My workbook considers three different definitions for compactness in logic. It says that it can be shown that these are equivalent, but what would be a strategy to show this? I'm familiar with showing equivalence of formulas by showing that they have the same truth-table, but what would be the strategy for definitions?
The definitions are the following:
(1) A logic can be considered as compact if and only if for every set of sentences Γ and sentence ϕ, if Γ |= ϕ, then, for some finite ∆ ⊆ Γ, ∆ |= ϕ.
(2) A logic can be considered as compact if and only if for every set of sentences Γ, if every finite ∆ ⊆ Γ is satisfiable, then Γ is satisfiable.
(3) A logic can be considered as compact if and only if for every set of sentences Γ, if Γ is unsatisfiable, then there is some finite ∆ ⊆ Γ which is unsatisfiable.
HINT:
For $(3)$ implies $(1)$: Suppose $(3)$ holds, and suppose $\Gamma\models\varphi$. Then consider the set $\Gamma\cup\{\neg\varphi\}$. This is unsatisfiable (why?) so some finite subset $\Delta\cup\{\neg\varphi\}$ is unsatisfiable by (3); but then $\Delta\models\varphi$ (why?).
For $(1)$ implies $(2)$: Suppose $(1)$ holds, and suppose $\Gamma$ is unsatisfiable. Then (fix some arbitrary $\varphi$) $\Gamma\models \varphi\wedge\neg\varphi$ (why?). By $(1)$, for some finite $\Delta\subset \Gamma$ we have $\Delta\models \varphi\wedge\neg\varphi$, so this $\Delta$ is unsatisfiable (why?).
For $(2)$ implies $(3)$: think about contrapositives . . .
Note that there's an assumption in the arguments above that the logic includes negation, $\neg$. This is in fact necessary: there are logics without negation which are compact in the sense of (2) and (3), but not (1). Coming up with such an example is a good exercise . . .