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What are the sub-sets of a null set? I don't get any other set than {}. Please help me out. Thanks.

Rasmus
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    Sanity check: a null set has $0$ elements, so its powerset had better have $2^0 = 1$ elements. – Qiaochu Yuan Jun 13 '11 at 14:23
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    @fahad: For such a little guy, the empty set sure can cause a lot of trouble! Visualize a set as a plastic bag, with "things" in it. One particularly simple set is the empty plastic bag. – André Nicolas Jun 13 '11 at 14:43
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    @fahad: And earlier, I forgot to mention. A plastic bag can have other plastic bags in it, so some or all of the elements of a set can be themselves sets. – André Nicolas Jun 13 '11 at 15:50
  • Perhaps he means "null set" in the sense of measure theory, that is: a null set is a set of measure zero. Not at all the same thing as the empty set. I suspect this because he says a null set. – GEdgar Jun 13 '11 at 17:05
  • @Qiaochu I don't see how the author's slip here implies anything about his sanity. – Doug Spoonwood Jun 13 '11 at 20:56
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    @Doug: "Sanity check" here means (what I thought was a well-established term for) "simple test to make sure that you're on the right track." See, for example, http://en.wikipedia.org/wiki/Sanity_testing . It was not meant to imply anything about the author's sanity and I have no idea why you think I would say such a thing. – Qiaochu Yuan Jun 13 '11 at 20:57
  • @Qiaochu Fair enough. I hadn't heard of that phrase used in that way before. – Doug Spoonwood Jun 13 '11 at 21:01

2 Answers2

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You are right: the empty set has precisely one subset: the empty set.

As a formula: $P(\emptyset)=\{\emptyset\}=\bigl\{\{\}\bigr\}$.

Rasmus
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    To emphasize a subtlety in Rasmus's answer, the power set of the empty set is NOT the empty set (i.e. $\emptyset$ or ${}$). It is the set CONTAINING the empty set (i.e. ${\emptyset}$ or ${{}}$). The difference is subtle and easy to overlook. – Austin Mohr Jun 13 '11 at 15:46
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I would like to use the definition of "subsets".

Definition: Let $A$,$B$ be sets. We say that $A$ is a subset of $B$, denoted $A\subset B$, iff every element of $A$ is also an element of $B$, i.e.
For any object $x$, $x\in A\Rightarrow x\in B$.

Now assume that $A\subset\emptyset$, i.e., for any object $x$, $x\in A\Rightarrow x\in \emptyset$. Since $x\in\emptyset$ is always false, $x\in A$ should also be always false. And thus $A$ has to be the empty set $\emptyset$.