2
$\begingroup$

Are the two statements \begin{align*} \exists x \, \forall y : P(x,y) \qquad \text{and} \qquad \exists x : P(x,y) \; \forall y \end{align*} equivalent?

Or is this more of a matter of what one understands by the second statement, which I think of as really $\exists x : (P(x,y) \, \forall y)$?

$\endgroup$
6
  • $\begingroup$ I would say that the second is poor grammar and it is the first that is more pedantically correct. In colloquial language, yes they may be equivalent but you wouldn't catch me saying the second, at least not when written like above. $\endgroup$
    – JMoravitz
    Commented Mar 17, 2021 at 16:08
  • $\begingroup$ @JMoravitz Are you sure you wouldn't say the latter? Eg. "there exists a $N_0$ such that $f_n \geq 0$ for all $n \geq N_0$" reads better than "there exists a $N_0$ such that for all $n \geq N_0$, $f_n \geq 0$." I understand that formally the first one in the post is better but in terms of mathematical writing? $\endgroup$
    – BBB
    Commented Mar 17, 2021 at 16:10
  • 1
    $\begingroup$ @BBB If you need to write something in predicate logic, it's generally because you need a degree of clarity that the economy of your current natural-language discourse doesn't support. But then what's the point of writing in runes for clarity, if you decide to tweak their syntaxis to look more like a natural language sentence you are trying to avoid? I think you are just defeating the purpose of what you are doing. $\endgroup$
    – user239203
    Commented Mar 17, 2021 at 16:36
  • 1
    $\begingroup$ @BBB As a side note, as non-English speaker I think that "there exists an $N_0$ such that, for all $n\ge N_0$, $f_n\ge 0$" would be the superior wording, if it weren't for the fact that neither natural language nor LaTeX have an elegant separator between the restricted universal quantifier and the quantified statement. $\endgroup$
    – user239203
    Commented Mar 17, 2021 at 16:39
  • $\begingroup$ @Gae.S. "...if you decide to tweak their syntaxis to look more like a natural language sentence you are trying to avoid?" That is a good point. $\endgroup$
    – BBB
    Commented Mar 17, 2021 at 16:51

2 Answers 2

Reset to default
3
$\begingroup$

Let $P(x,y)$ be some property on two variables $x$ and $y$.

Note that when one writes $\exists x \, \forall y \, \colon P(x,y)$, it means that “There is an $x$, such that for all $y$, $P(x,y)$ holds.”. On the other hand, when one writes $\exists x \, \colon P(x,y) \, \forall y$, it means “There is an $x$ such that $P(x,y)$ holds for all $y$.”.

They are not strictly identical (they are different expressions), but they are equivalente, in the sense of being two ways of expressing the same idea.

Although, most people don’t write $\exists x \colon P(x,y) \, \forall y$, in mathematical notation.

$\endgroup$
6
  • 1
    $\begingroup$ I think the expression $P(x,y) \forall y$ is pretty common in the literature. $\endgroup$
    – user657166
    Commented Mar 17, 2021 at 16:21
  • 1
    $\begingroup$ "Although correctly, no one writes $\exists x \colon P(x,y) \, \forall y$." I disagree. There exists an $x \in \mathbb{R}$ such that $f(x,y) = 0$ for all $y \in \mathbb{R}$ is very common. $\endgroup$
    – BBB
    Commented Mar 17, 2021 at 16:23
  • $\begingroup$ @ZPlaya that might be true. From my experience, I always have encountered the fist expression in the literature. But they stand for the same idea. $\endgroup$
    – Air Mike
    Commented Mar 17, 2021 at 16:25
  • $\begingroup$ @BBB note that is the way we say that using our natural language. $\endgroup$
    – Air Mike
    Commented Mar 17, 2021 at 16:26
  • $\begingroup$ You may consider editing the last line to: "Although, most people don't write $\exists x:\,P(x,y)\,\forall y$ in mathematical notation.", as some people may, who knows? Anyways, your answer is nice (+1) as it mentions that the statements are not equivalent. $\endgroup$
    – ultralegend5385
    Commented Mar 17, 2021 at 16:27
3
$\begingroup$

Unfortunately, sometimes people write:

there exists $x$, $y=x^2$, for all $y$

but it is ambiguous. It might mean

there exists $x$, for all $y>0$, $y=x^2$

which is false,
or it might mean

for all $y>0$ there exists $x$, $y=x^2$

which is true.

$\endgroup$
5
  • $\begingroup$ But it's clear that they meant the false statement though. $\exists x, P(x,y), \forall y$ can in no way be reinterpreted as $\forall y, \exists x, P(x,y)$, can it? $\endgroup$
    – BBB
    Commented Mar 17, 2021 at 16:27
  • $\begingroup$ I mean, doesn't the fact that I used a ":" (such that) matter? $\endgroup$
    – BBB
    Commented Mar 17, 2021 at 16:29
  • 1
    $\begingroup$ @BBB you’re correct, the original statement would mean the false one. In no way you can interchange the quantifiers, if they are different, as in this case $\endgroup$
    – Air Mike
    Commented Mar 17, 2021 at 16:31
  • 1
    $\begingroup$ @BBB the colon only means “such that”. It wouldn’t change the meaning if you write a comma, for example. $\endgroup$
    – Air Mike
    Commented Mar 17, 2021 at 16:34
  • 1
    $\begingroup$ Especially when written in words, it is ambiguous. $\exists x, P(x,y), \forall y$ can mean either $(\exists x, P(x,y)), \forall y$ or $\exists x, (P(x,y), \forall y)$. It is no question of "interchange". $\endgroup$
    – GEdgar
    Commented Mar 17, 2021 at 17:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .