Post by Eugene 2.0 on Mar 18, 2022 8:30:59 GMT
What's Wrong With Some Proofs That The Empt Set is A Subset of Any Set?
I've found this scheme in many textbooks and in many sites. Here is the scheme by which many proof of that the empry set is a subset of any set works:
The general idea:
1) let x belongs to the empty set
2) if this is so, then there's an x such that x belongs to A (using the next property: if p, then ∃xA)
3) the step #1 is obviously false
4) x belongs to A
5) the empty set belongs to A (using the next property: if x=y, then ∃yA)
Formally:
1) x є ∅ (assumption)
2) x є ∅ → ∃xA (existential instantiation)
3) ~(x є ∅) (definition of ∅)
4) ∃xA (?)
5) B (existential generalization)
Must be obvious that the step #4 is false, because there's no necessity for the consequent of #2 to appear. That's why this proof is a fuzzy one. I'd say there has to be another proof for this, because this kind of proof isn't a good one.
My point is that the empty set can be viewed as a subset of all sets only taking these rules into account:
R1) if A and B are sets, then it is fair that either B⊃A or A=B
R2) for any set X, there's a subset Y such that either Y⊃X or Y=X
Then, assuming that ∅ might be either in a case of B⊃A, or a case of A=B, and further assuming for B that it has Y, we can use R2 to prove that each set has ∅ as its subset.
I've found this scheme in many textbooks and in many sites. Here is the scheme by which many proof of that the empry set is a subset of any set works:
The general idea:
1) let x belongs to the empty set
2) if this is so, then there's an x such that x belongs to A (using the next property: if p, then ∃xA)
3) the step #1 is obviously false
4) x belongs to A
5) the empty set belongs to A (using the next property: if x=y, then ∃yA)
Formally:
1) x є ∅ (assumption)
2) x є ∅ → ∃xA (existential instantiation)
3) ~(x є ∅) (definition of ∅)
4) ∃xA (?)
5) B (existential generalization)
Must be obvious that the step #4 is false, because there's no necessity for the consequent of #2 to appear. That's why this proof is a fuzzy one. I'd say there has to be another proof for this, because this kind of proof isn't a good one.
My point is that the empty set can be viewed as a subset of all sets only taking these rules into account:
R1) if A and B are sets, then it is fair that either B⊃A or A=B
R2) for any set X, there's a subset Y such that either Y⊃X or Y=X
Then, assuming that ∅ might be either in a case of B⊃A, or a case of A=B, and further assuming for B that it has Y, we can use R2 to prove that each set has ∅ as its subset.