A vacuously existing function?

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In summary, the conjecture in this conversation states that for any real number a, real-valued function f defined on [a,a]=\{a\}, and real number x in [a,a], there exists a function phi defined by {\displaystyle \phi(t)=\frac{f(t)-f(x)}{t-x}\quad(a<t<a,t\neq x)}. This function has an empty domain and therefore is vacuously created. The purpose of this function is to solve problems regarding the derivative of a function defined on a singleton, but this may not be a necessary or useful concept in general mathematics.
  • #1
julypraise
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Homework Statement


Conjecture. Suppose [itex]a\in \mathbb{R}[/itex]. Suppose [itex]f[/itex] is a real-valued function defined on [itex][a,a]=\{a\}[/itex]. Suppose [itex]x\in [a,a][/itex]. Then there exists a function [itex]\phi[/itex] defined by [itex]{\displaystyle \phi(t)=\frac{f(t)-f(x)}{t-x}\quad(a<t<a,t\neq x)}[/itex].

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

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

Homework Equations



Relevant posts are:

https://www.physicsforums.com/showthread.php?t=585386
https://www.physicsforums.com/showthread.php?t=338366

The Attempt at a Solution



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

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

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

(ii) I think this function is simply [itex]\emptyset[/itex] because the domain is empty set.
 
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  • #2
julypraise said:

Homework Statement


Conjecture. Suppose [itex]a\in \mathbb{R}[/itex]. Suppose [itex]f[/itex] is a real-valued function defined on [itex][a,a]=\{a\}[/itex]. Suppose [itex]x\in [a,a][/itex]. Then there exists a function [itex]\phi[/itex] defined by [itex]{\displaystyle \phi(t)=\frac{f(t)-f(x)}{t-x}\quad(a<t<a,t\neq x)}[/itex].

There is no number [itex]t[/itex] satisfying [itex]a < t < a[/itex]. Therefore, there is no [itex]t[/itex] for which you have defined [itex]\phi(t)[/itex]. 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.
 
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  • #3
P.S. You can equally well have said the following. Let S be any set (even the empty set) and define

[tex]\phi : \emptyset \rightarrow S[/tex]

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

Yes, this function exists. It is the set of points (a, s) such that [itex]a \in \emptyset[/itex] and [itex]\phi(a) = s[/itex]. Since there is no [itex]a[/itex] satisfying [itex]a \in \emptyset[/itex], the function is simply the empty set, as you indicated.
 
  • #4
jbunniii said:
There is no number [itex]t[/itex] satisfying [itex]a < t < a[/itex]. Therefore, there is no [itex]t[/itex] for which you have defined [itex]\phi(t)[/itex]. 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.

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

https://www.physicsforums.com/showthread.php?p=3808228&posted=1#post3808228
 
  • #5
julypraise said:
Maybe I do not know clearly the meaning of conjecture. Anyway what I hope to achieve with this function is to solve some problems that I posted on this post:

https://www.physicsforums.com/showthread.php?p=3808228&posted=1#post3808228

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

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
jbunniii said:
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.

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.)

Anyway as for this post, please answer to that post.

You know, all I want to get is a proper definition of derivative that I can use on a function defined on a singleton (at an isolated point). But maybe (what I'm doing) it's something time-wasting probably because general mathematicians wouldn't care about the derivative of a function at an isolated point.
 

Related to A vacuously existing function?

What is a vacuously existing function?

A vacuously existing function is a function that does not have any inputs or outputs. It has no defined domain or range, and therefore does not perform any operations or calculations. It is essentially a function without any meaningful purpose or impact.

How is a vacuously existing function different from a null function?

A vacuously existing function and a null function are often used interchangeably, but there is a subtle difference. A vacuously existing function has no inputs or outputs, while a null function may have inputs but does not produce any outputs.

What is the significance of a vacuously existing function in mathematics?

In mathematics, a vacuously existing function can be used as a placeholder or a base case for more complex functions. It can also be used to define certain properties or relationships between functions.

Can a vacuously existing function be useful in real-world applications?

Not typically, as it does not perform any meaningful operations or calculations. However, in some cases, it may be used in theoretical or abstract mathematical concepts.

How is a vacuously existing function represented mathematically?

A vacuously existing function is often represented by an empty set or the Greek letter "phi" (φ). It can also be written as f(x) = ∅ or f(x) = φ.

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