# Subtracting empty set from another

I am considering two sets $A = \{\emptyset, \{\emptyset\}\}\setminus\{\emptyset\}$, $B = \{\emptyset, \{\emptyset\}\}\setminus\{\{\emptyset\}\}$. From the definition of the set difference, I have to consider only those elements of the first set that don't belong to the second one. So my guess is $B = \{ \emptyset \}$. $A$ is more problematic - the empty set doesn't have any elements. My guess is there are no elements to remove. Then, is $A = \{\emptyset, \{\emptyset\}\}$ the correct result?

• Edited to reflect the more common wording you suggested. – Zelazny Oct 31 '16 at 19:39
• $\{\emptyset\}$ is not the empty set, namely $\emptyset \in \{\emptyset\}$. – Sloan Oct 31 '16 at 19:43
• If $A$ had been defined as $\{\emptyset, \{\emptyset\}\} \setminus \emptyset$ then you would have been correct. – Nex Oct 31 '16 at 19:43
• If coming from a computer science background: "" != [""] (use any other convenient false/null/empty value). – moonwave99 Oct 31 '16 at 23:39

You are not removing the empty set. You are removing a set whose only element is the empty set, as $$\emptyset\ne\{\emptyset\}$$

So, $$A\setminus\{\emptyset\}=\{\{\emptyset\}\}$$

Your claim would be true if the set being removed were indeed empty, i.e., $$A\setminus\emptyset=A$$

• Sometimes people identify an object with the singleton containing it. For example, removing a from A could well mean constructing the set $A\setminus\{a\}$ instead of $A\setminus a$. In this sense the empty set is removed from $A$ when constructing $A\setminus\{\emptyset\}$. This identification (or abuse of language if you will) is confusing in the OP's case, but it can be useful to observe that this identification is sometimes made. – Joonas Ilmavirta Nov 1 '16 at 10:47
• @JoonasIlmavirta No, I have used the notation of $A\setminus\{a\}$. And by doing $A\setminus\{\emptyset\}$, you are removing an element from the set that is the empty set. By this notation, to get the set difference with the empty set, you would need to specify all element in the empty set in the braces, so maybe $A\setminus\{\}$ could be considered. – GoodDeeds Nov 1 '16 at 16:07

Put $\{\varnothing\}=a$ and $\varnothing=b$. Hence, $$A=\{b,a\}\setminus\{b\}=\{a\}=\{\{\varnothing\}\}$$ $$B=\{b,a\}\setminus\{a\}=\{b\}=\{\varnothing\}$$

• Good answer, but I'd suggest changing the first sentence to read "Put $\{\emptyset\}=a$ and $\emptyset=b$ and note that $a\ne b,$" since the identity $\{b,a\}\setminus\{b\}=\{a\}$ depends on $a\ne b.$ – bof Nov 1 '16 at 1:16

It is true that subtracting the empty set gives you back the set you started with: $$A - \emptyset = A.$$

But $\{ \emptyset \}$ is not the empty set; it has one element, namely, $\emptyset$.

Let $e:=\varnothing, f:=\{\varnothing\}=\{e\}$.

Then

$$A = \{e, f\}\setminus\{e\}=\{f\}=\{\{\varnothing\}\}\\ B = \{e, f\}\setminus\{f\}=\{e\}=\{\varnothing\}$$

$A = \{\varnothing, \{\varnothing\}\}\setminus\{\varnothing\}$

$B = \{\varnothing, \{\varnothing\}\}\setminus\{\{\varnothing\}\}$

I think that seeing the $\varnothing$ somehow confuses your perception of what is going on. Towards that end, I wonder if this will help.

Let $x = \varnothing$ and $y = \{\varnothing\}$.

Then $A = \{\varnothing, \{\varnothing\}\}\setminus\{\varnothing\} =\{x,y\} \setminus y = \{x,y\} \setminus \{x\} = \{y\} = \{\{\varnothing\}\}$

and

$B = \{\varnothing, \{\varnothing\}\}\setminus\{\{\varnothing\}\} = \{x,y\}\setminus\{y\} = \{x\} = \{\varnothing\}$

You could also try this.

$A = \{\varnothing, \{\varnothing\}\}\setminus\{\varnothing\} = (\{\varnothing\} \cup \{\{\varnothing\}\})\setminus\{\varnothing\} = \{\{\varnothing\}\}$.

$B = \{\varnothing, \{\varnothing\}\}\setminus\{\{\varnothing\}\} = (\{\varnothing\} \cup \{\{\varnothing\}\}) \setminus\{\{\varnothing\}\} = \{\varnothing\}$

Subtracting empty set means you got the null,But {null } is not the empty set; it has one element.