Big intersection operation ∅ ≠ A ⊆ B ⇒ ⋂ B ⊆ ⋂ A .In...

Wronsonia8g

Wronsonia8g

Answered

2022-07-10

Big intersection operation
A B B A .
In each case, the proof is straightforward. For example, in the last case, we assume that every member of A is also a member of B. Hence if x B, i.e., if x belongs to every member of B, then a fortiori x belongs to every member of the smaller collection A. And consequently x A.
I wonder if the last part is really a proof. It is okay to assume x B and arrive at x A. But can I use this reasoning "then a fortiori x belongs to every member of the smaller collection" in a proof exercise? It looks like just using intuition (it does not seem rigorous).

Answer & Explanation

esperoanow

esperoanow

Expert

2022-07-11Added 11 answers

Step 1
I do not agree. It is a very rigorous argument, and I cannot see how it could be made more formal. Indeed, the proof assumes that A B, which means that (I quote from the proof)
every member of A is also a member of B.
Step 2
From that and from the fact that "x belongs to every member of B", it follows immediately that, in particular, "x belongs to every member of A". It is just a syllogistic inference roughly of the form "if X implies Y and Y implies Z, then X implies Z", which does not require any further justification.
Alissa Hancock

Alissa Hancock

Expert

2022-07-12Added 4 answers

Step 1
In case it helps, here is a different style of proof of the same.
For all x, we have
x A definition of  S :: S A x S A B , so by the definition of  , T A T B  for any  T ; logic S :: S B x S definition of  x B
Step 2
Or equivalently, by the definition of , B A .
Note that weakening the antecedent of (from S A to S B ), causes the expression S A x S as a whole to be strengthened, and therefore the middle step uses , 'flipping' the direction of the implication.

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