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$\exists \, x \, \in \, \emptyset : P(x) $ will be false no matter what the statement $P(x)$ is. There can be nothing in $\emptyset$ that, when plugged in for $x$, makes $P(x)$ come out true, because there is nothing in $\emptyset$ at all! It may not be so clear whether $\forall \, x \, \in \,\emptyset : P(x) $ should be considered true or false ...

$\Large{1.}$ I then paused reading to puzzle out on my own whether $\forall \, x \, \in \,\emptyset : P(x) $ were true or false.

$\boxed{\text{My Intuition :}}$ From the above, because $\emptyset$ contains nothing, thus $x \, \in \,\emptyset$ is false.
Moreover, the statement $ \forall \, x \, \in \,\emptyset $ is false.
Since nothing has alluded to or let on about $P(x)$, thus $\forall \, x \, \in \,\emptyset : P(x) $ is false. $\blacksquare$

Then I continued reading and was floored by the book's proof that $\forall \, x \, \in \,\emptyset : P(x) $ is in fact true:

After expanding the abbreviation of the quantifiers, $\forall \, x \, \in \,\emptyset : P(x) \quad \equiv \quad \forall \, x \, \left[\, x \, \in \,\emptyset \Longrightarrow P(x)\right]. \tag{*}$
Now according to the truth table for the conditional connective, the only way this can be false is if there is some value of $x$ such that $x \, \in \,\emptyset $ is true but $P(x)$ is false. But there is no such value of $x$, simply because there isn’t a value of $x $ for which $x \, \in \,\emptyset $ is true.
Thus, (*) is (vacuously) true.

Although I understand this proof and the Principle of Explosion, I still do not understand the failure of my intuition. Where are the mistake(s)?

$\Large{2.}$ Without any regard to $\exists \;x \in D \; : P(x) \quad \mathop{\equiv}^{dfn} \quad \exists \;x \in D \; \; [ \;x \in A \wedge P(x) \; ]$,
how and why would the second box NOT beguile me into believing as a vacuous truth $\exists \, x \, \in \, \emptyset : P(x) $?

I referenced 1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11 & 12. Source: P69 in How to Prove It by Daniel Velleman.


Supplement to mercio's Answer

I understand $\forall x \in \emptyset,P(x) \; \mathop{\equiv}^{\text{dfn}} \; \forall x, (\color{#B22222}{x\in \emptyset}\implies P(x)) \; \equiv \; \forall x,\color{#B22222}{false}\implies P(x)$.

Is it identical and perfectly correct to prune the steps above (and your writing) into the following?
In $\forall\;\bbox[5px,border:2px solid #32CD32]{\, \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \;,P(x)} \;$, since the green box is a vacuously true implication thus this whole statement is vacuously true.


Supplement to mercio's Comment on Sep 2

Since you wrote "the comma is not a connective", are you intimating that $ \forall \; \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \; \color{#0073CF}{\huge{,}} \;P(x) \quad \equiv \quad \forall \; \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \; \color{#0073CF}{\huge{\require{cancel} \xcancel{,}}}\; P(x) \quad ? \tag{*}$

In the latter, without the comma, nothing joins "$\forall \; \color{#B22222}{{x\in\emptyset}}$" with $P(x)$. So I can compass to understand the latter's fatuity. I don't know (feel free to surmise) why, but I believe: $ \forall \; \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \; \color{#0073CF}{\huge{,}} \;P(x) \quad \equiv \quad \forall \; \color{#B22222}{\underbrace{{x\in\emptyset}}_{false}} \; \color{#0073CF}{{\huge{\text{,}}} \text{we have that}} \; P(x). \tag{**}$
This is why the blue (and green box) comes across as a valid statement to me?

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@LePresentiment There's a typo in the gray box. It should be $\forall x(x\in \varnothing\implies P(x))$ on the right of $\equiv$. Regarding question 2, $\exists x\in \varnothing (P(x))$ is short for $\exists x(x\in \varnothing \land P(x))$. Knowing this I suppose the answer to 2 is evident. –  Git Gud Aug 28 '13 at 9:10
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I don't understand where $A$ comes from. The formula $\exists x\in X(P(x))$ is just short for $\exists x(x\in X\land P(x))$. Does this help? –  Git Gud Aug 28 '13 at 9:36
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@LePressentiment Ohh, I get it now. You don't need to have bounds on the quantification. The quantifiers range over all the universe and that's OK. It's probably safe to assume that the universe is the class of all sets, but this is several orders of magnitude above what is intended with that book, with regards to detail. Can you go on thinking it's just the universe of all things? –  Git Gud Aug 28 '13 at 9:44
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@LePressentiment (in which $U$ is the universe and $A$ is a set). Because $\in$ is a symbol which is used only in relations betweens sets and since $U$ isn't a set, you can't use $x\in U$ formally. But of course, informally, you can think about elements belonging to the universe. As I said before this is several orders of detail-magnitude above what is intended with that book. And all I said above is about 'regular' mathematics. It could very well be that sometimes the universe is a set and you can talk about $x\in U$, (but this wouldn't be in $\sf ZFC$). –  Git Gud Aug 29 '13 at 8:08
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@LePressentiment All in all, you just don't need to write $x\in U$ because if the universe is whatever it is, then by definition the quantification is already bounded by it. –  Git Gud Aug 29 '13 at 8:29

