There is still a question in the air: why **B** does not belong to **(a, b)**? We had stated that this was a demonstration that **( a, b) ≠ {a, b}. **Similarly, we could say that

**does not belong to**

*The***(**. But to see this very clearly, we must write:

*a, b*)**(**and then conclude that both

*a, b*) = {{*The*}, {*a, b*}}**how**

*The***do not belong to**

*B***(**In the case of the set

*a, b*).**It seems easy to conclude. If we go to the last consequences, we find that this is not so simple, and we face the problem of the existence of sets with "infinite" parentheses. That**

*The***"aberration"**comes from the hypothesis that

**“A set belongs to itself”**. So reasoning with the whole

**, we soon come to the idea of a set belonging to itself, and therefore to the idea of a set with infinite parentheses. But what if we tried to see clearly why**

*The***not belong to**

*B***(**?

*a, b*)Thinking of the expression **( a, b) = {{The}, {a, b}}**,we reason that

**must be equal to**

*B***{**or

*The*}**must be equal to**

*B***{**. We seem to be in trouble: how to produce an aberration from these two ideas? There seems to be no problem in the idea that

*a, b*}**butthe idea that**

*b =*{*The*}**leads us to the idea that**

*b =*{*a, b*}**belongs to**

*B***.That is, the second idea is what we want, but we still have to deal with the possibility that**

*B***. What to do?**

*b =*{*The*}Let's take a quick look: We can demonstrate that an aberration appears if we admit that ** The** belongs to the ordered pair

**(**. We would like to know if an aberration also appears by reasoning that

*a, b*)**belongs to the ordered pair**

*B***(**. But the situation does not seem to be

*a, b*)**“Symmetrical”**. Now it would be very strange to be able to demonstrate that

**(**just thinking with the set

*a, b*) ≠ {*a, b*}**. Although in the definition of ordered pair there is a**

*The***"asymmetry"**natural as you can easily see from the expression that defines the ordered pair. Well, we can't surprise that fact since

**“Ordered pair”**just means that the pair is ordered, that is, there is an order between their sets

**and**

*The***. If the pair is ordered, then of course there is no problem if it contains an asymmetry.**

*B*But, in terms of demonstration, we are bothered by the idea that to demonstrate our thesis we necessarily have to reason only with the whole ** The**. The way out of this discomfort lies in the symmetry of a mathematical concept. In general, a mathematical concept contains an intrinsic symmetry, that is, internally, in its own definition. For example, when we set the number

**1**there is no way to do it differently than

**2**in terms of concept. It would be strange, and even intolerable, that in defining the number

**2**different conceptual properties would appear than those required for the number

**1**. Yes, they are different numbers, but both are equally numbers and therefore conceptually as numbers have to be

**“Symmetrical”**. We will return to this question of number definition soon, as soon as the development of our history of set theory permits. We could already define them, for example,

**zero is the empty set**, but we still don't know if there is any set, including the empty set.

It's time for us to solve our puzzle today: to demonstrate that **( a, b) ≠ {a, b}**using the set

**, just consider the symmetry**

*B***{**. It is a very simple symmetry, but it solves our puzzle. Through this symmetry we can reason the following: To demonstrate that the ordered pair is different from the pair, we assume not. Then we have:

*a, b*} = {*b, a*}**{**, since we are assuming that the ordered pair is equal to the even set. Now just repeat with

*b, a*}={{*B*}, {*b, a*}}**the reasoning we use with the set**

*B***previously.**

*The*Back to columns

<