I think I have a fundamental misunderstanding of the definition of an union set; the following is from the notes published by Prof David Makinson at LSE for a further logic course:
Prove that from its definition that tautological implication satisfies monotony/monotonicity of Tarski condition (Whenever $A ⊨ \alpha$ then $A\cup B ⊨ \alpha$).
Solution: Suppose $A ⊨ \alpha$; we want to show that $A\cup B \vDash \alpha$. Let $v$ be any valuation and suppose $v$ makes all formulae in $A\cup B$ true. Then it makes all formulae in $A$ true, so by the first supposition it makes $\alpha$ true.
$v(\forall x(x\in A\cup B))=T$ leads to $v(\forall x (x\in A))=T$ because $A\cup B=\forall x(x\in A \lor x\in B) $, therefore it means everything that is either a member of $A$ or $B$ is true. From that all formulae in A are true.
But this doesn't seem to make sense if I look at the definition: IF (and this is a BIG if) I am understanding correctly, all formulae in $A\cup B$ are still true under $v$ if $v(\forall x (x\in A))=T$ but $v(\forall x (x\in B))=F$, because of the inclusive disjunction in the definition of $A\cup B$. (In particular $x\in A\cup B$ still holds if $x\in A$ but $\lnot x\in B$)
But if so, 'All formulae in $A\cup B$ are true. Then it makes all formulae in $A$ true', if I replace the 2nd sentence with 'Then it makes all formulae in $B$ true', it doesn't make sense anymore because in the current truth assignment, we are assuming all formulae in A are true but all formulae in B are false.
I am sure it is my understanding of union set that went wrong (2nd paragraph); but I can't tell how. Could anyone tell me where I went wrong please?
The union $A\cup B$ is defined as $\{x|x\in A \vee x\in B\}$. So, $c\in A\cup B$ iff it is true that "$c\in A$ or $c\in B$."
For example, if $A=\{1,2,3\}$ and $B=\{2,3,4,5\}$, then $A\cup B = \{1,2,3,4,5\}$. Using the definition, $2\in A\cup B$ because it is true that "$2\in A$ or $2\in B$" ... so are other elements of $A\cup B$.
For now, we make the numbers "represent" propositions. Suppose a valuation $v$ makes all formulae in $A∪B$ true. That is, $v(1)=T, v(2)=T,..., v(5)=T$.
With $A=\{1,2,3\}$, do we have $v(1)=T, v(2)=T$, and $v(3)=T$ ?
Edit:
Another example: Let $A=\{1,2,3\}$ and $B=\{4,5\}$. Then, $A\cup B=\{1,2,3,4,5\}$.
Now, suppose we have a valuation $v$ such that all formulae in $A$ are true but all formulae in $B$ are false. So we have $v(1)=v(2)=v(3)=T$ and $v(4)=v(5)=F$.
With $A\cup B=\{1,2,3,4,5\}$, do we have $v(1)=v(2)=...=v(5)=T$ (or are all formulae in $A\cup B$ true)?