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 $n(A)<n(B)$. Same can be said about $B$ and $C$: $n(B)<n(C)$. Combining the two inequalities: $n(A)<n(C)$ 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"

  • $\begingroup$ @MauroALLEGRANZA $‘\subsetneq’ ≠ ‘\not\subseteq’$ $\endgroup$
    – k.stm
    Jul 30, 2020 at 7:37
  • $\begingroup$ @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. $\endgroup$
    – Simplex1
    Jul 30, 2020 at 7:38
  • $\begingroup$ @MauroALLEGRANZA It's not $\subseteq$ it's $\subsetneq$. $\endgroup$ Jul 30, 2020 at 7:38
  • 1
    $\begingroup$ Your second method is only valid for finite sets. $\endgroup$ Jul 30, 2020 at 7:39
  • 1
    $\begingroup$ @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 $z$? $\endgroup$
    – Simplex1
    Jul 30, 2020 at 7:44

1 Answer 1


I really don't think the second solution is correct because is not a bimplication. It works only one way. If A is a subset of C, then n(A)<n(C). Not the other way round.

Coming to the first one, I think this one is correct. Perhaps you don't even need to prove using an element which doesn't exist in A. It is simply shown by first few lines of your solutions that for all elements which are present in A, they are also present in C. Thus A is a subset of C.

  • $\begingroup$ The second argument does not work in the direction you are mentioning, because $A\subsetneq B\subsetneq C$ is perfectly consistent with $n(A)=n(B)=n(C)$. $\endgroup$
    – user239203
    Jul 30, 2020 at 7:58

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