"$c \implies p$"

"Definition 7 Let $\mathscr F$ be an arbitrary family of sets. The intersection of sets in $\mathscr F$ denoted by $\bigcap_{A\in \mathscr F} A$ or $\bigcap \mathscr F$ , is the set of all elements that are in A for all $A\in \mathscr F$. That is,

$\bigcap\limits_{A\in \mathscr F} A$ = {$x \in U$: $x \in A$ for all $A\in \mathscr F$}

Here "$x \in A$ for all $A\in \mathscr F$" may be expressed alternately as "$A\in \mathscr F \to x \in A $." The latter expression has an advantage in proving theorems, as we shall see in the coming Theorem 7.

If the family $\mathscr F$ is indexed by the set $\Gamma$, the following alternate notation may be used:

$\bigcap \limits_{r \in \Gamma}A_{r}$ = {x $\in U$ : $x \in A_r$ for all $r \in \Gamma$}"

"Theorem 7 (b) Let {$A_r$|r$\in \Gamma$} be the empty family of sets; that is, $\Gamma = \emptyset$. Then $$ \bigcap \limits_{r \in \emptyset}A_{r}=U$$

[proof] (b) We shall prove that $x \in \bigcap \limits_{r\in\emptyset}A_r$ for all x in U. Observe that

$$ x \in \bigcap \limits_{r\in \emptyset}A_r$$ $$\equiv x \in A_r \space\space for \space all\space r \in \emptyset \space\space Def\space 7$$ $$\equiv r \in \emptyset \to x \in A_r$$"

Source: Set Theory by You-Feng Lin, Shwu-Yeng.T

I don't know why $r \in \emptyset \to x \in A_r$ leads to U. I know it's deduced from $ x \in \bigcap \limits_{r\in \emptyset}A_r$ and it's true because $r \in \emptyset$ is a contradiction, but it's just another expression of $ x \in \bigcap \limits_{r\in \emptyset}A_r$ by applying Def 7, and the proof didn't reach U. If $r \in \emptyset \to x \in A_r$ leads to U, then all intersections of family of sets can lead to U.

That is, $\bigcap \limits_{r \in \Gamma}A_{r}$ =U by developing the logical step as $\bigcap \limits_{r \in \Gamma}A_{r} \equiv r \in \Gamma \to x \in A_r$ the same way as, $\bigcap \limits_{r \in \emptyset}A_{r}=U$, which is absurd.



$\emptyset$ is the empty set; thus : $r \in \emptyset$ is false for any $r$.

This means that, by the truth table for the conditional, that :

$r \in \emptyset \to x \in A_r$

is true for any $r$, and thus $x \in \bigcap \limits_{r \in \emptyset}A_{r}$.

But this is true for any $x$, i.e. for all $x \in U$.

In conclusion, $x \in U \to x \in \bigcap \limits_{r \in \emptyset}A_{r}$ and this, by definition of inclusion, means :

$U \subseteq \bigcap \limits_{r \in \emptyset}A_{r}$.

Obviuosly : $\bigcap \limits_{r \in \emptyset}A_{r} \subseteq U$, and thus we have proved $=$.

  • $\begingroup$ If $ \bigcap \limits_{r \in \emptyset}A_{r}=U$ true, then can $\bigcap \limits_{r \in \Gamma}A_{r}$ =U also be true? $\endgroup$ – buzzee Jan 8 '16 at 8:26
  • $\begingroup$ @buzzee - of course no. The proof above relies on the fact that $r \in \emptyset$ is always false; with a $\Gamma \ne \emptyset$, $r \in \Gamma$ is true for suitable $r$ and thus the conditional $r \in \Gamma → x \in A_r$ is not true for any $x$. $\endgroup$ – Mauro ALLEGRANZA Jan 8 '16 at 8:35

It's just a detail of predicate logic. Since $r\in\emptyset$ is always false it will imply anything, especially $x\in A_r$.

If you want to use the word contradiction about $r\in\emptyset$ you can prove this by reductio ad absurdum:

First you assume the premise $r\in\emptyset$.

Under that assumption you assume $x\notin A_r$ and conclude since $r\in\emptyset$ follows which is an contradiction you would conclude $x\in A_r$.

Now since that was under the premise $r\in\emptyset$ it would follow that $r\in\emptyset\Rightarrow x\in A_r$.

You could also consider the semantics of what an implication says. It says that under the condition that the LHS of the implication is true the RHS is true, but if the LHS is not true it nothing is said about the truth of the RHS. Consequently if the LHS is never true, nothing is said about the truth of the RHS (by the implication).

  • $\begingroup$ Then can $\bigcap \limits_{r \in \Gamma}A_{r}$ =U also be true? If "$r \in \emptyset \to x \in A_r$" leads to U, we can develop a logical step as $\bigcap \limits_{r \in \Gamma}A_{r} \equiv r \in \Gamma \to x \in A_r$, and $r \in \Gamma \to x \in A_r$ can lead to U as we did in the intersection of sets indexed by empty set. $\endgroup$ – buzzee Jan 8 '16 at 8:45
  • $\begingroup$ No, because $r\in\Gamma$ need not be generally false. It's the falsehood of $r\in\emptyset$ that makes the implication valid. $\endgroup$ – skyking Jan 8 '16 at 8:47
  • $\begingroup$ Oh! The generality of falsehood! I didn't notice that. That clears up my doubts. $\endgroup$ – buzzee Jan 8 '16 at 8:49

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