Questions about proving that "Φ is a subset of any set" Most of the books use the counter proof method (assuming that the existing element x belongs to Φ, but X does not belong to set a, then "Φ is not a subset of any set". But because there is no element in Φ, the previous assumption that "Φ is not a subset of any set" is wrong, that is, the conclusion that "Φ is a subset of any set" is correct) But I don't think this proof is right. Another way of thinking, if you want to prove that "Φ is a subset of any set", as long as you can prove that (element x belongs to Φ, and X belongs to set a), but because there is no element in Φ, it is wrong to assume that (element x belongs to Φ, and X belongs to set a), that is, "Φ is not a subset of any set". Is it true that "Φ is not a subset of any set"? Therefore, I think it is not feasible to prove that "Φ is a subset of any set" by using the two "tools" of the definition of Φ (there is no element in Φ) and the definition of subset (if the element x belongs to a and X belongs to B, then a is a subset of B), Therefore, I think that "Φ is a subset of any set" should be an artificial definition, rather than a theorem deduced by reverse proof Definition of subsets [if (all) elements x belong to set a, then x belongs to set B if and only if a is a subset of B] Let proposition p: (all) elements x belong to set a; Q: X belongs to set B; Z: A is a subset of B; The definition of subsets can be transformed into propositional formula: ZP - > Q (1) Let a: the element x belongs to Φ; b: The element x belongs to any set a; c: Φ is a subset of any set a; It is impossible to bring in the above proposition formula (1) to prove that it is a permanent truth formula, that is, proposition a = t, B = f, because the truth value of proposition a is always = t (the definition of Φ), so proposition C is always = t, that is, "Φ is a subset of any set a"... Proof over! Is that right?

Questions about proving that "Φ is a subset of any set" Most of the books use the counter proof method (assuming that the existing element x belongs to Φ, but X does not belong to set a, then "Φ is not a subset of any set". But because there is no element in Φ, the previous assumption that "Φ is not a subset of any set" is wrong, that is, the conclusion that "Φ is a subset of any set" is correct) But I don't think this proof is right. Another way of thinking, if you want to prove that "Φ is a subset of any set", as long as you can prove that (element x belongs to Φ, and X belongs to set a), but because there is no element in Φ, it is wrong to assume that (element x belongs to Φ, and X belongs to set a), that is, "Φ is not a subset of any set". Is it true that "Φ is not a subset of any set"? Therefore, I think it is not feasible to prove that "Φ is a subset of any set" by using the two "tools" of the definition of Φ (there is no element in Φ) and the definition of subset (if the element x belongs to a and X belongs to B, then a is a subset of B), Therefore, I think that "Φ is a subset of any set" should be an artificial definition, rather than a theorem deduced by reverse proof Definition of subsets [if (all) elements x belong to set a, then x belongs to set B if and only if a is a subset of B] Let proposition p: (all) elements x belong to set a; Q: X belongs to set B; Z: A is a subset of B; The definition of subsets can be transformed into propositional formula: ZP - > Q (1) Let a: the element x belongs to Φ; b: The element x belongs to any set a; c: Φ is a subset of any set a; It is impossible to bring in the above proposition formula (1) to prove that it is a permanent truth formula, that is, proposition a = t, B = f, because the truth value of proposition a is always = t (the definition of Φ), so proposition C is always = t, that is, "Φ is a subset of any set a"... Proof over! Is that right?

Element x belongs to a = > element x belongs to B (1) is equivalent to: element X does not belong to B = > element X does not belong to a (2). So as long as we prove that for any set B, as long as X does not belong to B, then x does not belong to Φ

Can we prove that an empty set is a subset of any set

In the set part of the six-year high school mathematics textbook, it is stipulated that "an empty set is a subset of any set". In addition, in some pamphlets, some people regard "an empty set is a subset of any set" as a theorem. The proof is as follows: "assuming that, on the contrary, an empty set must not be a subset of set a, then at least one element in it must not belong to a, which is contradictory to the definition of an empty set, so it must be red

Why is an empty set a subset of all sets

Because every nonempty set is an empty set after deleting its elements
Non empty set = empty set + element
The former of course includes the latter