Vishtasb

Two at a time. That's my mean.

Sorry but It's not my fault.😓 Ask the google translator.

Let me ask the question in other words. Assume a persion who wants to climbe n stairs. But this persion can only climb stairs one by one or decussate. There is how many ways that this guy can climb these stairs?

Sorry because of the miss understandig. I meant that to climb stairs one by one or decussate.

If we can climb one stair, one stair, or two stairs, two stairs, in what ways can we climb n stairs?

But this method is about countable sets. Existence of countable sets can be proved by using infinity axiom. Because they're a kind of inductive sets (or actually can be isomorphism by inductive sets). But My real question is about uncountable sets. As I know there are two ways to make a new mathematical object. One method is 'constructive method' and the other is 'axiomatic approach'. For example in constructive method we defone 0 . Which is the empty set. And any other number is set of all before numbers. But in axiomatic approach we assume that the set of integers 'exist'. So if we use the constructive method we can not just say that the set of all real numbers exist. But we can construct it by using some ways such as ordered fileds or Dedkind? Cuts.

infinity is not a number or even a value! and i think there is no one can explain what "infinity" is. we just know how it behave.

In the set theory we have inductive sets which are those sets like \( A \) which \( \emptyset \in A\) and \( \forall a \in A (a^{+} \in A. \) and \( a^{+}= a \cup \{a\} \). Then we have the infinity axiom which says there exists an inductive set: \[ (\exist A) [\emptyset \in A \and (\forall a \in A) a^{+} \in A]. \] Now my question is how do we can conclude than infinity sets, specially those sets that are uncountable such as \( \mathbb{R} \) exists?