We already know how to proliferate the objects of the world of sets. If there is a set C, then there is a pair {C, C} and therefore there is the set {C} and therefore there is the set {C, {C}} and therefore there is the set… So already We are able to obtain “a multitude of sets” by proliferating new sets from just one set.

Now let's proliferate new sets "inside" a given set. That is, let's “pop up” new sets from a given set, but inside it. It is Axiom 6, the axiom of the parts of a set that allows us to increase the population of objects in the mathematical universe that we have so far.

It is an intuitive notion **the notion of the parts of a set**. But why can we “**think about them**”? It is precisely the **Axiom 6** which allows us to suppose that the parts of a set are legitimate sets for our thinking.

**Axiom 6**

For each set C, there is a set P (C) such that if A is contained in C, then A belongs to P (C).

For example, what are the parts of the set C = {0, 1, 2, 3}? Since the empty set Æ is contained in any set, so it is contained in C. Therefore, the empty set Æ is a set that belongs to the parts of C, that is, Æ belongs to the set P (C). Well, what are all the other sets that belong to the parts of C? Let us first enumerate all those that have only one set: {0}, {1}, {2}, {3}. Now let's enumerate all those sets that have two sets: {0, 1}, {0, 2}, {0, 3}, {1, 2}, {1, 3}, {2, 3}. Now let's enumerate all those sets that have three sets: {0, 1, 2}, {0, 2, 3}, {1, 2, 3}, {0, 2, 3}. Finally, let's enumerate all those sets that have four sets: {0, 1, 2, 3}.

Did you notice that we use the notion of "one", "two", "three", "four" to solve the problem of finding all parts of set C? Let's admit, for now, that we already know what these "entities" are. We will soon see that these entities are nothing more than natural numbers, that is, the first numbers that emerge "naturally" early in our investigation of a set theory. On the other hand, you were able to fully understand our “reasoning” to obtain the set of parts of set C.

We obtained “16” parts for the set C = {0, 1, 2, 3}. If our set C were set {0, 1, 2, 3, 4} we would get "32" parts. Would you know why this? Let's leave this challenge for you to solve until the next column: * when a set has n sets then its set of parts has the power * 2

^{no}, that is, the number of sets of P (C) is “two raised to

*no*" In our example, there will be “two to fifth” parts in the set {0, 1, 2, 3, 4}. It was by calculating this "potency" that we became more comfortable with the result obtained in the case of the sixteen parts because we became convinced that we had not forgotten any part.

It is interesting to note that this power “law” for testing whether the number of parts obtained is correct is also true for the extreme case of the empty set. We ask: how many and what are the parts of the empty set? Answer: two raised to 0! But, on the other hand, we easily know that the empty set has only one part, namely the empty part. That is why we say that 'two raised to zero is one'. But this brings us an intriguing problem: how much is the power “*no* raised to zero ”?

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