I have to prove a few different statements.
The first is if $A \subset B$ and $B \subset C$ then prove $A \subset C$. This one is fairly straight forward, but I'm stuck on how the next one differs.
Prove that if $A \in B$ and $B \in C$ then $A \in C$.
I don't really understand how to put this in logical symbols. I've only seen $a \in A$ written out but never "a set $A$ is an element of a set $B$".
Here's what I have for a proof at this point, assuming I understand what "a set $A$ is an element of a set $B$" means: suppose $A \in B$ and $B \in C$. Then $A \in C$.
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$\begingroup$The implication $(A\in B) \wedge (B\in C) \implies A\in C$ is false. Just take $B=\{A\}$ and $C=\{B\} = \{\{A\}\}$ to have a counterexample.
$\endgroup$ 8 $\begingroup$Take $A=\varnothing$, $B=\{\varnothing\}$ and $C=\{\{\varnothing\}\}$.
Then clearly $A\in B\wedge B\in C$, but $A\in C$ implies $\varnothing=\{\varnothing\}$ wich cannot be true.
This because $\{\varnothing\}$ has elements and $\varnothing$ has not.
We conclude that the implication is false.
$\endgroup$ $\begingroup$Set membership is not like putting stuff in a box.
If you put a box of chocolates into a drawer then it's certainly true that the drawer contains chocolates, but the concept of set membership is different from this. The membership only considers the direct containment, so by the mathematical language the drawer only contains a box (and in that box there's chocolates).
And that's where your reasoning is wrong (mathematically).
$\endgroup$ 2 $\begingroup$You can't prove that if $A \in B$ and $B \in C$ then $A \in C$ because it is not always true.
Let $B = \{A, 1 \}$. Then $A \in B$
Let $C = \{ B,2\} = \{ \{A, 1 \}, 2 \}$
The members of $C$ are $\{A, 1 \}$ and $2$. So $A \notin C$.
Yes $A$, is in $\{A, 1 \}$, but it is not in $C$.
$\endgroup$ 1 $\begingroup$There are sets for which the relation $A\in B$ is transitive: ordinals sets, which are defined as sets on which the relation $x\in y$ is a strict well-ordering.
Furthermore for such sets, one has $(A\in B)\Rightarrow(A\subset B)$.
Example: $X=\bigl\{\varnothing,\{\varnothing\},\{ \varnothing,\{\varnothing\}\}\bigr\}$.
Added:
To answer directly the question, the main difference, from a logical point of view, is that, while $(A\in B)\wedge(B\in C)\Rightarrow (A\in C)$ involves individual elements-sets, $(A\subset B)\wedge(B\subset C)\Rightarrow (A\subset C)$ involves universal quantifiers: $$\forall x(x\in A\Rightarrow x\in B)\wedge \forall x(x\in B\Rightarrow x\in C)\Rightarrow\forall x(x\in A\Rightarrow x\in C).$$
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