Real Number Continuum Why real number has number of elements equal to power set of natural number??

I can't think of an intuitive answer for this.
Please suggest some insight to this.
Intuitively speaking, you can describe a bijection as follows. Warning: some cheating is involved.

First, identify a real $x$ in $[0,1]$ with its expansion base $2$, $x = 0.x_1x_2\dots$. For instance, $0 = 0.000\dots$, $1/2 = 0.100\dots$, $1 = 0.111\dots$, and so on. (It's not quite honest what I did, because some numbers have more than one binary expansion - but trust me that it won't change much).

Next, a sequence $x_1,x_2,\dots$ of $0$s and $1$s can be identified with a set of natural numbers. Simply look at the set $A = \{ n \in \mathbb{N} \ | \ x_n = 1 \}$. You can convince yourself that this correspondence is indeed a bijection between all sequences and all sets.

Finally, combining the two steps, you find a bijection between $[0,1]$ and $\mathbb{N}$. If you want to replace $[0,1]$ by $\mathbb{R}$, just find any bijection between the two (hint: think of $\tan x$).
Because $\mathbb{R}$ contains irrational numbers (which are uncountable) and $P(\mathbb{N})$ contains an infinite number of combinations of numbers.