1
$\begingroup$

Trying to show that the empty set $ \emptyset \subseteq A $, for any set $ A $.

Let $ x \in \emptyset $, then by definition, $ x \in \emptyset \iff (x \neq x) $.

$ x \in \emptyset \implies (x \neq x) \lor P $ where $ P $ is any statement

Let $ P $ be $ (x \in A) $, then $$ x \in \emptyset \implies (x \neq x) \lor (x \in A). $$ But $ (x \neq x) $ is false, then we can write $$ x \in \emptyset \implies (x \in A),$$ which is equivalent to $ \emptyset \subseteq A $.

Is this correct?

$\endgroup$
  • $\begingroup$ While nothing you have written is wrong, it seems to take an unusual detour. Why not say, "false implies true and false implies false are always true", so $x \in \emptyset \to x \in A$ is always true. $\endgroup$ – Mees de Vries Nov 10 '16 at 17:51
  • $\begingroup$ You should define $\forall x: x\notin \emptyset$. $\endgroup$ – Dan Christensen Nov 10 '16 at 17:54
  • $\begingroup$ It really bugs me when people say "by definition" without actually instantiating a definition. $\endgroup$ – DanielV Nov 10 '16 at 18:04
  • $\begingroup$ "DanielV", just look carefully, the definition is there. ;-) $\endgroup$ – Joseph Nov 10 '16 at 18:14
2
$\begingroup$

It is easier to note that $x \in \emptyset$ is false, hence $x \in \emptyset \Rightarrow x \in A$ is true for an arbitrary $x$. Every "$\Rightarrow$"-conclusion that you make based on a false premise is true trivially. This is easy to prove with the help of a truth table.

$\endgroup$
2
$\begingroup$

The "normal" proof is "vacuous" as they say.

If $X \not \subset A$ then there is some $x \in X$ and $x \not \in A$

So, is there any $x \in \emptyset$ and $x \not \in A$ ? No there isn't (because there is no $x \in \emptyset$ ), so $\emptyset \subset A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.