Ethen Blackwell

2022-07-15

Discrete math prove sets
I am wondering if my answer to this problem is correct.
Statement: $\mathrm{\forall }A,B,C:\left[A\subseteq B\subseteq C\right]\wedge \left[C×B\subseteq B×A\right]\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left[A=B=C\right]$.
Question: is the above statement true or false?
Proof: for some which gives us $a\in A\subseteq B\subseteq C⇒a\in B$ and $a\in C$ also we get $b\in B\subseteq C⇒b\in C$ så we get $x=\left(a,b\right)\in C×B$ and $x\in B×A$.
Is this proof enough or even correct?
Can someone tell me if I'm right or wrong?

Dominique Ferrell

Expert

Step 1
The statement is not true.
If $A=B=\varnothing$ then the conditions are satisfied for every set C.
The statement is true under the extra condition that $B\ne \varnothing$.
Let it be that ${b}_{0}\in B$.
If $c\in C$ then $⟨c,{b}_{0}⟩\in C×B\subseteq B×A$ and we conclude that $c\in B$.
Step 2
Proved is now that $C\subseteq B$ and combining this with $B\subseteq C$ we conclude that $B=C$.
If $b\in B$ then $b\in C$ (see above) so $⟨b,b⟩\in C×B\subseteq B×A$ and we conclude that $b\in A$.
Proved is now that $B\subseteq A$ and combining this with $A\subseteq B$ we conclude that $A=B$.

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