Is $|P(\mathbb{N})| = |\mathbb{R}|$?

Question

If so, what is the argument? I know that the cardinality of the power set greater than the cardinality of the natural numbers, and I assume that the cardinality of the power set of the natural numbers is smaller than that of the power set of the real numbers, though I have not been able to prove this to myself definitively (it simply seems intuitive that the power set of a countable set is smaller than the power set of an uncountable set).

The two approaches that I have considered have been:

(1) Create a bijection between $P(\mathbb{N})$ and $\mathbb{R}.$

(2) Show that there exists no uncountable set smaller than the power set of the real numbers.

By the way, it is not necessarily true that if $A$ has strictly smaller cardinality than $B$ then $\mathcal P(A)$ has strictly smaller cardinality than $\mathcal P(B),$ although of course it cannot be greater. In fact it is consistent with ZFC that $|\mathcal P(\aleph_1)| =|\mathcal P(\mathbb N)| =|\mathbb R|$ where $\aleph_1$ is the smallest uncountable cardinal (and in fact for any cardinality less than the reals). — spaceisdarkgreen, Jun 14 '18 at 18:00

score 1 · Accepted Answer · answered Jun 14 '18 at 17:29

One argument used to count $\mathbb{R}$ is to express real numbers between $0$ and $1$ using binary decimals. Those correspond to subsets of $\mathbb{N}$ - think about the decimal places with value $1$. (You have to be a little careful with the few reals = countably many - that have two binary decimal representations).

Is $|P(\mathbb{N})| = |\mathbb{R}|$?

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