The set of all natural numbers, , is a countably infinite set. In fact, . Now, let G be the set of all even natural numbers. Thus, G = {2, 4, 6, 8, 10, 12, …}. Note that G is only part of . Mathematically speaking, G ⊂ . The question now is: “Which has more elements, G or ?” At first glance, G has less elements than . Is it correct? Not necessarily! We shall look at these sets in another way such that we arrive at a conclusion that the number of elements of G is equal to the number of elements of . How is it possible that the whole has as many elements as G, which is only part of ?

The question is about the equivalence of two sets in set theory. Here is the definition of the equivalence of two sets. **“The set A is said to be equivalent to the set B, denoted by A ~ B, if there is a one-to-one correspondence between the sets A and B.”**

As an example of equivalent sets, let K = {1, 2, 3, 4, 5} and L = {8, 9, 10, 11, 12}. Since K and L have a finite number of elements, we can easily determine the number of elements of each set. It turns out that they have an equal number of elements, i.e. 5. By the definition of * cardinality* of finite sets, we can conclude that K and L have an equal

*cardinality*, that is 5. A theorem in set theory states that if A and B are finite sets then: “

**A and B have an equal cardinality if and only if A and B can be put into a one-to-one correspondence**.” As a direct consequence of the theorem, we can conclude that there is a one-to-one correspondence between K and L. To construct the one-to-one correspondence (or bijection), let f be a function from K to L defined by f(x) = x + 7 for every x ∈ K. (It can be easily proved that f is a bijective function.) Thus, f = {(1,8), ( 2,9), (3,10), (4,11), (5,12)}. See Figure 1 below.

**Figure 1**

Furthermore, by the definition of equivalence of two sets, it follows that K is equivalent to L.

The “message” here is …

**“A is equivalent to B” ⇔ “A and B have an equal number of elements” **………………………………………………….. (*)

** on condition that A and B are finite sets**.

What would happen if we removed the condition above and adopted (*) as the meaning of “A and B have an equal number of elements.”? If we did so, then G and would have an equal number of elements. To prove this, define a bijection h from to G by h(x) = 2x for every x ∈ . Thus, h = {(1,2), (2,4), (3,6), (4,8), …}. See Figure 2 below.

By defining h as above, we have proved that there is a one-to-one correspondence between and G. Referring to the definition of the equivalence of two sets above, we conclude that G is equivalent to . Moreover, by the “new” definition (*) we can deduce that G and have an equal number of elements.

Similarly, by the “new” definition, the intervals [0,1] and [0,2] have an equal number of elements despite the fact that [0,1] is a proper subset of [0,2]. By defining a bijection t from [0,1] to [0,2] by t(x) = 2x for every x ∈ [0,1], it can be inferred that [0,1] and [0,2] have an equal number of elements.