Theorem. Let $X$ denote an arbitrary set such that $|X|=n$. Then $|\mathcal P(X)|=2^n$.

Proof. The proof is by induction on the numbers of elements of $X$.

For the base case, suppose $|X|=0$. Clearly, $X=\emptyset$. But the empty set is the only subset of itself, so $|\mathcal P(X)|=1=2^0$.

Now the induction step. Suppose $|X|=n$; by the induction hypotesis, we know that $|\mathcal P(X)|=2^n$. Let $Y$ be a set with $n+1$ elements, namely $Y=X\cup\{a\}$. There are two kinds of subsets of $Y$: those that include $a$ and those that don't. The first are exactly the subsets of $X$, and there are $2^n$ of them. The latter are sets of the form $Z\cup\{a\}$, where $Z\in\mathcal P(X)$; since there are $2^n$ possible choices for $Z$, there must be exactly $2^n$ subsets of $Y$ of which $a$ is an element. Therefore $|\mathcal P(Y)|=2^n+2^n=2^{n+1}$. $\square$

Image that replaced text.

From the above explanation, I don't understand why the set that contains $\{a\}$ will contain $2^{|n|}$ elements when it should clearly be $2^{|1|}$.

The construction of a new set $S$ is the union of the old set with cardinality $n$ and an new element $\{a\}$, therefore the set that do not contain $\{a\}$ still has cardinality $n$ and the set that contains $\{a\}$ is just $\{a\}$, one element...

Can someone please elucidate?

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    $\begingroup$ It is the sets that contain $a$. $\endgroup$ – André Nicolas Sep 14 '14 at 16:45

You misread the proof. Since set $X$ has $n$ elements, the induction hypothesis tells us that $|\mathcal{P}(X)| = 2^n$. The set $Y = X \cup \{a\}$ has $n + 1$ elements. It subsets are either subsets of $X$, of which there are $2^n$ by the induction hypothesis, or the union of a subset $Z$ of $X$ with $\{a\}$. By the induction hypothesis, there are $2^n$ subsets $Z$ of $X$. Hence, there are $2^n$ subsets of the form $Z \cup \{a\}$ of the set $Y$. Hence, $Y$ has $2^n$ subsets that do not contain $a$ and $2^n$ subsets that do contain $a$ for a total of $2^n + 2^n = 2 \cdot 2^n = 2^{n + 1}$ subsets of $Y$, which is what the author wants to show.

  • $\begingroup$ Sorry, you lost me on the fourth sentence. Can you show me explicitly what is $Z$? $\endgroup$ – Carlos - the Mongoose - Danger Sep 14 '14 at 17:00
  • $\begingroup$ Oh I think I see it now, could you elaborate whether if I take the union of two sets say {A,B} U {C}, do I get {A, B, C} or {A, B, {A,C}, {B,C}}? $\endgroup$ – Carlos - the Mongoose - Danger Sep 14 '14 at 17:11
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    $\begingroup$ @ Aåkon I think it's the former. He introduced $Z$ ($2^{n}$ times) to show that $\mid \mathcal{P}(Y) \setminus \mathcal{P}(X) \mid = \mid \mathcal{P}(X)\mid$ $\endgroup$ – Chandran Goodchild Feb 9 at 13:02

Here is an another, combinatorial proof. Let $X$ be a set with $n$ elements. To form a subset of $X$, we go over each element of $X$ and exercise a choice of whether or not to include in the subset. Every sequence of choices gives a different subset.

Since for every element, there are 2 choices, for $n$ elements, there are $2 \times 2 \times ...$ $n$ times, choices.

Therefore, there are $2^n$ distinct subsets of $X$.

  • $\begingroup$ this explanation was very clear to understand $\endgroup$ – Sujal Mandal Jul 23 '18 at 4:54

Suppose you've already shown that $X=\{1,2\}$ has $2^2=4$ subsets, namely ${\cal P}(X)=\{\emptyset,\{1\},\{2\},X\}$. Now you add a new element $a=3$ to get $Y=X\cup \{a\}=\{1,2,3\}$. The four subsets of $X$ are also subsets of $Y$, but you get new subsets - those which contain $a$, e.g. $\{1,3\}$. But each of these is one of the ones you already had together with the new $a$, e.g. $\{1,3\}$ is $\{1\} \cup \{3\}$.

So, the new ones are $\emptyset \cup \{3\}$, $\{1\} \cup \{3\}$, $\{2\} \cup \{3\}$, and $X \cup \{3\}$ and those are exactly as much as you already had - four of them. Which implies that ${\cal P}(Y)$ has twice as much elements (the old ones and the new ones) as ${\cal P}(X)$, so $2\times2^2=2^3=8$.

$$ \begin{array}{|c|c|} \hline \text{Subsets of }X&\text{New subsets}\\ \hline \emptyset&\{3\}\\ \hline \{1\}&\{1,3\}\\ \hline \{2\}&\{2,3\}\\ \hline X&Y\\ \hline \end{array} $$


You say : The construction of a new set S is the union of the old set with cardinality n and an new element {a}, therefore the set that do not contain {a} still has cardinality n and the set that contains {a} is just {a}, one element...

Consider the old set $S$ of cardinality $n$ whose power set has cardinality $2^n$. What is the power set of $S \cup \{a\}$. Each set $T \in \mathcal P (S)$ gives rise to two sets in $\mathcal P(S \cup \{a\})$, one which contains $a$ and one which does not. Thus, you have $2 \times 2^n$ sets in $\mathcal P(S \cup \{a\})$.


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