Theorem. Show that if $\beta$ and $\beta'$ are two elements of a set $I$ and to each $\alpha \in I$ we assign a set $A_{\alpha}$, then $$\{ x \in A_{\beta}: x \in A_{\alpha} \text{ for all } \alpha \in I \} = \{ x \in A_{\beta'}: x \in A_{\alpha} \text{ for all } \alpha \in I \}.$$

Proof.

Since $\beta, \beta' \in I$ we have $(\forall \alpha \in I) \: x \in A_{\alpha} \rightarrow ((\exists \beta \in I) \: x \in A_{\beta}) \wedge ((\exists \beta' \in I) \: x \in A_{\beta'})$. This implies $$((\exists \beta \in I) \: y \in A_{\beta}) \wedge ((\forall \alpha \in I) \: y \in A_{\alpha} ) \leftrightarrow ((\exists \beta' \in I) \: y \in A_{\beta'}) \wedge ((\forall \alpha \in I) \: y \in A_{\alpha}).$$ According to the axiom of separation (3.5) this is equivalent to saying $$y \in \{ x \in A_{\beta} : x \in A_{\alpha} \forall \alpha \in I \} \stackrel{(3.5)}{\longleftrightarrow} y \in A_{\beta} \wedge y \in \{ x : x \in A_{\alpha} \forall \alpha \in I \}.$$ $\Box$

2015 FEB 01 (v1.0)
Contact me: m.herrmann followed by an -at- followed by blaetterundsterne.org.

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