# Is a power set null because it contains an empty set?

All power sets contain an empty set, so does this make the power set itself an empty set? Or am I mistaken?

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A student I TA'ed called $\{\{\}\}$ once the almost empty set. – Raphael Nov 15 '10 at 18:52
If you have an empty bag inside a second bag, is the second bag empty? – Arturo Magidin Nov 15 '10 at 19:16
You should avoid calling the empty set "null," by the way. Null usually means measure-zero, which is completely different from being empty. – asmeurer Dec 22 '10 at 20:51
null for empty is widely used in certain branches of mathemact. But (as you note) not in measure theory. – GEdgar Aug 22 '11 at 14:44

The short answer to your question is "A power set is never null (or empty), even if it is the power set of the empty set."

A power set $\mathcal{P}(X)$ is the set of subsets of a given set $X$. The elements of $\mathcal{P}(X)$ are themselves sets, trivial or not. Since the empty set $\emptyset$ -- the set containing no elements -- is a subset of every set $X$, then $\emptyset \in \mathcal{P}(X)$ as a set. Thus, every power set is non-trivial and has at least one element, namely the empty set. For example, if $X =$ {$a, b,c$}, then $\mathcal{P}(X)$ has 8 elements: the empty set $\emptyset$, the singletons {$a$}, {$b$} and {$c$}, all pairs {$a,b$}, {$a,c$} and {$b,c$} and the entire set $X =$ {$a, b,c$}. In fact, if $X$ has $|X|$ elements, one can prove that $|\mathcal{P}(X)| = 2^{|X|}$, even in the case that $X$ is infinite. When $X$ is finite, the proof follows from the identity \begin{eqnarray} \sum_{k = 0}^{n} \binom{n}{k} = 2^{n}, \end{eqnarray} which is the sum of all ways to choose $k$ objects from $n$ objects as $k$ increases from $0$ to $n$. I'll leave it as an exercise to show how this counts $|\mathcal{P}(X)|$ with $|X| = n$.

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You seem to be saying that if a set contains the empty set, it is itself empty. This not true. In fact the opposite is true:

A set is empty if and only if it contains no elements. If a set contains the empty set, it has at least one element (the empty set) and is thus, by definition, not empty.

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An empty set contains nothing. A power set of some set $A$, $\mathscr{P}(A)$ always contains something, at least $A$ itself, thus $\mathscr{P}(A)$ is not empty.

The reason I decided to add my two cents is to point out to the frequent source of confusion which is associating empty set with zero, and then associating zero with "nothing". Then, as the (faulty) reasoning goes, the set that contains only the empty set contains nothing, and therefore it is the empty set. The major cause of such confusion is mostly due to notation: designating the empty set as $\emptyset$, which looks almost like $0$.

My rule of thumb: if in doubt, replace $\emptyset$ with $\{ \, \}$ and then think of $\{ \, \}$ as a container that contains nothing. Empty container. Clearly, the empty set. But, say, we have some other set $A$ that contains only the empty set. Again, $A$ is a container, but now it contains another container that contains nothing: $\{ \ \{ \, \} \ \}$. Clearly, a set $A$ contains something, so it's not empty. However writing the same thing as $\{ \ \emptyset \ \}$ may cause some confusion to those exploring the set theory: is that zero the set $A$ contains? Looks like zero, anyway! But zero is nothing! So the set contains nothing. So it is the empty set.

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To add on the previous answers, I just want to clarify that as it was remarked before the power set of $A$, usually denoted by $\mathcal{P}(A)$ is the collection of all the subsets of $A$.

Namely, if $A = \emptyset$ then it has only one subset - itself, as the empty set is a subset of every given set. Therefore, $\mathcal{P}(\emptyset) = \{\emptyset \}$, and again as remarked before - it means that it has one element, and more specifically that it is non empty.

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