Wronsonia8g

2022-07-10

Big intersection operation
$\mathrm{\varnothing }\ne A\subseteq B\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}⇒\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\bigcap B\subseteq \bigcap A.$
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 $x\in \bigcap B$, 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 $x\in \bigcap A$.
I wonder if the last part is really a proof. It is okay to assume $x\in \bigcap B$ and arrive at $x\in \bigcap A$. 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).

esperoanow

Expert

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 $A\subseteq B$, 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

Expert

Step 1
In case it helps, here is a different style of proof of the same.
For all x, we have

Step 2
Or equivalently, by the definition of $\phantom{\rule{thickmathspace}{0ex}}\subseteq \phantom{\rule{thickmathspace}{0ex}}$, $\phantom{\rule{thickmathspace}{0ex}}\bigcap B\subseteq \bigcap A\phantom{\rule{thickmathspace}{0ex}}$.
Note that weakening the antecedent of $\phantom{\rule{thickmathspace}{0ex}}⇒\phantom{\rule{thickmathspace}{0ex}}$ (from $\phantom{\rule{thickmathspace}{0ex}}S\in A\phantom{\rule{thickmathspace}{0ex}}$ to $\phantom{\rule{thickmathspace}{0ex}}S\in B\phantom{\rule{thickmathspace}{0ex}}$), causes the expression $\phantom{\rule{thickmathspace}{0ex}}S\in A\phantom{\rule{thickmathspace}{0ex}}⇒\phantom{\rule{thickmathspace}{0ex}}x\in S\phantom{\rule{thickmathspace}{0ex}}$ as a whole to be strengthened, and therefore the middle step uses $\phantom{\rule{thickmathspace}{0ex}}⇐\phantom{\rule{thickmathspace}{0ex}}$, 'flipping' the direction of the implication.

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