# Understanding why the empty set is closed

Definition. A set is called closed if its complement in $\mathbb{R}$ is open.

In my lecture notes it says: $\emptyset$ is closed because $\emptyset = \emptyset \setminus \mathbb{R}$ and $\mathbb{R}$ is open. I think there is a typo because $\emptyset \neq \emptyset \setminus \mathbb{R}$, right? It should be $\emptyset = \mathbb{R} \setminus \mathbb{R}$. Can you please check this?

• $(\phi)^c = \mathbb R \setminus \phi = \mathbb R$ , which is open in $\mathbb R$ .so $\phi$ is closed according to your definition. – hiren_garai Sep 10 '17 at 15:05
• @HirenGarai What is the $\phi$? – PozcuKushimotoStreet Sep 10 '17 at 15:10
• Its the Empty set. – hiren_garai Sep 10 '17 at 15:11
• @HirenGarai: $\phi$ is a lower-case Greek letter phi. If you squint at it it may look slightly like $\varnothing$, which is the symbol for the empty set, but it is not that symbol. – Henning Makholm Sep 10 '17 at 15:14
• @HenningMakholm And here I though $\varnothing$ looks more like the diameter symbol than like $\emptyset$ ;) – Hagen von Eitzen Sep 10 '17 at 15:36

It's both correct and a typo.

That is:

• The useful statement is "$\mathbb{R}=\mathbb{R}\setminus\emptyset$": since $\mathbb{R}$ is open, this means the complement of $\emptyset$ (in $\mathbb{R}$) is open - so $\emptyset$ is closed. This is (presumably) what the author meant to write.

• However, it is true that $\emptyset=\emptyset\setminus\mathbb{R}$; it's just not helpful here. Remember that "$A\setminus B$" is the set of all things in $A$ which aren't in $B$. Well, there are no things in $\emptyset$ which aren't in $\mathbb{R}$ (in fact, there are no things in $\emptyset$ at all!), so $\emptyset\setminus\mathbb{R}=\emptyset$. (I'm pointing this out because you ask whether $\emptyset\setminus\mathbb{R}\not=\emptyset$, at the end of your question.)

• For the first bullet the definition that you seem to use is (retaining OP's phrasing of the definition of 'closed' set), "A set is called closed if it can be expressed as the complement of some set $A$ which is open in $\mathbb{R}$." So I don't think that the author meant to write what you suggest him/her to (presumably) write simply because the definition that OP gives doesn't agree exactly in its intention with what I have just written (although it can be proved). – user 170039 Sep 10 '17 at 15:31
• @user170039 Whoops, read too fast - good catch. – Noah Schweber Sep 10 '17 at 15:31
• OP suggested the from $\emptyset = \mathbb{R} \setminus \mathbb{R}$ which is equivalent to the form $\mathbb{R} = \mathbb{R} \setminus \emptyset$ in this answer. – JiK Sep 11 '17 at 10:35

The complement of $\emptyset$ in $\mathbb{R}$ is $\mathbb{R}\setminus \emptyset$ which is equal to $\mathbb{R}$. And $\emptyset$ is closed in $\mathbb{R}$ because $\mathbb{R}$ is open in $\mathbb{R}$.

Furthermore, even though $\emptyset=\emptyset\setminus \mathbb{R}$, that doesn't let's us conclude that $\emptyset$ is closed in $\mathbb{R}$.

Yes, it is probably a typo. It should be $\emptyset=\mathbb{R}\setminus\mathbb R$.