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.
 A: Note that I'm assuming in the following that $0\notin\mathbb{N}$ (because I think that way it's a bit easier to write the bijection). Of course adjusting for $0\in N$ is trivial.
Step 1: Define a bijection from $\mathbb{R}$ to the interval $(0,1)$. This is easy, e.g
$$ f(x) = \mathrm{e}^{-\mathrm e^x} $$
Step 2: Define a bijection from $(0,1)$ to $(0,1]$ by the function
$$F(x) = \begin{cases}
\frac{1}{n-1} & x=\frac1n, n\in\mathbb{N}, n>1\\
x & \text{otherwise}
\end{cases}$$
Step 3: For each number $y\in(0,1]$, take the ternary expansion. Always choose the expansion which ends in "$222\ldots$" instead of "$000\ldots$".
Now define
$$g(y) = (\mathcal{X}, \mathcal{Y})$$
where
$$\begin{align}
\mathcal{X} &= \{n\in\mathbb{N}: \text{The $n$-th digit of the ternary representation of $y$ is $2$}\}\\
\mathcal{Y} &= \{n\in\mathbb{N}: \text{The $n$-th digit of the ternary representation of $y$ is not $0$}\}
\end{align}$$
Quite obviously $\mathcal{X}\subset\mathcal{Y}\subset{N}$. However, by construction, we only get infinite $\mathcal{Y}$ (because finite $\mathcal{Y}$ would correspond to a ternary expansion ending in period $0$). On the other hand, all pairs of arbitrary $\mathcal{X}$ and infinite $\mathcal{Y}$ with $\mathcal{X}\subset\mathcal{Y}$ uniquely define an $y\in(0,1]$.
Proof that $g$ is bijective:
For any $y\in (0,1]$, there's (thanks to the disambiguation rule to never take the periodic 0 representation) exactly one ternary representation, and each such representation there's exactly one $y$.
The sets $\mathcal{X}$ and $\mathcal{Y}$ encode exactly the sequence of digits (digit $n$ is $0$ iff $n\notin\mathcal{Y}$, digit $n$ is $1$ iff $n\in\mathcal{Y}\setminus\mathcal{X}$, digit $n$ is $2$ iff $n\in\mathcal{X}$). As long as $\mathcal{Y}$ is infinite, the corresponding ternary representation will not be finite (end in period $0$), and vice versa. Thus we have an one-to-one mapping between the set pairs with infinite $\mathcal{Y}$ and the infinite ternary representations, and thus the real numbers in the interval $(0,1]$.
Step 4: Now all that's left is to map the pairs of sets with the subset property where $\mathcal{Y}$ is infinite to the set of pairs of possibly finite sets with the inclusion property.
This bijection from the set of all pairs to the set of infinite pairs (note: that's the opposite direction we need, but I think it's easier to write down than the reverse) is given by
$$h(X,Y) =
\begin{cases}
(X,Y) & \text{neither $Y$ nor $\mathbb{N}\setminus X$ is finite}\\
(X+1,Y+1) & \text{$\mathbb{N}\setminus X$ is finite}\\
\left(\{1\}\cup\left((\mathbb{N}\setminus Y)+1\right),\{1\}\cup\left((\mathbb{N}\setminus X)+1\right)\right) & \text{Y is finite}
\end{cases}$$
Here $X+1 = \{n+1: n\in X\}$.
The complete bijection is therefore $h^{-1}\circ g\circ F\circ f$.
A: 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”.
