Before all, I'm apologize if my question is too common here. I'm only want to know if my proof is correct or need some adjustments.

I'm reading the Terence Tao's Analysis book as I mentioned in my other question. The problem in the book says:


*My approach for the first part, was:

Let I be the set of all the natural numbers between 1 and n: $$ I:= \left \{ i \in N : 1 \leq i\leq n \right \} $$

(=>) So, if we assume that the both ordered n-tuple are equal: $$ \left (x_{i} \right)_{i\in I} = \left (y_{i} \right)_{i\in I} $$

As the two functions are equal (by hypothesis). So for each i in I, we have that both are equal (using the definition of equality of function given in previous chapter by the author). Therefore: $$ \left (\forall i \in I \right) \left ( x \left ( i \right)= y \left ( i \right) \right) $$

(<=) On the other hand, if we assume that: $$ \left (\forall i \in I \right) \left ( x_{i}= y_{i} \right) $$ As both function have the same domain and range that means both are equal: $$ \left (x_{i} \right)_{i\in I} = \left (y_{i} \right)_{i\in I} $$

*For the second part, I have some troubles but here is my approach:

First we write the Cartesian product as the author do:

$$ \prod_{i\in I} X_{i} := \left \{ \left (x_{i} \right )_{i \in I}:\left(\forall i \in I \right) \left (x_{i} \in X_{i} \right) \right \} $$

I form the set of all the functions from the set of natural numbers called I to a target set X, which I defined in the next way: $$ X^{I} \text{ where } X:= \bigcup_{i \in I} X_{i} $$

Then, using the axiom schema of specification we can form the set:

$$ g\in \left \{ f\in X^{I} : \left (\forall i \in I \right) \left (f \left ( i \right) \in X_{i} \right) \right \} \leftrightarrow \\ \left(\text{$g: I \rightarrow X$ is a mapping and } \left (\left (\forall i \in I \right) \left (f \left ( i \right) \in X_{i} \right) \right ) \right ) $$

So, using the definition given by the author only we need to show that g is onto which means that define an ordered n-tuple (right?).

$$ x \in X \leftrightarrow x \in \bigcup_{i \in I} X_{i}\leftrightarrow \left (\exists i \in I \right )\left ( x \in X_{i} \right ) $$

g is surjective because for each x in the target set always we can find a i in the domain. Then, g is an ordered n-tuple. And that means g lies in collection of all n-tuples defined at the first place.

On the other hand:

$$ x\in \prod_{i\in I} X_{i}\leftrightarrow \left (x = \left (x_{i} \right )_{i \in I} \right )\text{ and } \left(\forall i \in I \right) \left (x_{i} \in X_{i} \right) $$

So, that means $x$ is an n-tuple and by hypothesis, $x$ is a surjective function from the set $I$ to $X$. Then, $x$ lies in set formed above.

Hence, the Cartesian product is a set.

(I'm sorry for my grammatical mistakes, English is not my first language).

  • $\begingroup$ So, that's OK the prove? $\endgroup$ – Jose Antonio Jul 4 '13 at 20:12

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