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Let's suppose that we need to prove a statement of the form $\forall x(x\in A\longrightarrow\phi)$, where $\phi$ is any property (for example when we need to prove that $A\subseteq A\cup B$). Is it really necessary to consider the case when $A=\emptyset$?

I mean deductively we are only concerned when $x\in A$ is true. This means that we can directly consider the fact $x\in A$ as true and then deduce $\phi$ and we don't care about the case $A=\emptyset$ so we can forget about it. But then I'm confused here becase I've read in some books and seen some of my teachers saying something like

"if $A=\emptyset$ then the case is trivialy true because.... Let's now suppose that $\exists x\in A$, then....".

In the case of my example I would simply write something like:

Let's suppose an arbitrary $x$ such that $x\in A$. Then $x\in A \vee x \in B$. This means $x\in A\cup B$, by definition. Therefore $\forall x(x\in A\longrightarrow x\in A\cup B)$, which means $A\subseteq A\cup B$.

So, my question here is if my proof is really complete or if I'm missing the case when $A=\emptyset$.

In a more general case I can think of proving somegthing of the form $\forall x(\phi\longrightarrow \psi)$. Then I would just write as a proof by using deduction something like

"Let's suppose an arbitrary $x$ such that $\phi$ then... but this implies that... therefore $\psi$. This means $\forall x(\phi\longrightarrow \psi)$".

But here I don't know if I have to consider the case when $\phi $ is not satisfied by $x$ and mention that in the proof to be complete.

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  • $\begingroup$ Technically, your thoughts are completely correct. Although $\forall\ A (\text{bla})$ seems like a typo ($\forall\ x$) $\endgroup$ – AlexR Sep 10 '13 at 19:59
  • $\begingroup$ @AlexR Thanks. Yes, sorry, that was a typo. I made the correction already. So, basically you would say that when the "trivial" case is made explicit it's just to make enphasis or something like that, but it's redundant? $\endgroup$ – Daniela Diaz Sep 10 '13 at 20:15
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    $\begingroup$ If you try to prove something for all elements in a set, yes. Usually however, the "emphasis" is needed - if you prove something for the set, such as that it is disjoint to another set and you want to have an element from the set, you'll have to exclude $A\neq\emptyset$ before you do that... $\endgroup$ – AlexR Sep 10 '13 at 20:18
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If you want to prove a statement of the form $$\forall x: \quad (\Phi(x) \Rightarrow \Psi(x))$$ You can assume $x$ with $\Phi(x)$ to exist; in other words you need not require $$\{x : \Phi(x)\} \neq \emptyset$$ for the proof. (So all statements hold for the elements of the empty set ;-) )
However
If you want to prove $$\Psi(A)$$ for some set $A$ and want to use an element $x\in A$ for the proof, you first need to show that $$A \neq \emptyset$$ or make a branch for this as a special case.

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    $\begingroup$ $$\Psi(A) : A = \mathbb{R}$$ Showing that $A$ is open and closed doesn't suffice, because $A$ also needs to be nonempty... Sorry I can't make up a better example from scratch. $\endgroup$ – AlexR Sep 10 '13 at 20:33

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