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I came across this question:

Show that $A \cap \emptyset= \emptyset$

The solution is that since (earlier in the book I am following) $A\cap B\subseteq B$, this proves that $A\cap \emptyset\subseteq\emptyset$.

Also that the empty set is a member of all sets, so that proves $\emptyset\subseteq A\cap\emptyset$, by the axiom of extensionality, this proves that $A\cap\emptyset=\emptyset$.

But I found this weird: for all $x, A\cap B$ iff $x\in A$ and $x\in B$. So surely $A\cap\emptyset\subseteq\emptyset$ means that $x\in\emptyset$, which contradicts with the fact that there is no element in the empty set!

Could anyone help please?

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  • $\begingroup$ Your "$\in$" should be "$\subseteq$", no? $\endgroup$ – Shalop May 16 '15 at 5:55
  • $\begingroup$ $A\cap B\subseteq B$, but $A\cap B$ is not an element of $B$. $\endgroup$ – Brian M. Scott May 16 '15 at 5:55
  • $\begingroup$ $\subset \neq \in$. $\endgroup$ – copper.hat May 16 '15 at 5:58
  • $\begingroup$ My apology, the signs should be correct now $\endgroup$ – Daniel Mak May 16 '15 at 6:00
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I'll write the empty set as {}.

As you stated, $x\in A\cap B \iff x\in A \wedge x\in B$.
If we accept $A\cap$ {} = {},
$x\in$ {} $\iff x\in A \wedge x\in$ {}

For any x you could substitute in, you have a false statement on both sides.
So the double implication is true for all x.

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$(\forall x\in\varnothing : x\in A)$ is true because you can find no counterexamples in the empty set.   There are no element in the empty set which is not in $A$.

$$\forall x\in \varnothing : x\in A \quad\iff\quad \neg\exists x\in\varnothing:x\notin A$$ $$\forall x (x\in \varnothing \to x\in A) \quad\iff\quad \neg\exists x(x\in\varnothing\wedge x\notin A)$$ This is what is known as a vacuous truth.   Everything is true about the elements of an empty set; because there are none where it might be false.

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  • $\begingroup$ Excuse me if I am wrong, but my understanding is that vacuous truths are usually where the antecedent of a conditional is false (i.e. ∀x(x∈∅)), thus making the conditional true. When I was trying to prove A∩∅⊆∅, I would need to suppose any x∈A and x∈∅, but x∈∅ seems to contradict directly with the antecedent of the said conditional, i.e. we are supposing an x∈∅, but it contradicts with the falsity of ∀x(x∈∅)! $\endgroup$ – Daniel Mak May 16 '15 at 15:03
  • $\begingroup$ @DanielMak Yes, and $A\cap \varnothing \subseteq \varnothing$ means $\forall x \Big(\big((x\in A)\wedge(x\in \varnothing)\big) \to x\in \varnothing\Big)$. Since the antecedant is always false, then the implication is vacuously true. $\endgroup$ – Graham Kemp May 16 '15 at 16:04
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First of all you can see here, and here, difference of $\in$ and $\subseteq$. Then you can say $\emptyset$ has no element, so $ A \cap \emptyset$ has no element which means $A \cap \emptyset= \emptyset$. (Or you can use the other argument).

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