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In a book1 I have encountered the following:

To check your understanding of [the importance of quantifier order], consider the following two statements. One is true, and the other is false. Which is which? $$ \exists y > 0\ \text{such that}\ \forall x > 0, y < x $$ $$ \forall x > 0\ \exists y > 0\ \text{such that}\ y < x $$

(The type of $x$ and $y$ aren't specified, but I've assumed they are real numbers)

I think I understand how the two statements are different. However, to me, both statements appear to be false:

Start by assuming either statement is true. Since $y < x$, we can state that $y = x - d$, where $d > 0$. Let us consider the case that $x = d$. Substituting $x$ for $d$ gives us: $y = d - d \implies y = 0$. This contradicts the requirement in both statements that $y > 0$. Both statements must be false.

Alternatively stated: Whatever the value of $y$, there is a value of $x$ where $x = y$, contradicting the requirement that $y < x$.

Has the author made a mistake or am I making one? If so, please explain my mistake and help me understand which one is true, and why (with emphasis on the order of the quantifiers if possible).

1How to Study for a Mathematics Degree by Lara Alcock, p73

  • You can say that there's some $d$ such that $y=x-d$ - but you're 'assigning' $d$ (that is, effectively assigning a value to $y$) before assigning $x$. In fact, for the truth of the second statement you (of necessity) have a different $y$ (and a different $d$) for each $x$. – Steven Stadnicki Jan 28 '15 at 05:58
  • Statement 1 says you may fix $y$ first. It is easy to see that that does not work. Statement 2 says choose $x$ first and then search for $y$ (which may depend on $x$). You can take $y=x/2$. – Thomas Jan 28 '15 at 06:00
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    The second can be read as follows: If you give be an $x\gt 0$, I can produce a $y\gt 0$ such that $y\lt x$. Sure can. – André Nicolas Jan 28 '15 at 06:01
  • Thanks everyone, I understand how my disproof can't work for statement 2 now. I now realise what was confusing me: I was mistakenly thinking that there'd be a smallest possible value of $x$ (i.e. if you can set $y$ to $x/2$ then you can set $x$ to the same value and eventually there won't be some $y < x$). I didn't think about the fact that you can keep halving reals indefinitely. What should I do with this question? Write an answer? Delete? –  Jan 28 '15 at 06:18
  • @curiousinternals It would be good if you wrote an answer, no need to delete - other people may have the same question later – Peter Woolfitt Jan 28 '15 at 06:22
  • Given that it appears to me to already be a duplicate (see my note above), deleting it might be in order if you're amenable to doing such; it's certainly your choice. – Steven Stadnicki Jan 28 '15 at 06:53
  • @StevenStadnicki The linked post doesn't address my second misunderstanding about there being a smallest possible $x$ where $x > 0$, so I'm going to go ahead and write an answer (might take a while). I'll probably reference that post in my answer though. –  Jan 28 '15 at 07:27
  • @StevenStadnicki After further thought, I don't think the question is a duplicate at all. In the end I realised I wasn't confused about the quantifiers (Thank goodness! I'd already used those two years ago): See my answer for details. –  Jan 28 '15 at 12:43

1 Answers1

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The first statement is false, but the second is true.

Proofs

First Statement

The first statement is false if for every value of $y$, there is at least one value of $x$ that is not greater than $y$. A couple of examples are $y/2$ and $y$. Neither of these will ever be less than or equal to zero, so are valid values for $x$.

Second Statement

$x/2$ is always less than $x$ but greater than zero, thus there is always a suitable value for $y$ to take.1

Why I Was Incorrect

Self Contradiction

Assuming the first or second statement to be true, for $y = x - d$ to be true, $d > 0$ and $d < x$ must also be true. Under these conditions $x \ne d$. In other words, my statement contradicted the other two statements; The other statements did not contradict themselves. A proper proof by contradiction needs to show how the assumption leads to a contradiction, not how arbitrary statements can contradict the assumption. I could have stated $y = 0$ straight away and it would have been just as invalid.

Finding the Smallest Positive Real

Another mistake I made was thinking that there can be some smallest value of $x$ that is still greater than zero. This is why I was confused in the first place. Since $x > 0$, there is no smallest possible value:

Let's suppose there is a smallest positive number, say $s$. So what is ${s \over 2}$?

1Thanks to Thomas and André Nicolas for pointing this out in the comments.

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