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How can it be that the empty set is a subset of every set but not an element of every set?

I've understood that the empty set must be a subset of every set because if it were not a subset of every set then the statement $\exists x : x \in \emptyset \land x \notin M$, would need to be ture, but since the empty set has no elements this statement is a contradiction.

But how can it be that the empty set is not element of every set, if it is a subset of every set?

I also understood that it cannot be an element of itself, otherwise it wouldn't be the empty set anymore, but why is it not an element of every non-empty set? If the empty set is subset of every set, than it also would need to be a subset of itself ($\emptyset \subset \emptyset$), how can that be?

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    $\begingroup$ It seems like there is a fundamental misunderstanding you have between the definition of an element of a set and a subset of a set. For example, do you recognise that the sets $\{1\}, \{\{1\}\}, \{\{\{1\}\}\}$ are three different and distinct sets, each being an element of but not a subset of the next? $\endgroup$ – Dan Rust Dec 4 '20 at 15:20
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    $\begingroup$ A "bag analogy" is useful here. The empty set is like an empty bag. It is vacuously true that the things in an empty bag are in all bags whatsoever (this is the "bag version" of $\emptyset$ being a subset of every set); however, of course it is not the case that every bag has an empty bag inside it (this is the "bag version" of $\emptyset$ not being an element of every set). The "bag analogy" is of limited use, but this is a situation where it works well. $\endgroup$ – Noah Schweber Dec 4 '20 at 15:22
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    $\begingroup$ Think of sets as boxes. The empty set is an emtpy box. Box a is a subset of box b if after you dump the contents of box a into box b, but not putting in things that are already there, box b 's contents do not change. c is an element of a box if the box has c inside it. So, the empty set is a subset of any set (any box contains all elements of an empty box); but not all boxes contain an empty box (not all sets contain the empty set). There is a difference between an empty box and a box containing an empty box. Perhaps this is a silly analogy... $\endgroup$ – David Mitra Dec 4 '20 at 15:24
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    $\begingroup$ Because it satisfy the definition of subset of. $\endgroup$ – Mauro ALLEGRANZA Dec 4 '20 at 15:28
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    $\begingroup$ The set of real numbers is "made of" real numbers. The set of rationals is a subset of the set of real but it is not itself a number. Thus, element and subset are two concepts that do not coincide. $\endgroup$ – Mauro ALLEGRANZA Dec 4 '20 at 15:38
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  • There might be versions of set theory where the requirement " the empty set is an element of every set" is satisfied. What I mean is that it does not seem absurd prima facie. For example, in the set theoretic consruction of natural numbers, number zero ( that is : the empty set) is an element of every ( natural) number greater than 0 , and these numbers are sets. ( for example , $1=\{\emptyset\}= \{0\}, 2= \{\emptyset, \{\emptyset\}\}=\{0,1\} , 3= \{0,1,2\} $.

  • However, the question " is every set a member of every set ( different from itself)?" can be settled as a pure matter of fact. Any counter-example would do; Consider, for example, the set : $\{1, 2,3\}$.

  • I think the question is : why does it seem plausible that, if a set is a subset of every set, then it should also be an element of every set? Maybe one could try to reconstruct the reasoning that produces this false appearence :

(1) The empty set s a subset of every set, say, of set S

(2) Therefore, all the elements of $\emptyset$ are also elements of S.

(3) Therefore the totality of the elements of $\emptyset$ is an element of S.

(4) But this totality is $\emptyset$ itself .

(5) Therefore $\emptyset\in S$.

  • The mistake is hidden in steps (3) and (4).

a) as to (3) : though it is true distributively that every element $\emptyset$ is an element of S, it is not true collectively

b) as to (4) the totality of the elements of $\emptyset$ is nothingness ( nil, nothing) , and therefore is not the same thing as the empty set which is something ( namely, a set).

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    $\begingroup$ Thank you so much for this answer @FloridusFloridi. $\endgroup$ – NilsK Dec 5 '20 at 8:16
  • $\begingroup$ @Nisk. You're welcome :) $\endgroup$ – Floridus Floridi Dec 5 '20 at 11:39

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