Are universal quantified statements defined for inequalities even if the inequality is undefined?

The following universally quantified statement has an undefined inequality when $x = 1$:

$∀x∈ℝ \dfrac 1{(x−1)^2}>0$

Is such a statement false or undefined?

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3 Answers

I would not write down something like that (so I'd consider the statement ill-typed), but funny enough I would write down $$\forall x \in {\mathbb R}. x \neq 1 \rightarrow \frac{1}{(x-1)^2} > 0.$$

(It's possible to give a foundational justification in type theory for this, where the expression $\frac{1}{(x-1)^2} > 0$ has a so-called dependent type, but I'm not really thinking of it that way.)

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Why is it ill-typed? – user1644677 Dec 12 '13 at 2:06
Because the expression $\frac{1}{(x-1)^2}$ is not defined for all elements in the domain of $x$. – Magdiragdag Dec 12 '13 at 8:09

Here is an attempt to add something to this tricky question.

I'll simplify a bit, like nik in comments of another answer. Let's consider the function $x \to \frac 1 {x^2}$

Everybody will agree, I guess, that it is defined on $\Bbb R^*$.

Wait. What is a function?

Here is the definition of Wikipedia:

A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output

But, in at least one textbook, I can find two definitions, closely related:

1. A function $f: E \to F$ is given by two sets $E$ and $F$, and a relation $\mathcal R$ on $E\times F$ such that $x \mathcal R y \wedge x\mathcal R y' \implies y=y'$. It need not have an "image" (in $F$) for each element of $E$: the domain of the function, $\mathrm{Dom} (f)$, is the set of elements of $E$ which have an image. A function is really a relation $x \mathcal R y$ such that an $x$ may have at most one image.

2. Now, an application is a function $f: E \to F$ with $\mathrm{Dom} (f) = E$.

Usually one does not pay much attention to the detail of these definition. But here it seems to me there is something important.

Of course, our function $f : \Bbb R \to \Bbb R$ defined by $f(x)=\frac 1 {x^2}$ has $\mathrm{Dom} (f)=\Bbb R^*$.

Now, what does it really mean to write $\forall x \in \Bbb R, f(x) > 0$ ?

We can rewrite this with our relation,

$$\forall x \in \Bbb R, \forall y \in \Bbb R, x \mathcal R y \implies y>0$$

And this is actually true. For $x=0$, the implication is still true, because there is no $y$ such that $x \mathcal R y$.

Simply, there is a point of $E=\Bbb R$ where $f$ is not defined, so writing $\forall x \in \Bbb R, f(x) > 0$ is a bit misleading: one could easily imagine it implies $f$ is an application, and I would even agree that in most interpretations of this sentence, it's what one really means.

It's just that an application is a function with an additionnal property regarding the domain (the property "$\mathrm{Dom}(f)=E$"), and here this property cannot be true. The property itself is perfectly defined, it states $\forall x \in E, \; \exists y \in F, \; x \mathcal R y$. And it's wrong for our $f$.

So my opinion, after some thought, about this sentence: "$\forall x \in \Bbb R, \frac{1}{x^2}>0$", is it's true.

I think what is deeply misleading in this example is that nobody ever goes back to the very definitions of mathematical objects, using only ZF axioms. It would be far too cumbersome (and here, I didn't). One prefers to get a good intuition about mathematical things, but sometimes, intuition bites :-)

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I must say I "like" the way of thinking that leads to "Everybody understands $f \colon E \to F$ as meaning that $f$ is a function with domain $E$ and range contained in $F$, so let's go and redefine it". – Daniel Fischer Dec 22 '13 at 15:37
@DanielFischer ??? – Jean-Claude Arbaut Dec 22 '13 at 16:55
The textbook definition using the word function and the notation $f\colon E \to F$ in an uncanonical way and calling "application" what (almost) everybody calls function. – Daniel Fischer Dec 22 '13 at 18:25
@DanielFischer Since coming to MSE, I'm slowly losing the habit to say "everybody does this". It appears to be almost always wrong. But I'll have a look at Bourbaki, just for fun. Not everybody does like Bourbaki, for sure, but it will be enough for me. And of course, I may have misinterpreted or misused the definition I found for this answer, I don't claim to have "the truth". Still, this question of definition of a function is a bit subtle and the intuitive definition of a function is not same as defined using ZF axiom and only $\emptyset$ and $\in$ – Jean-Claude Arbaut Dec 22 '13 at 19:49
Yes, "everybody this or that" tends to be wrong, there are (almost?) always exceptions. I didn't mean to convey any criticism of you. I just take issue with the terminology the textbook uses. Everywhere else I have seen, $f\colon E \to F$ is used as an alternative notation for $f \in F^E$. But well. As long as people don't start saying compact when they mean quasicompact ... – Daniel Fischer Dec 22 '13 at 20:04

Yes the "forall" statement is false. Note that this doesn't make $\exists x\in\mathbb R\colon \frac1{(x-1)^2}\le 0$ true however.

To elaborate a bit: The rules of inference for "forall" (which ultimately convex the "meaning" of $\forall$) would allow us to infer from $\forall x\in \mathbb R\frac1{(1-x)^2}>0$ then $\frac1{(1-x)^2}>0$. However, $>$ is a relation (in this context) $\subseteq \mathbb R\times\mathbb R$ and whatever $\frac1{(1-x)^2}$ may be, it is not $\in\mathbb R$, hence the inference would allow us to conclude something wrong - horrifying! So we better agree that $\forall x\in \mathbb R\frac1{(1-x)^2}>0$ is not true. Instead, $\forall x\in \mathbb R\setminus\{1\}\colon\frac1{(1-x)^2}>0$ is fine. Now some confusion may arise as falseness of $\forall x\in \mathbb R\frac1{(1-x)^2}>0$ implies trueness of $\exists x\in \mathbb R\colon \neg\left(\frac1{(1-x)^2}>0\right)$. Just as we neither have $i>0$ nor $i\le 0$ in $\mathbb C$, we should keep in mind that this is different from the (false) statement $\exists x\in \mathbb R\colon \frac1{(1-x)^2}\le0$. For a strict treatment, one should be careful whenever one uses partial functions such as division.

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I don't think this answers the question. My interpretation of the question is the following: $\forall x\in \mathbb R\left(\dfrac{1}{(x-1)^2}>0\right)$, well, $1\in \mathbb R$, so what happens then? – Git Gud Dec 11 '13 at 19:39
@GitGud It happens that for $x=1$ the statement is false, since the quotient is meaningless. That is $\forall x \in \Bbb R \;...$ is false. – Jean-Claude Arbaut Dec 11 '13 at 21:55
@nik Good point! I didn't think about excluded middle. – Jean-Claude Arbaut Dec 11 '13 at 22:02
@arbautjc The so called 'statement' is not false for $x=1$, as you say, it's meaningless. It's not true and it's not false. – Git Gud Dec 11 '13 at 22:08
@arbautjc: Actually it's more complicated than what I thought, because $\forall x \in \mathbb{R} \frac{1}{x^2} > 0 := \forall x (x \in \mathbb{R} \Rightarrow \frac{1}{x^2})$, so its negation isn't quite $\exists x \in \mathbb{R} \frac{1}{x^2} \leq 0$ (I'm tired of typing $(x-1)$). – Najib Idrissi Dec 11 '13 at 22:09