Object in Domain: Dealing with Set Theory

Question

Consider the following Proposition:

• Let $A\subseteq B$. Also, let $B\subseteq C$. Thus, $A\subseteq C$.

Proof:

• Let $A\subseteq B$. Also, let $B\subseteq C$.

  What goes here? Assume $x\in A$. As $x\in A$ and $A\subseteq B$, $x\in B$. As $x\in B$ and $B\subseteq C$, $x\in C$.

Is it that we should let $x$ be an arbitrary element in the domain? But what domain?

Answer 1

Proof:

• Let $A\subseteq B$. Also, let $B\subseteq C$. Assume $x\in A$. Thus, $x=a$ for some $a\in A$. As $x\in A$ and $A\subseteq B$, $x\in B$. As $x\in B$ and $B\subseteq C$, $x\in C$.

Two comments:

First, there is no need for the:

Thus, $x=a$ for some $a\in A$.

line in your proof. Indeed, notice that you don't use this $a$ later in your proof anyway. Instead, can do it all by reference to $x$, i.e. you can do:

Proof:

• Let $A\subseteq B$. Also, let $B\subseteq C$. Assume $x\in A$. As $x\in A$ and $A\subseteq B$, $x\in B$. As $x\in B$ and $B\subseteq C$, $x\in C$.

Second, to get the universal in there, simply do:

• Let $x$ be some arbitrary object from the domain. Let $A\subseteq B$. Also, let $B\subseteq C$. Assume $x\in A$. As $x\in A$ and $A\subseteq B$, $x\in B$. As $x\in B$ and $B\subseteq C$, $x\in C$. Since $x$ was arbitrary, we thus have that for any $x$: if $x \in A$ then $x \in C$. Hence, $A \subseteq C$

Answer 2

Why are you talking of domains? There's no functions involved.

These are sets. Sets are collections of elements[*].

$A \subset B$ means all the elements of $A$ are elements of $B$. so if $A \subset B$ and $B\subset C$ then all the elements of $A$ are elements of $B$. And as they are elements of $B$ and $B\subset C$ they must also be elements of $C$. So all elements of $A$ are elements of $C$. So $A \subset C$.

That's all there is to it.

Now when we do a proof by "element chasing". The entire idea is that if we pick an arbitrary element then because it was arbitrary with no particular special powers, whatever we can conclude about it, we can conclude about all elements.

So if $x \in A$ then ... $x \in B$ because all elements of $A$ are elements of $B$. That's what subsets mean. And if $x \in B$ then $x \in C$ because all elements of $B$ or elements of $C$. That's what subsets mean.

And if $x \in C$ and there was nothing special or distinguishable about $x$ then all elements of $A$ are in $C$.

So $A \subset C$.

That's all there is to it.

....

[*] And $x$ is in the set $A$. If you need to consider something to be a "domain" (what does "domain" mean in this context? If it means anything, it means the set that is the subject of our statement) it is $A$. $A$ is the "domain". (This isn't wrong. $A$ is the subject of our statement. But I find this language about "domains" to be obfuscating at best.)