# 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. Oct 31, 2016 at 19:39
• $\{\emptyset\}$ is not the empty set, namely $\emptyset \in \{\emptyset\}$. Oct 31, 2016 at 19:43
• If $A$ had been defined as $\{\emptyset, \{\emptyset\}\} \setminus \emptyset$ then you would have been correct.
– Nex
Oct 31, 2016 at 19:43
• If coming from a computer science background: "" != [""] (use any other convenient false/null/empty value). Oct 31, 2016 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. Nov 1, 2016 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. Nov 1, 2016 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, 2016 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.