Take a set $S$. Why is the empty set part of the power set of $S$?

Intuitively speaking, the power set is the collection of all possible subsets of $S$. How is $\emptyset$ such a subset then? Why is not selecting any element and still forming a subset allowed?

  • $\begingroup$ If you disallow the empty set, you should similarly rule out the full set (to avoid an exception to "the complement of every subset in the powerset is also in the powerset"). $\endgroup$ – Yves Daoust Sep 24 '14 at 11:17
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    $\begingroup$ There are no elements in the empty set that are not elements of $S$. Even stronger: there are no elements at all in the empty set. $\endgroup$ – drhab Sep 24 '14 at 11:35

Suppose $B$ is some set.

We could change the rules and say that $\emptyset$ is not a subset of $B$. But it would be inconvenient. Here are just two of many examples.

Suppose that $P$ and $Q$ are subsets of $B$. We would like to be able to conclude that $P\cap Q$ is a subset of $B$. This is intuitively appealing, because if every element of $P$ is in $B$, and every element of $Q$ is also in $B$, then surely every element that is in both $P$ and $Q$ is in $B$.

But with your proposal, we cannot say that $P\cap Q$ is a subset of $B$. Because $P\cap Q$ might be empty.

Here is another example. It is tempting to say that $X$ is always a subset of $X\cup Y$. Because surely anything in $X$ is also in $X\cup Y$, which is defined to be the set of things that are in $X$ or that are in $Y$.

But with your proposal, we cannot say this, because $X$ might be empty.

Instead of making a lot of annoying and useless exceptions like those, we find that it's simpler all around to agree that the empty set is a subset of every set.


Simply put, because the empty set is a subset of every set.

The definition of $A \subseteq B$ is as follows:

$\forall x (x \in A \rightarrow x \in B)$

Where $A = \emptyset$ this holds trivially; there are no $x \in \emptyset$ and so it satisfies the conditional vacuously.

Hence $\emptyset$ is a subset of every set, and hence it's in the powerset of every set.


This is a very natural usage.

Consider the set {a, b}. You form its subsets from all combinations of the rules "take a/leave a" and "take b/leave b".

There are four combinations: {a, b} (Ta/Tb), {b} (La/Tb), {a} (Ta/Lb) and... {} (La/Lb).

This is why a powerset has $2^N$ elements ($2^N$ combinations).

You can look at it differently, by using the binary notation, one bit per element: $11$, $01$, $10$, $00$. These are all binary numbers of two bits, there is no reason to exclude $00$.

  • $\begingroup$ I don't think I agree with what you have written. The binary formulation that you have mentioned becomes clear only in retrospect. When you say "create all subsets of this set", you think "in how many ways can I select one element? In how many ways can I select two?..." One never thinks "in how many ways do I select none?" $\endgroup$ – user67803 Sep 24 '14 at 14:54
  • $\begingroup$ Okay, leave the empty set out and undo the symmetry everywhere it needs. $\endgroup$ – Yves Daoust Sep 24 '14 at 15:09

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