Is an empty set equal to another empty set? [duplicate]

I have a definition that claims that two sets are equal A = B, if and only if:

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

An empty set contains no elements. If I define the sets:

A = $\emptyset$

B = $\emptyset$

then can I conclude that A $\neq$ B? Since they contain no elements? Or would I have to show that an elements exists in one but not the other? To me it seems that these are not equal, but I'm struggling to wrap my head around this.

Thanks,
Paul

To show two sets are equivalent, you should show that $A\subseteq B$ and $B\subseteq A$. This implies that $A=B$. If $A=\varnothing$ and $B=\varnothing$, then try an element-chasing proof to show that $A=B$.

($\to$): If $x\in A$, then $x\in B$. Thus, $A\subseteq B$. $\qquad$[Vacuously true]

($\leftarrow$): If $x\in B$, then $x\in A$. Thus, $B\subseteq A$.$\qquad$ [Vacuously true]

Thus, by mutual subset inclusion, we have that $A=B$.

This conclusion is pretty lame though, as it is an example of a so-called vacuous truth. The implication $p\to q$ is only false when $p$ is true and $q$ is false. Thus, assuming anything to be in an empty set will give you all sorts of bizarre conclusions.

Addendum: Some of the confusion seems to be rooted in what it means to be a subset as opposed to an element. Thus, I am going to list several claims where the goal is to figure out whether or not the claim is true or false (hopefully this may help the OP and some other users). Answers will be provided on the side of each claim.

Claims:

(a) $0\in\varnothing\qquad\qquad$ [False]

(b) $\varnothing\in\{0\}\qquad\qquad$ [False]

(c) $\{0\}\subset\varnothing\qquad\qquad$ [False]

(d) $\varnothing\subset\{0\}\qquad\qquad$ [True]

(e) $\{0\}\in\{0\}\qquad\qquad$ [False]

(f) $\{0\}\subset\{0\}\qquad\qquad$ [False]

(g) $\{\varnothing\}\subseteq\{\varnothing\}\qquad\qquad$ [True]

(h) $\varnothing\in\{\varnothing\}\qquad\qquad$ [True]

(i) $\varnothing\in\{\varnothing,\{\varnothing\}\}\qquad\qquad$ [True]

(j) $\{\varnothing\}\in\{\varnothing\}\qquad\qquad$ [False]

(k) $\{\varnothing\}\in\{\{\varnothing\}\}\qquad\qquad$ [True]

(l) $\{\varnothing\}\subset\{\varnothing,\{\varnothing\}\}\qquad\qquad$ [True]

(m) $\{\{\varnothing\}\}\subset\{\varnothing,\{\varnothing\}\}\qquad\qquad$ [True]

(n) $\{\{\varnothing\}\}\subset\{\{\varnothing\},\{\varnothing\}\}\qquad\qquad$ [False]

Note: Below, $x$ is meant simply to denote a letter, not a set (which is often indicated by writing a capital letter, as was done in the initial explanation). For (t), if $x$ did denote a set, then $x=\varnothing$ would make (t) true as opposed to false.

(o) $x\in\{x\}\qquad\qquad$ [True]

(p) $\{x\}\subseteq\{x\}\qquad\qquad$ [True]

(q) $\{x\}\in\{x\}\qquad\qquad$ [False]

(r) $\{x\}\in\{\{x\}\}\qquad\qquad$ [True]

(s) $\varnothing\subseteq\{x\}\qquad\qquad$ [True]

(t) $\varnothing\in\{x\}\qquad\qquad$ [False]

(u) $\varnothing\in\varnothing\qquad\qquad$ [False]

(v) $\varnothing\subseteq\varnothing\qquad\qquad$ [True]

In your example, you would conclude that $A=B$ because the two conditionals are vacuously true. There are no elements in $A$, so the hypothesis of $x\in A\rightarrow x\in B$ is always false. Therefore, the conditional is always true. The other direction is similar.

By definition two sets are equal if and only if they have exactly the same elements. If $A$ is empty and $B$ is empty, then $A$ and $B$ have exactly the same elements, so $A=B$. There is just one empty set, and it’s the same set whether you describe it as $\{\}$, as the set of integers that are both odd and even, or as the set of living human beings over $20$ feet tall.