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This question already has an answer here:

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

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marked as duplicate by user147263, Claude Leibovici, user149792, Daniel W. Farlow, Martin Sleziak Feb 27 '15 at 13:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Certainly the statement holds for all $x$, since there are no $x$s to begin with. $\endgroup$ – Uncountable Feb 27 '15 at 2:04
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    $\begingroup$ Wow, thanks for all the answers, did not expect them so quickly! Allow me to digest :) $\endgroup$ – Paweł Czopowik Feb 27 '15 at 2:16
  • $\begingroup$ Ok, it all makes sense. All the responses were very helpful. I think I was just stuck in a mental loop :) $\endgroup$ – Paweł Czopowik Feb 27 '15 at 5:18
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    $\begingroup$ If you know equality between sets is a transitive relation, then if you suppose $A=\emptyset$ and $\emptyset=B$, you are allowed to conclude $A=B$ from that, by transitivity. So maybe your question is more like: what if there are two empty sets, say $A=\emptyset_1$ and $B=\emptyset_2$, can we know these are equal? Once that is proved, we can agree to let $\emptyset$ denote the unique empty set. $\endgroup$ – Jeppe Stig Nielsen Feb 27 '15 at 12:06
  • $\begingroup$ See also Vacuous truth at Wikipedia. $\endgroup$ – Martin Sleziak Feb 27 '15 at 13:34
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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]

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  • $\begingroup$ But $A\nsubseteq B$ since A and B = {} and a subset of {} is {$\varnothing$} and those are not equal, correct? $\endgroup$ – Paweł Czopowik Feb 27 '15 at 2:14
  • $\begingroup$ @PawełCzopowik $\{\}$ is very different from $\{\varnothing\}$ [the first is the empty set while the second is the set with the empty set as an element]. In general, $A\subseteq A$; thus, $\varnothing\subseteq\varnothing$ or, similarly, $\{\}\subseteq \{\}$. I think you are mixing up what it means to be an element of a set and a subset of a set. $\endgroup$ – Daniel W. Farlow Feb 27 '15 at 2:17
  • $\begingroup$ That I understand. My book says that an empty set {} has only one subset which is itself, thus $\{\varnothing\}$. Hence, $A\nsubseteq B$. $\endgroup$ – Paweł Czopowik Feb 27 '15 at 2:27
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    $\begingroup$ @PawełCzopowik You are incorrect. Every set is a subset of itself. $\endgroup$ – Daniel W. Farlow Feb 27 '15 at 4:28
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    $\begingroup$ @PawełCzopowik Every set is a subset of itself but no set is a proper subset of itself. Maybe that will clarify things for you. $\endgroup$ – Daniel W. Farlow Feb 27 '15 at 4:46
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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.

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    $\begingroup$ This was exactly what I wanted to point out, that both $x \in A$ and $x \in B$ have the same truth value when we're talking about $A = \emptyset = B$. I was also going to throw in a comment about the transitivity of '=', but that's neither here nor there. $\endgroup$ – pjs36 Feb 27 '15 at 2:08
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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.

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Here's a proof by contradiction. Assume that $A=\emptyset$ and $B=\emptyset$ are not equal. Thus, either there is an element in $A$ that is not in $B$ or there is an element in $B$ that is not in $A$. But there are no elements in $A$ or in $B$, a contradiction.

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Since $A=\varnothing$ no $x$ can be found such that $x\in A$.

That implies that no $x$ can be found such that $x\in A$ and $x\notin B$.

Conclusion: for every $x\in A$ we will have $x\in B$ or equivalently $A\subseteq B$.

Likewise we find: $B\subseteq A$.


If a box does not contain any frogs then it is true that every frog in the box is blue. Simply because no frog can be found in it that is not blue. Sortlike we find that every element in $\varnothing$ belongs to set $B$.

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$A=B\iff A\subseteq B$ and $B\subseteq A$.

Are all the elements of $A$ contained in $B$?

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By definition, the relation "$=$" is transitive; if $A=C$ and $C=B$ then $A=B$. now take $C:=\varnothing $


I fact there is only one empty set (in each universe); the set that has no element. People show it by $\varnothing $. you can name it $A$ or $B$ or ... . But they are just different names for one creature.

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  • $\begingroup$ I have to think in only one universe? ;) Thanks for that definition. $\endgroup$ – Paweł Czopowik Mar 1 '15 at 5:58
  • $\begingroup$ you are welcome. one universe is not sufficient for you? :) $\endgroup$ – user 1 Mar 1 '15 at 7:44

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