2 Answers 2

You are right when you say that forall $x$, the statement $x \in \emptyset$ is false. This means that $\forall x, \neg (x \in \emptyset)$, which is equivalent to $\forall x, x \notin \emptyset$. Perfectly true.

Then you say "the statement $\forall x \in \emptyset$ is false". $\forall x \in \emptyset$ is NOT a statement, it's an incomplete sentence. Either you write "$\forall x, P(x)$", either you write "$\forall x \in X, P(x)$", which is a shorthand for "$\forall x, (x \in X \implies P(x))$". "$\forall x \in \emptyset$" is not a statement. It can't be true or false.

$\forall x \in \emptyset, P(x)$ is a shorthand for $\forall x, (x \in \emptyset \implies P(x))$, which is equivalent (since $x \in \emptyset$ is always false) to $\forall x, false \implies P(x)$. After looking at the truth table for $\implies$, this is equivalent to $\forall x, true$ (whatever $P(x)$ may be), which is $true$.

If you want to disprove $\forall x \in \emptyset, P(x)$ you have to show me an $x \in \emptyset$ such that $P(x)$ is false. Well you will never find an $x \in \emptyset$.

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@LePressentiment : your green box doesn't mean anything. The comma is not a connective like $\land$ or $\lor$, it is an integral part of the quantification "$\forall x \in \emptyset,$" or "$\forall x,$". What you are doing is like analyzing a sentence by splitting it into small chunks and in doing so, splitting a word in two. It looks like you will get horribly confused with $\exists x \in \emptyset, P(x)$ because there seem to be the same "green box" there even though its meaning (if any) is completely different. –  mercio Sep 2 '13 at 10:21
    
@LePressentiment : the $\in$ in the "$\forall x \in XA, $" is just notation. the $x \in AX$ is just notation. You can't replace it with $false$. When writing a $\forall$ in everyday mathematics, if you don't write "$\forall x \in X, something$" for some name $x$ and some set $A$, you are writing nonsense. In first-order logic, $\in$ may not even be part of the language, so there we write $\forall x, something$. Notice that in any case, there is never any way someone can read $\forall false, something$ and give it any meaning. –  mercio Sep 8 '13 at 7:47
    
and as I said, you cannot replace $\exists x \in \emptyset, P(x)$ with $\exists false, P(x)$ and conclude that it is true because this is just nonsense. The translation of $\exists x \in X, P(x)$ in first-order logic is $\forall x, (x \in X \land P(x))$. If you don't want to use that, then you have a problem. A proof of a $\exists x \in A, P(x)$ usually starts by pointing out some element $x \in A$, then you set out to prove $P(x)$ for this particular element. When $A = \emptyset$ you will never be able to point out an element $x \in A$, so you can't even complete the first step. –  mercio Sep 8 '13 at 7:52
    
Many thanks again. I now see that the red part by itself doesn't make sense. You wrote "if you don't write "$∀x∈X,$ something" for some name $x$ and some set $A$, you are writing nonsense." Thus, discounting all the $\color{#B22222}{\text{red falses}}$ in my 2nd supplementary, could you please expound on equations $(*)$ and $(**)$? Is the blue what you weighed in on and what I have? –  LePressentiment Sep 9 '13 at 3:52
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May I please ask if you could kindly respond to my comment on Sep 9? –  LePressentiment Oct 3 '13 at 11:06

IMO, it's our grasp of natural language that leads us astray here. Natural language is not about explicitly expressing precise ideas; there is a lot of ambiguity and implicit inference involved.

In particular, one doesn't speak about "all of something" unless there is (possibly hypothetically) actually something to speak about.

As you are trained to make this inference, when you hear "$\forall x \in \varnothing:P$", you mentally add the implicit hypothesis "$\exists x \in \varnothing$", which is where your intuition goes awry.

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