. Well, of course that should touch, but that would be 1 and 1. I'm creating these little pairs from these two sets of real numbers. Which brings me to one more thing. Of course the x elements will be all of this. So my x is my set of all real numbers and that I call my domain. y is also the set of all real numbers and that is my co-domain. But just think about it, we will never get one of these negative x values. So y is never going to be less than 0. So y is never going to be less than 0. And that makes y larger than or equal to 0 my range. So there is this difference between the co-domain and the range. Just for you to remember. So this is the product set, but what we have really done is a mapping. You map from one set to another set. You give me an element in the first set, in the domain, and I map it to an element in the co-domain. That's all we've ever done, that's all you did with algebra. Now there are three types of mapping. Injective, which we also call onto. We have surjective, 1 to 1 I should say. Surjective, which we call onto and then bijection. Now let's start with an injective function. Now I don't want to write down all of the real numbers. That would be impossible. So I'm going to have much smaller sets. My set A sub 1, A sub 2, A sub 3. A sub 1 in this first instance. It only has three elements. 1, 2, 3. A sub 2 has four elements. A, B, C, D. A sub 3 has four elements. P, Q, R, S. Now what is an injective mapping? An injective mapping said every element in the co-domain has at most one element in the domain mapping to it. Now there are other ways to say this as well. But I want you to remember the concept because you've got to use it again and again. So look at this. A only has one element and A sub 1 mapping to it. B only has one. C only has one. And D has nothing. So it says at most one. So one or nothing. One, one, one, nothing. One to one function. One to one mapping. An injective mapping. Let's look at A sub 2 to A sub 3. One mapping to it. One mapping to it. Two mapping to it. Cannot be one to one. What about mapping A sub 1 to A sub 3 through this A sub 2? So one goes to P. Two goes to Q. Three goes to R. So P has only one element mapping to it. Q only has one element mapping to it. R only has one element mapping to it. It has no element mapping to it. So A sub 1 through A sub 2 mapping to A sub 3 is one to one. Let's look at a surjective mapping or an onto mapping. That says every element in the co-domain the second one must have at least one mapping to it from elements in the domain. So A has one. B has one. C has two. D has nothing. That cannot be surjective or onto. Let's look at the mapping of A sub 2 to A sub 3. One, one, two. So every element in the co-domain has at least one element in the domain mapping to it. So certainly this A sub 2 maps to A sub 3 is a surjective or onto mapping. What about this not being surjective, this being surjective, is this to that surjective? So one goes to Q. Two goes to R. Three goes to S. Four goes to S. So every element here in my co-domain has at least one element in the domain mapping to it. So A sub 1 through A sub 2 to A sub 3 that mapping is surjective. Now what is a bijection then? This is the important one because we're going to see that it's only bijections as mappings that can have inverses. So bijection is both if it's one to one and onto. So let's look at one goes to A. Two goes to B. Three goes to C. Nothing goes to D. So this is one to one. A goes to P. B goes to Q. C goes to R. And D goes to R. Onto or surjective. Let's look at A to A3. One to P. Two to Q. Three to R. So one element has in this domain has one element that it maps to here. That is both. One to one and onto. So that would be a bijection.