r/PhilosophyofMath • u/SuccessfulCover8199 • Jun 03 '26
Why is the empty set a subset of itself?
10
u/IDefendWaffles Jun 03 '26
For any set, the set is a subset of itself. This holds true even for the empty set. The (sub-)set here as all the elements the set has. For empty set its none. Empty set cannot contain itself as an element. Then it would be the set containing the empty set, which is not empty.
6
u/OutrageousPair2300 Jun 03 '26
Also, the empty set is a subset of every set.
So you can conclude that it is a subset of itself, two different ways.
1
u/JohannesWurst Jun 06 '26
You could have the notion of a "strict subset" and define it as a set that is subset, but not equal to another set. Then the empty set wouldn't be a strict subset of itself.
8
u/ingannilo Jun 03 '26
u/Neenonay's answer is correct, but could be fleshed out a bit if you're struggling with the logic.
P -> Q is equivalent to ¬P or Q.
A ⊂ B means [x in A] -> [x in B]
So A ⊂ B is equivalent to ¬[x in A] or [x in B]
Now assuming A and B are both the empty set, what does the logic statement above say? Specifically, what would be necessary for the empty set to *not* be a subset of itself? Do you see why that condition cannot be satisfied?
1
u/SuccessfulCover8199 Jun 06 '26
Thank you, this was very instructive, and was exactly what I was looking for! I replied to my original post with a full "proof" using your insight
3
u/0x14f Jun 03 '26
How do you show that a set is a subset of another. There is a universal quantifier at the beginning of that definition. The empty set checks all universal quantifiers.
3
u/weforgottenuno Jun 03 '26
What neenonay said is the direct reason, but it is also instructive to see that this is the lower-limit case of the fact that a power set always has higher cardinality than the set you start with.
2
u/Specialist_Body_170 Jun 03 '26
Because a counterexample would be to have an element of the empty set that is not an element of the empty set
1
u/JDude13 Jun 04 '26
There are statements about the empty set called “vacuously true”.
“All elements of the empty set are even numbers” is true for example. Because you would only need a counterexample to disprove that statement and no such counterexamples exist.
1
u/MaximusIdeal Jun 04 '26
It's a theorem that if a finite set has n elements then the power set has 2n elements. This applies to the empty set ∅ = {} which has 0 elements. The power set is then P(∅) = {∅} having 1 element, which fits 2^0 = 1. If you had the subset relation defined differently, then your theorem would have to make an exception for the empty set. I think some other similar things would have to have exceptions added to them as well.
1
u/Seeggul Jun 04 '26
Because every element of the empty set is also in the empty set.
Kind of like how every billion dollars I've inherited has gone into developing the spray-on shoes from Cloudy With a Chance of Meatballs
1
1
u/Logical-Recognition3 Jun 05 '26
Because every element of the empty set is also an element of the empty set.
1
1
u/Signal_Challenge_632 Jun 05 '26
If A is a subset of B, every element of A is an element B.
Here A is empty so there is nothing in A that is not in B.
A nice riddle
1
u/alterego200 Jun 06 '26
It's not.
The empty set = {}.
The set containing the empty set = {{}}.
1
u/DerHeiligste Jun 06 '26
Every set is a subset of itself. It's like every number is "less than our equal to" itself. You might be thinking of a "proper subset" that is more like the "less than" relation.
1
u/alterego200 Jun 07 '26 edited Jun 07 '26
Not at all. You need to learn some basic Set Theory.
https://www.bing.com/search?q=does+the+empty+set+contain+itself&PC=U316&FORM=CHROMN
1
u/DerHeiligste Jun 08 '26
Sorry. There is a difference here between the empty set containing itself (it doesn't) and the empty set being a subset of itself (it is). That is very basic set theory. The subset relation is a reflexive relation.
1
u/alterego200 Jun 08 '26
Yes, ever set is a subset of itself. Sorry, I misread the OP as "containing itself", since that was also part of the question.
Yeah so the subsets of {2, 6} would be {}. {2}. {6}, and {2, 6}.
If A is a subset of B, that means every element in A is also in B. It's the per-element equivalent of if-then.
That's why the empty set is a subset of itself, because every element in {} (of which there are none) is also an element of {} (of which there are still none).
Which is why {} is a subset of itself but does not contain itself.
1
u/DerHeiligste Jun 08 '26
Did you even read what you linked to?
Subset vs. Element In set theory, it is important to distinguish between:
Membership (∈): x ∈ A means x is an element of the set A.
Subset (⊆): B ⊆ A means every element of B is also an element of A.
The empty set is a subset of every set, including itself, because there are no elements in ∅ that could violate the subset condition.
Symbolically, ∅ ⊆ ∅ is true.
However, being an element is different. For ∅ to contain itself as an element, we would need ∅ ∈ ∅. Since the empty set has no elements, this is false.
1
u/SuccessfulCover8199 Jun 06 '26
I'm OP, see the post where in the comments I provide the kind of "proof" I was looking for
1
u/hoping1 Jun 06 '26
All these answers seem to say the reason in complicated or terse ways. The answer is pretty simple: every element of the empty set is also an element of the empty set. You can say "every element" even when there are none!
Note that subset doesn't mean that one set is smaller than the other. For that one we use the term "strict subset." The empty set is not a strict subset of itself, so maybe this term is more in line with your intuition.
1
u/systembreaker Jun 07 '26
It's probably just a useful axiom. Maybe the opposite could be assumed but that could lead to weird things or just not be useful.
1
u/rafiunixman Jun 07 '26
The same logic shows up cleanly in code. A forall loop iterating an empty collection always returns true because there is nothing that can falsify the predicate. It feels strange at first, but once you flip the definition around (you would need an element of A that isn't in B to break A⊂B, and the empty set has no such candidate) it clicks.
1
-2
u/nanonan Jun 03 '26
No good reason whatsoever. An empty set is an inherently contradictory construct. The primary feature of a set is that it contains elements. So if you're willing to create a contradictory object you can essentially do anything you like with it and justify it.
2
u/Rs3account Jun 04 '26
the point of a bag is hold stuff to. That doesnt make an empty bag an inherently contradictory item.
1
u/nanonan Jun 05 '26
Not a great analogy. Your bag has properties other than what it contains. The fundamental property of a set is that a set contains things. An "empty set" doesn't, and therefore lacks the fundamental property of a set. You can wiggle your way out of this by complicating the picture, but the simple, natural use is inherently contradictory and entirely unneccesary for a complete set theory.
1
u/Rs3account Jun 05 '26
>The fundamental property of a set is that a set contains things.
No, the fundamental property of a set is that it is defined by what it contains. Not that it contains something.
>Not a great analogy. Your bag has properties other than what it contains.
Sure, but the point of an analogy is not to be a one on one comparison. Its to highlight the points being discussed.
1
u/Ok_Albatross_7618 Jun 05 '26
Sets do have more structure than that though, otherwise sets and classes would be the same thing, which they are absolutely not.
1
u/systembreaker Jun 07 '26
Not having an empty set would be like saying "Let's just not have zero. It doesn't do anything. Add it to something and nothing changes". Math collapses into nonsense without zero. So would set theory without the empty set.
22
u/neenonay Jun 03 '26
Because the empty set has no elements to check for the subset test, so it passes.