I got the constant function $\mathbb{R}->\mathbb{R}$ $f(x) = c$

which could be expressed as $\forall x \in \mathbb{R} \exists!c \in \mathbb{R}:f(x) = c $

But after I negated this term, therefore the described function is not constant, I get: $\neg(\forall x \in \mathbb{R} \exists!c \in \mathbb{R}:f(x) = c) $

$\leftrightarrow \exists x \in \mathbb{R} \neg (\exists!c \in \mathbb{R}:f(x) = c))$

$\leftrightarrow \exists x \in \mathbb{R} \neg (\exists c \in \mathbb{R}:(f(x)=c \wedge \forall c' \in \mathbb{R}:(f(c) = c' => c = c')))$

$\leftrightarrow \exists x \in \mathbb{R} \forall c \in \mathbb{R}:(\neg f(x) = c \vee \exists c' \in \mathbb{R}:(f(x) = c' \wedge c \neq c')) $

$\leftrightarrow \exists x \in \mathbb{R} \forall c \in \mathbb{R} (f(x) = c => \exists c' \in \mathbb{R} (f(x) = c' \wedge c \neq c))$

which does not make a lot sense to me as it sais "There is a x, for all c, f(x) = c implies that there is a c' for which f(x)=c' implies that c is not c', which sounds like there is a x for which f(x) does have two results c and c' which wouldn't be a funtion anymore. Is my math bad or am I misunderstanding the resulting term?

  • 3
    $\begingroup$ Your original quantified statement is not correct. $\endgroup$ – Ink Jan 13 '13 at 23:52
  • $\begingroup$ well ok, then my math is wrong. I'd appreciate the correct term. I'd accept that as an answer then $\endgroup$ – Sim Jan 13 '13 at 23:53
  • $\begingroup$ This is the quantification you want. $\exists! c \in \mathbb{R} \forall x \in \mathbb{R}:f(x) = c $ $\endgroup$ – Ink Jan 13 '13 at 23:55
  • $\begingroup$ I think you made a mistake on the last assertion. The $\wedge$ became $\Longrightarrow$. $\endgroup$ – Git Gud Jan 13 '13 at 23:56
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    $\begingroup$ $\forall x \exists ! c\in\mathbb{R}:f(x)=c$ is true for any well-defined function. $\endgroup$ – Eckhard Jan 13 '13 at 23:57

This is because you somewhat messed up the negation process. This is a reason not to favor the "There is exactly one ..."

Let me take a different route than you.

$f\colon\Bbb{R\to R}$ is constant if and only if $\forall x\forall y(f(x)=f(y))$. Therefore the negation of that is:

$$ \lnot(\forall x\forall y(f(x)=f(y)))\iff\\ \exists x\lnot(\forall y(f(x)=f(y)))\iff\\ \exists x\exists y\lnot(f(x)=f(y))\iff\\ \exists x\exists y(f(x)\neq f(y))\qquad $$

Namely $f$ is not constant if and only if there are $x$ and $y$ such that $f(x)\neq f(y)$. The reason you had a problem is that $\exists!$ hides the following formula:

$$\exists!x\varphi(x)\iff\exists x(\varphi(x)\land\forall y(\varphi(y)\rightarrow x=y))$$

So when we apply a negation to $\exists!$ we need to apply the negation to the above formula, you seem to have made a mistake in applying the negation to the inner formula which is what caused the mistake, but generally I find it better to express things without "exists a unique ...", especially if we wish to negate them later.

Also as pointed out in the comments, you switched the order of the quantifiers. $\forall x\exists!y f(x)=y$ is the definition of a function; but a constant function would be $\exists!y\forall x(f(x)=y)$. In fact you don't even need the uniqueness, $\exists c\forall x(f(x)=c)$ suffices to show that $f$ is constant.


$$\forall x \in \mathbb{R}\, \exists!c \in \mathbb{R}:f(x) = c$$ says that for every $x$ there is a unique $c$ such that $f(x) = c$: and that only says that $f$ is total and single-valued (not multi-valued). It plainly does not say that there is one and only one value which $f$ takes for every argument, i.e. that $f$ is a constant function, which is $$\exists!c \in \mathbb{R\,}\forall x \in \mathbb{R}:f(x) = c.$$ The order of quantifiers matters crucially!


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