Let A$\subset$B$\subset$C , and S a property. Can S implie c $\in$ B, and b $\in$ A given c,b satisfying S, and c$\in$C, b$\in$B?

Question

Given S a singgle property, and A$\subset$B$\subset$C all proper containment.

Does it ever happend in proof that utilizing every single bit of S proves c $\in$ B for any c $\in$ C satisfying S. Then we use the same statement S to show that all b $\in$ B satifying S is actually in A. Notice I used all aspects of S to show that S every c $\in$ C satisfying S is in B. Then I use the exact same aspects of S to show that again all b $\in$ B satifying S is actually in A.

This way, I show that all c $\in$ C satisfying S is actually in A.

Is it possible at all? Last time I run into similar problems, but my proof turned out to be wrong.

Answer 1

In some cases, YES!

For obvious example suppose an element satisfies $S$ iff it was in $A$!

But you should consider more properties maybe.

Answer 2

Not so clear to understand precisely your question...

Let me provide an answer based on what I understood.

First It is true that if $(\forall c \in C) \ S(c)$ implies that $(\forall b \in B) \ S(b)$. This whatever the statement $S(x)$ is.

Second It may be that $(\forall c \in C) (S(c) \implies c \in B)$. But this is really dependent on your statement $S$.

Same thing regarding $A \subsetneq B$.