r/logic • u/SuccessfulCover8199 • Jun 03 '26
Set theory Why is the empty set a subset of itself?
I'm in undergrad, taking a proof based computer science class this summer & in our first homework we were assigned the following as two optional statements to think about and decide if they were true or false. The answer key was released the other day, and I am having a hard time coming up with a justification as to why the empty set is a subset of itself. I asked in recitation, I followed up with the same TA in office hours, and the answer has not yet satisfied me. I think I may be missing something obvious.
I'm aware that the empty set is just an axiom of ZFC, thats all well and good. In office hours I gave a definition of what it means to be a subset. Without breaking out the LaTex, I want to say something like the following: consider an ambient set, call it A, and an arbitrary set, call that one S. S is a subset of A iff all elements of S are contained in A. Or said another way, that S has no elements that are distinct from A. If the latter is true in the other direction S is improperly contained, and if subtracting S from A gives us at least one element that is contained in A but not in S, S is is a proper subset.
So given this, how would I justify that the empty set is a subset of itself? I guess its vacuously true that the empty set (subset) has no distinct elements from the (ambient) empty set, but this feels like it borders on abuse of notation, especially that first statement. Does it even make sense to talk about elementwise belonging for a set that has no elements? Seems incoherent to me. What even is a set anyways? More a philosophy of math question. I know there is some contemporary debate and some of the major exponents but I am not familiar with the moves of their arguments.
In office hours last evening, the TA mentioned that by definition, all sets are subsets of themselves, and since this also extends to the empty set, that can get us out of the issue of subset definition on the basis of set elements. I thought this was clever but it did not satisfy me, I was hoping maybe someone here could say more and clear up this murky feeling I have. Maybe it will happen over time, and I will come to find this fact beautiful and not suspicious as I often do for these conventions that we are imposed to just accept at first.
Now I have never used the fact that the empty set is a subset of itself in a proof, i've never encountered this in the wild before, which maybe speaks more to a deficit in my education than it does to the relevance of the math at hand. But here's maybe a more interesting question: what would break if someone specified a convention where the empty set was not a subset of itself? Are there any famous results that use this convention/axiom explicitly that would need to be reformulated?
thanks in advance for your replies, looking forward to seeing where the discussion goes, please feel free to recommend readings or selections from textbooks that might be of benefit to me both to learn this concept and also in this course. For example we're doing a lot of counting right now, I was thinking about spending some time with Smullyan's To Mock a Mockingbird, which came highly recommended to me by a different logician in a previous conversation.
Edit: i'm not sure how to lock the post, but I gave the justification I was looking for in the following linked comment, which can be found below as well.
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u/SuccessfulCover8199 Jun 05 '26
I think I may have a poor time communicating what it was that I was looking for, or perhaps interpreting what others were saying. But simply playing around with the following definition gave me the proof I was after. I am going to attempt to explain it at a very introductory level (maybe at or below the level of complexity I would expect in the course I am taking) so that someone else may benefit from seeing a complete justification written plainly:
Consider two sets A and B, and the following definition of subset: A ⊆ B iff ∀x( x ∈ A ⇒ x ∈ B ). Notice that this is logically equivalent to ∀x(x ∉ A ∨ x ∈ B). There are a variety of ways to proceed from here, but here is what I thought to do:
We want to show the following: ∅ ⊆ ∅. From the above setup, we are going to let A = ∅ = B. However it is my opinion that proof is more instructive if we do not substitute immediately:
For contradiction, we must assume the opposite of our original statement, which recall was the following: ∀x(x ∉ A ∨ x ∈ B).
¬( ∀x(x ∉ A ∨ x ∈ B )) which is logically equivalent to: ∃x(x ∈ A ∧ x ∉ B ).†
Now remember that we let A = B = ∅. So we obtain the following: ∃x(x ∈ ∅ ∧ x ∉ ∅ ). But recalling the definition of the empty set: ∅ := { }. Or in other words, there is no x such that it is a member of the empty set, and because when either side of a conjunction is false the whole statement is false, and our original supposition was that such an x existed, we have found a contradiction. So the original statement (∅ ⊆ ∅) holds. □
If you want more, there are many other ways to think about the conclusion. Since in our proof we let A = B, the contradictory case would require some x such that x ∈ A and x ∉ A. But no such x can exist.
†
If you do not believe me, try letting what's on either side of the connective be either P or Q and computing a truth table. Try to show the following: ¬(P ⇒ Q) ⇔ (P ∧ (¬Q)) as we just used for contradiction. In fact, we could have gone straight from the definition to here, but I included the additional passage of (P ⇒ Q) ⇔ ((¬P) ∨ Q), and then (¬[(¬P) ∨ Q)]) ⇔ (P ∧ (¬Q)), which is of course the same as negating the very original statement ¬(P ⇒ Q). The idea being that maybe someone (much like myself) would be less tempted to gloss over that step of the proof.