(studying analysis 1 by terrence tao)
A1: Set is an unordered collection of objects.
A2: $x \in A \text{ if x is an element, } x\notin A \text{ if not}$.
A3: Set A and B are equal iff $\forall x, x \in A \iff x \in B$.
A4: Set is empty $\emptyset \text{ if } \forall x, x \notin \emptyset$.
Lemma 3.1.6 (something we want to prove)
$ A \neq \emptyset \to \exists x, x \in A $
$\neg p \lor q $.
$A = \emptyset \text{ or } \exists x, x \in A$.
$\forall A(\forall x, x \notin A \text{ or } \exists x, x \in A)$
set A has nothing or something.
It doesn't seem to exclude A can have nothing and can have something where ~p is true and q is true at the same time.
Whereas contrapositive
$\neg q \to \neg p$
$\forall x, x \notin A \to A = \emptyset$
reiterates the definition of $\emptyset$ and it's contrapositive and our original $p \to q$ says,
For a set A other than $\emptyset$, $\exists x, x \in A$
So this statement does seem to exclude the possibility of A can have something and nothing at the same time.
Q1. So my question is, how should I interpret the statement to avoid my logical misunderstanding?
ie, if I interpret the lemma as $\neg p \lor q$, I would say, look set A is either empty or has something. what is there for me to prove? it's obvious.
(You didn't even ask me to prove if A can have nothing and something at the same time).
That's not what the $p \to q$ seems to ask to proof, so proving $\neg p \lor q$ doesn't seem to be equal to proving $p \to q$.
$$\neg p \lor q \iff p \to q$$
Textbook proof: proof by contradiction
Suppose ~q: $\forall x, x \notin A $
by A4 (definition of empty set) we also know there is $\emptyset$ which has the same property, ie $\forall x, x \notin \emptyset$
by A3 (set equality) $(\forall x, x \in A \iff x \in \emptyset) \iff A=\emptyset$
Q2. minor question, where the contrapositive didn't use A3 (set equality condition) to say $A=\emptyset$, the textbook did use A3.
We are defining emptyset (not sure if axiom and definition are different), and we found something that has the same property as the definition. (normally this would suffice)
Then we took extra step to show that $A=\emptyset$
What would it be that requires the extra step here?
(analogy) It's as if after we define $0$ as $a + 0 = a$, then we happen to find $a + b = a$ then do extra step to show $b = 0$
I found where the error in my reasoning is.
For a set A other than $\emptyset$, $\exists x, x \in A$
So this statement does seem to exclude the possibility of A can have something and nothing at the same time.
This implicitly assumes that a set having $\exists x, x \in A$ property can't have $\forall x, x \notin A $ at the same time.
But it doesn't say so, as the $\neg p \lor q$ version doesn't say so.
The fact that a set can't have the two property at the same time comes from A3.
So this fixes the gap of two interpretations.
Possibly related question is..
probably I'm getting confused about what a mathematical object we can express.
(I have not much physics knowledge but for analogy)
If an experiment says x is particle or wave and it's not consistent.
I guess in math, we say definition of equality should be something that doesn't use the particle_or_wave function
Because it violates the axiom of substitution.
And this confuses me..