Bijection between $\mathbb{R}$ and $\{(X,Y), X\subset Y\subset \mathbb{N}\}$ Does anyone know how to construct a bijection between $\mathbb{R}$ and $\{(X,Y), X\subset Y\subset \mathbb{N}\}$?($\subset$ means "lies in or equals to") 
I'll  appreciate any help!
Thanks in advance.

No, you can't give an "explicit" bijection. What you can do is showing that your set $S=\{(X,Y):X\subset Y\subset \mathbb{N}\}$ has the same cardinality as $\mathcal{P}(\mathbb{N})$ (the set of subsets of $\mathbb{N}$ which is known to be equipotent to $\mathbb{R}$.
You have an obvious injection $S\to \mathcal{P}(\mathbb{N})\times \mathcal{P}(\mathbb{N})$, so that $|S|\le|\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N})|=|\mathcal{P}(\mathbb{N})|$.
You have also an injection $\mathcal{P}(\mathbb{N})\to S$, defined by $Y\mapsto(\emptyset,Y)$, so $|\mathcal{P}(\mathbb{N})|\le|S|$.
Showing an explicit bijection $S\to\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N})$ would already be quite complicated, but the Cantor-Schröder-Bernstein theorem gives the conclusion.
Some slight modifications are needed if with $\subset$ you mean “strict inclusion”.
