# Homework Help: A vacuously existing function?

1. Mar 10, 2012

### julypraise

1. The problem statement, all variables and given/known data
Conjecture. Suppose $a\in \mathbb{R}$. Suppose $f$ is a real-valued function defined on $[a,a]=\{a\}$. Suppose $x\in [a,a]$. Then there exists a function $\phi$ defined by ${\displaystyle \phi(t)=\frac{f(t)-f(x)}{t-x}\quad(a<t<a,t\neq x)}$.

(i) Before proving (or disproving this) does this conjecture make sense in the first place?

(ii) If make sense, does it truely exist?

2. Relevant equations

Relevant posts are:

3. The attempt at a solution

(i) If I kinda restate this conjecture, it becomes:

Conjecture. Suppose $a\in \mathbb{R}$. Suppose $f$ is a real-valued function defined on $[a,a]=\{a\}$. Suppose $x\in [a,a]$. Then there exists a function ${\displaystyle \phi:\{t\in \mathbb{R}: a<t<a\} \to \mathbb{R} : t \mapsto \frac{f(t)-f(x)}{t-x}}$.

So it seems make sense in the ground of first order language and ZFC. Isn't it?

(ii) I think this function is simply $\emptyset$ because the domain is empty set.

Last edited: Mar 10, 2012
2. Mar 10, 2012

### jbunniii

There is no number $t$ satisfying $a < t < a$. Therefore, there is no $t$ for which you have defined $\phi(t)$. So you can say that you have vacuously created a function whose domain is the empty set. I'm not sure why it merits being called a "conjecture" or what you hope to achieve with this function.

Last edited: Mar 10, 2012
3. Mar 10, 2012

### jbunniii

P.S. You can equally well have said the following. Let S be any set (even the empty set) and define

$$\phi : \emptyset \rightarrow S$$

As the domain is empty, you don't need to specify any "formula" for how to "evaluate" $\phi$.

Yes, this function exists. It is the set of points (a, s) such that $a \in \emptyset$ and $\phi(a) = s$. Since there is no $a$ satisfying $a \in \emptyset$, the function is simply the empty set, as you indicated.

4. Mar 10, 2012

### julypraise

Maybe I do not know clearly the meaning of conjecture. Anyway what I hope to acheive with this function is to solve some problems that I posted on this post:

5. Mar 10, 2012

### jbunniii

OK, to answer your question in post #3 of that thread, yes, your quotient definition defines an empty function, and that empty function exists (in the same sense in which the empty set exists). There's no logical issue that I can see with your definition.

It's also true that your statement in post #5 of that thread is vacuously true for any value L.

However, the (usual) definition of a limit isn't merely that statement. I don't have Rudin here with me, so I can't check his definition, but most authors define the limit of a function at a point x as follows:

"Let $f : A \rightarrow B$ be a function, and let $x$ be an accumulation point of $A$. Then we write $\lim_{t \rightarrow x} f(t) = L$ if for every $\epsilon > 0$..."

i.e. the notion of a limit is defined only at accumulation points of the domain. Since the empty set has no accumulation points, the notion of a limit of an empty function is undefined.

6. Mar 13, 2012

### julypraise

Yes, what you say is exactly true and I agree with this to a full extent. But if you read my post carefully, you will see that the problem I proposed arises exactly because of what you said, that is, because the notion of limit is undefined at a point which is not a limit point, I can't use Rudin's definition to get the deriviative of a function defined at a singleton. (The problem is not of limit-definition but derivative-definition.)