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I have some reasons for why the empty set is a subset of every set,are they correct? i think they are.correct me if I'm wrong,or tell me more

I think the empty set being a subset of every set is like a notion of the idea that if everything in the universe is seen just as number of things, then the number representing collection of smallest number of things would be there in the number representing any collection of all things.

or another way of looking at it is if any collection of things is considered as a universe itself,then of course you could keep taking things away from it to make the collection smaller,and conclude this collection of things was inside what you originally had and conclude that nothing must be in it by the end of the process.

hence nothing is in everything.

One more idea is if it is not a subset of some set, then there must be something in it that is not there in the other set.

there is nothing in it.

hence the contradiction.

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  • A set X is a subset of a set Y if every element of X is an element of Y. For any set Y whatever, every element of the empty set is vacuously an element of Y. In other words there are no elements of the empty set to falsify the proposition. Therefore the empty set is a subset of every set. There is no other reason. It follows from the definition of subset.
    – user4894
    Jul 16, 2022 at 21:57
  • You can simply regard subset inclusion relation as part-whole mereology relation, then perhaps it's much easier to see empty set is a subset (as an empty part) of every set (as a whole), while it's not an element of every set... Jul 17, 2022 at 2:48
  • @Double Knot by nothing in everything i do not mean the empty set is in everything,i mean the contents of empty set are in every collection of things.since almost everything in the universe can be seen as a collection,it could be that nothing is in everything.well atleast on macroscopic everyday life scales,or other ones too,idk i haven't studied till that level.think its true,or a sign why it works in math.well is it so wrong to share my views. :3 i don't tell you to accept them,i just ask do you think so too?if yes good,if not,well..its just a post not the end of the world. Jul 18, 2022 at 17:51
  • It's a basic knowledge in modern standard ZF(C) set theory, without (weak) large cardinal axioms there's not a single model acting as a universe you happen to live inside ever pinned down except inner models such as Godel's constructible L as a transitive class, thus whether I agree with your proposition assuming I understand it fully or not, it perhaps won't be conclusive since per Godel's incompleteness consistency of ZFC is unsure so even proving it may be in vein. It seems you're confused about subset inclusion relation and membership relation which is the core of any type of set theory... Jul 18, 2022 at 22:06
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    @DoubleKnot I do not know you,but you seem to be a person who has seriously read a lot and i have respect for that surely. Mar 16, 2023 at 15:24

1 Answer 1

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I think the empty set being a subset of every set is like a notion of the idea that if everything in the universe is seen just as number of things, then the number representing collection of smallest number of things would be there in the number representing any collection of all things.

This shows that the empty set is smaller than the set of everything. It's a necessary condition, but not a sufficient one, for the subset relationship you're trying to show.

(∃x, ⊤) ⇒ (∀A, (∀x, x∈A) ⇒ |∅| ≤ |A|)

or another way of looking at it is if any collection of things is considered as a universe itself,then of course you could keep taking things away from it to make the collection smaller,and conclude this collection of things was inside what you originally had and conclude that nothing must be in it by the end of the process.

This is a proof by induction – except backwards. This only really works to show that the empty set is a subset of all finite sets.

  • ∀(A: Set), A=∅ ∨ ∃x, x∈A
  • ∀(A: Set), ∀x, x∈A ⇒ (A∖{x}) ⊂ A
  • ∴ ∀(A: Set), A=∅ ∨ ∃(B: Set), B⊂A
  • Therefore, by induction and transitivity of ⊂, ∅ ⊂ A.

One more idea is if it is not a subset of some set, then there must be something in it that is not there in the other set.

there is nothing in it.

hence the contradiction.

This one works for infinite sets as well as finite sets, but only if you take the law of the excluded middle.

  • ∀(A: Set), ∀(B: Set), B⊆A ⇔ ∀x, x∈B ⇒ x∈A
  • ∴ ∀(A: Set), ∀(B: Set), B⊈A ⇔ ∃x, x∈B ∧ x∉A
  • ∀x, x∉∅
  • ∴ ∀(A: Set), ∅⊈A ⇒ ∃x, x∈∅ ∧ x∉A
  • ∴ ∀(A: Set), ∅⊈A ⇒ ∃x, x∈∅
  • But we know that ∀x, x∉∅ – therefore ∃x, x∈∅ is false!
  • ∴ ∀(A: Set), ∅⊈A ⇒ ⊥
  • ∴ (by the law of the excluded middle) ∀(A: Set), ∅⊆A
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  • thank you,i just wished to know if there is some actual maths behind what i said. Jul 16, 2022 at 18:28
  • @RishabhNarula Oh, okay. I'll try to formalise your arguments then.
    – wizzwizz4
    Jul 16, 2022 at 18:28
  • i really dislike the third one, because the contradiction just sort of puts it in words and logic but doen't really visually show it like 2nd one could.why won't that work for infinite sets? Jul 16, 2022 at 18:32
  • @RishabhNarula If you have an infinite set, you might not even be able to select one! (See: the axiom of choice.) Also, my intuition says you can do something forever and still not exhaust an infinite set.
    – wizzwizz4
    Jul 16, 2022 at 18:41
  • thanks for answering,maybe someday i'll sit down and try to understand the rigor version. Jul 17, 2022 at 0:48

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