# (Verification) If $A \subsetneq B$, and $B \subsetneq C$, then $A \subsetneq C$

Statement to be proved: If $$A \subsetneq B$$, and $$B \subsetneq C$$, then $$A \subsetneq C$$.

First method: Inequality: If $$A \subsetneq B$$ then there exists an $$x\in B$$ such that $$x\notin A$$. Let $$z:z\in B \wedge z\notin A$$

Similarly, if $$x\in B\implies x\in C$$, then $$z\in C$$, but if $$z\notin A$$, then there is an element in $$C$$ not in $$A$$, so $$A\ne C$$.

$$(x\in A\implies x\in B, x\in B \implies x\in C) \implies (x\in A \implies x\in C)$$. Thus at best $$x\subset C$$. But since $$A\ne C$$, then $$A\subsetneq C$$. Q.E.D.

Second method (I couldn't use this in the exercise as cardinality came later):

$$A\subsetneq B$$ means that all elements in $$A$$ are in $$B$$, but some elements in $$B$$ are not in $$A$$. So $$n(A). Same can be said about $$B$$ and $$C$$: $$n(B). Combining the two inequalities: $$n(A) so it follows that $$A\neq C$$.

The fact that all elements in $$A$$ are in $$C$$ is proven the same way as in the first method.

I think this is correct, but I'm not entirely sure as to the logical soundess. I also think there is a faster way of doing it (but I mainly want to know if it is correct).

EDIT: I use $$\subsetneq$$ to mean "A is a subset of B, but A is not equal to B"

• @MauroALLEGRANZA $‘\subsetneq’ ≠ ‘\not\subseteq’$ Jul 30, 2020 at 7:37
• @MauroALLEGRANZA I use the symbol $\subsetneq$ to be 'A is a subset of B, but A is not equal to B'. The book I'm reading (Tao's Analysis I) uses this notation. Jul 30, 2020 at 7:38
• @MauroALLEGRANZA It's not $\subseteq$ it's $\subsetneq$. Jul 30, 2020 at 7:38
• Your second method is only valid for finite sets. Jul 30, 2020 at 7:39
• @MauroALLEGRANZA But for a $z$ in $B$, the same $z$ is in $C$, and due to that, that very $z$ which is also in $C$ is not in $A$. Why can't we have the same $z$? Jul 30, 2020 at 7:44

• The second argument does not work in the direction you are mentioning, because $A\subsetneq B\subsetneq C$ is perfectly consistent with $n(A)=n(B)=n(C)$.