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I am working with the following set of axioms:

1) There is a set.

2) Two sets are equal iff they contain equal elements.

3) if $A$ is a set and $p(x)$ is a set property (e.g. $x = x$), then $\{a \in A : p(a)\}$ is also a set.

4) if $A$ and $B$ are sets, then there is a set which contains only elements of $A$ and $B$.

5) if $\mathbb A$ a set, then $\bigcup \mathbb A :=\{a | \exists A \in \mathbb A : a \in A\}$ is also a set.

6) For each set there is its power set.

I my lecture it was claimed that if $\mathbb A \ne \emptyset$, then $\{a \in \bigcup \mathbb A | \forall A \in \mathbb A\ : a \in A \}$ is also a set.

I am not sure why do we require that $\mathbb A \ne \emptyset$.

Indeed, the fact that $\{a \in \bigcup \mathbb A | \forall A \in \mathbb A\ : a \in A \}$ is a set follows directly from 3 and 5 without the requirement.

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    $\begingroup$ This answer and its comments may be useful. $\endgroup$ – Renan Maneli Mezabarba Apr 1 '17 at 14:17
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    $\begingroup$ @SergeyZykov Take $A=\emptyset$ and $B=\{X,Y\}$. Then $A\subset B$, but with your definition, $\bigcap A=\emptyset$ while $\bigcap B=X\cap Y$. So, in this case $\bigcap B\subset \bigcap A$ holds iff $X\cap Y=\emptyset$. $\endgroup$ – Renan Maneli Mezabarba Apr 1 '17 at 14:31
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    $\begingroup$ @RenanManeliMezabarba I see, so to avoid this situation we do not define $\bigcap \mathbb \emptyset$, correct? $\endgroup$ – Sergey Zykov Apr 1 '17 at 14:36
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    $\begingroup$ That's correct. It's like dividing by zero in elementary algebra/calculus, we simply don't do that. $\endgroup$ – Renan Maneli Mezabarba Apr 1 '17 at 14:37
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    $\begingroup$ @RenanManeliMezabarba nope, existence of $\{A,B\}$ follows from the axioms above as follows: $\{A\}=\{x \in P(A): x = A\}$ (ax. 6) and $\{A,B\}=\{A\} \cup \{B\}$ (ax.4). $\endgroup$ – Sergey Zykov Apr 1 '17 at 14:58
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If $\mathbb{A} =\emptyset$, the property $\phi(a):= \forall A \in \mathbb{A}: a \in A$, holds for any $a$ (a universal statement over an empty domain holds,also see wikipedia). But here we are working inside $\cup \mathbb{A}$ which is $\emptyset$ by the definition if $\mathbb{A} = \emptyset$. This is a good thing, because otherwise we would have defined a set of all sets, which cannot exist by Russell's paradox.

So we'd just be defining the empty set instead of the intersection $\cap \mathbb{A}$, which we define here for non-empty $\mathbb{A}$. This is normally not what you want e.g. normally if $\mathbb{A} \subseteq \mathbb{B}$ implies $\cap \mathbb{B} \subseteq \cap \mathbb{A}$, but this would then fail for this case. etc.

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