Universe of discourse in $A \subseteq B$

Question

In the following logical analysis:

$A \subseteq B $

$\forall x(x \in A \implies x \in B)$

Is the universe of discourse for the above logical form is A since the above logical form will be only true for A or am I missing something here ?

Answer 1

We can try rewriting the above formula - step by step - in "pure" first-order language.

First of all, $a \subseteq b$ is an abbreviation, defined trough the formula :

$∀x(x \in a \rightarrow x \in b)$;

thus, we have to consider it universally quantified, i.e. as :

$\forall a \forall b \forall x (x \in a \rightarrow x \in b)$.

Now, $\in$ is a binary predicate; to be "formal", we have to use it as follows : $\in (x,y)$.

Thus, our formula is :

$\forall a \forall b \forall x (\in(x, a) \rightarrow \quad \in(x, b))$.

Now we have a "pure" first-order formula; to interpret it we must choose a domain $D$, the elements of which are sets.

The variables $x, y, a, b, \ldots$ take values in the domain, i.e. they must denote objects in $D$, i.e. sets.

The (binary) predicate letter $\in$ must be interpreted with the relation that holds between objects of the domain and $\in (x,y)$ holds iff the object denoted by $x$ belongs to oject denoted by $y$ (the same holds for the binary relation $<$ in arithmetic, where $x, y$ take values in the domain $\mathbb N$ of natural numbers and $x < y$ holds iff the number denoted by $x$ is less then the number denoted by $y$).

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This is the correct approach to the interpretation of the formula : $a \subseteq b$.

All the variables denote objects of the domain, which is not specified into the formula.