I am having problems understanding this mark: I have this set defined as follows:
$$A_n=\{(i,j)\in Z^2: (i,j)=1, 0\le i,j \le n\} $$ 
What does  (i,j)=1 means?
Thanks in advance
Now that we know it means gcd(i,j)=1, How can I calculate the size of this set?
 A: As a guess I would say that it means that $gcd(x,y)=1$ (the gcd part often gets dropped as in Apostol "Introduction to Analytic Number theory"), but it is not possible to give a proper answer without more context.
Was this in a book or in some course notes? What are you studying in this case?
EDIT: Hence this would be the set of coprime pairs in $Z^2$, as others have commented.
EDIT 2: We need to sum the number of coprimes to each  $i,j \le n$. This is the totient summation, $\Phi(n)$.
A: Let $\displaystyle A_N=\{(i,j)\in{\mathbb{Z}}^2: \gcd(i,j)=1, \ 0\leq i,j\leq N \}$. Prove the existence of 
$$\lim_{N\to\infty}\frac{|A_N|}{N^2}$$
and compute that limit.
Perhaps we start with reasoning the convergence. We do that by taking this question to the probability area; asking ourselves what $\frac{|A_N|}{N^2}$ actually means. If we fixing $N=K$, then $\frac{|A_K|}{K^2}$ is the probability that two integers $i$ and $j$, where $0 \leq i,j\leq K$, no common factor. Therefore, 
$$\lim_{N\to\infty}\frac{|A_N|}{N^2}$$
answers the question: For two random integers $i$ and $j$, what is the probability that they have no common factor?
Now, we want to compute that limit. Here we have different approaches solving the problem:
Approch $1$: With probability
Two $i$ and $j$ integer are coprime if there is no prime number $p$ divides both $i$ and $j$. For a particular prime $p$, the probability that $i$ is divisible by $p$ is $\frac{1}{p}$, and the same for $j$, so the probability that both are divisible by $p$ is $\frac{1}{p^2}$, and so the probability that $p$ does not divide both of them is $1-\frac{1}{p^2}$. The probability that no prime divides both $i$ and $j$ is therefore the Euler's product($s=2$):
$$\prod_{p\in\mathbb{P}}(1-\frac{1}{p^2})$$
Now, with the help of famous identity,
$$\zeta(2)=\sum_{s=1}^{\infty}\frac{1}{n^2}=\prod_{p\in\mathbb{P}}\frac{1}{(1-\frac{1}{p^2})}$$
And the fact that,
$$\zeta(2)=\frac{\pi^2}{6}$$
After combining these all, we will get,
$$\prod_{p\in\mathbb{P}}(1-\frac{1}{p^2})=\frac{6}{\pi^2}$$
Thus,
$$\lim_{N\to\infty}\frac{|A_N|}{N^2}=\prod_{p\in\mathbb{P}}(1-\frac{1}{p^2})=\frac{6}{\pi^2}$$
Approch $2$: Number theoretical:
Let us define three new sets:
$$P_N=\{(i,j)\in{\mathbb{Z}}^2: \gcd(i,j)=1, \ 0\leq i>j\leq N \}$$
$$O_N=\{(i,j)\in{\mathbb{Z}}^2: \gcd(i,j)=1, \ 0\leq i=j\leq N \}$$
$$N_N=\{(i,j)\in{\mathbb{Z}}^2: \gcd(i,j)=1, \ 0\leq i<j\leq N \}$$
Now, noticing that 
$$|P_N|=|\{(i,j)\in{\mathbb{Z}}^2: \gcd(i,j)=1, \ 0\leq i>j\leq N \}|=\varphi(i)$$
$$|O_N|=|\{(i,j)\in{\mathbb{Z}}^2: \gcd(i,j)=1, \ 0\leq i=j\leq N \}|=1$$
$$|N_N|=|\{(i,j)\in{\mathbb{Z}}^2: \gcd(i,j)=1, \ 0\leq i<j\leq N \}|=\varphi(j)$$
where $\varphi$ is Euler's totient function.
Therefore,
$$|A_N|=1+2\sum_{n=1}^n\varphi(n)$$
Using, Martens theorem($1874$) 
$$\sum_{n\leq x}\varphi(n)=\frac{3}{\pi^2}x^2+\mathcal{O}(x\log x) $$
Then,
$$\lim_{N\to\infty}\frac{|A_N|}{N^2}=\lim_{N\to\infty}(\frac{1}{N^2}+\frac{6}{\pi^2}+\mathcal{O}(\frac{\log N}{N}))=\frac{6}{\pi^2}$$
