
Topsyturvy continuity
Background. As you would now, the continuity^{12} of a function f at x_{0}, a point in the domain of f, is defined using a condition in f's codomain first ("given ε > 0
∣f(x)  f(x_{0})∣ < ε") and then proceeds with a condition in f's domain ("
then there exists a δ > 0
∣x  x_{0}∣ < δ").
Once a friend of mine remarked that he felt that the definition of continuity is unnatural in that it starts in a function's codomain instead of in its domain. I remember that I had the same feeling myself when I learned about continuity. I guess that the natural approach, as indicated by my friend, was adopted first, and that the standard definition came later and proved to be more fruitful than the natural approach in some way. (This seems indeed to have been the case^{2}.)
I thought that it might be fun to see what such a natural approach would look like. (If you disagree this is the place to stop reading this thread.) So I put my mathematical explorer's boots on and dived into the jungle of mathematics. I wanted to see if I could build myself a natural definition of continuity and compare this concept with the existing definition of continuity. I decided to call this natural definition of continuity topsyturvy continuity because it does things in the opposite way as compared to the existing definition of continuity. The result follows below.
Definition. A function f is said to be topsyturvy continuous at a point x_{0} if for a given δ > 0 a k > 0 exists such that whenever x  x_{0} < δ, it is true that f(x)  f(x_{0}) ≤ k x  x_{0}.
Comments:  This is just a simple definition of topsyturvy continuity.  Note that k may depend on x_{0} and δ.  Note that ≤ is used instead of <. This is to cover the case x = x_{0} (a < would have given 0 < 0).
Now we must check that this definition relates to the standard definition of continuity in a proper way. A topsyturvy continuous function should be continuous. However, the topsyturvy continuity concept may be weaker than standard continuity so that a continuous function may not be topsyturvy continuous at all points. Therefore, topsyturvy continuity should imply continuity. Indeed it does, as is stated in the following theorem.
Theorem. A topsyturvy continuous function f is continuous. Proof: Let ε > 0 be given. Choose some point x_{0} in the domain of f. Since f is topsyturvy continuous at x_{0}, a positive k exists so that when x  x_{0} < ε it is true that f(x)  f(x_{0}) ≤ k x  x_{0}.
Define δ = min (ε, ε/k). We observe that δ > 0. Since we have δ ≤ ε, the topsyturvy continuity condition f(x)  f(x_{0}) ≤ k x  x_{0} still holds with the same constant k when x  x_{0} < δ.
When x  x_{0} < δ it is also true that x  x_{0} < ε/k since δ ≤ ε/k, and multiplying both sides by k yields k x  x_{0} < k*ε/k = ε. Using this in f(x)  f(x_{0}) ≤ k x  x_{0} gives f(x)  f(x_{0}) < ε still maintaining x  x_{0} < δ.
We conclude that for an ε > 0 a δ > 0 exists so that f(x)  f(x_{0}) < ε whenever x  x_{0} < δ, and this means that f is continuous at x_{0}. QED.
References: 1 Tom M. Apostol, Calculus, Volume 1, OneVariable Calculus, with an Introduction to Linear Algebra, Second edition, 1967, p. 130131. 2 The wikipedia page for Continuous function.
I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway.
 bob bundy
 Moderator
Re: Topsyturvy continuity
hi Ivar,
Have you considered the converse theorem?
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Topsyturvy continuity
Where did you get the definition in the first paragraph from, because I see a different one on Wikipedia...
The limit operator is just an excuse for doing something you know you can't. It's the subject that nobody knows anything about that we can all talk about! ― Richard Feynman Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
Re: Topsyturvy continuity
No, Bob, I haven't. I suppose you are thinking about whether continuity implies topsyturvy continuity. I don't think this is true so a counterproof would be called for. I believe the problem is connected to points in f's domain where f has an infinite derivative. Let's have a closer look.
Let f have an infinite derivative at a point x_{0}. This means that (f(x)  f(x_{0})) / (x  x_{0}) approaches infinity as x approaches x_{0}, or, in other words, that for every M > 0 a δ > 0 exists so that (f(x)  f(x_{0})) / (x  x_{0}) > M whenever x  x_{0} < δ and x ≠ x_{0}. (We need the condition x ≠ x_{0} to avoid 0/0, which is undefined.) Now we take the absolute value of the left hand side of this inequality and multiply both sides by x  x_{0}, and we get f(x)  f(x_{0}) > M x  x_{0} whenever x  x_{0} < δ and x ≠ x_{0} (1)
Now, topsyturvy continuity at x_{0} means: given a δ' > 0, a k > 0 exists so that f(x)  f(x_{0}) ≤ k x  x_{0} whenever x  x_{0} < δ' (2)
Let us try to prove that f is not topsyturvy continuous at x_{0}. We do that by proposing (2) to be true and see if we can derive a contradiction.
We choose M = k + 1 in (1). Regard the point x_{1} = x_{0} + min(δ',δ)/2. Since x_{1}  x_{0} < δ and x_{1}  x_{0} < δ' x_{1} fits into both (1) and (2). From (1) we get: f(x_{1})  f(x_{0}) > (k + 1) x_{1}  x_{0}, and from (2) we get: f(x_{1})  f(x_{0}) ≤ k x_{1}  x_{0}. These two inequalities are contradictory to each other, and we conclude that our proposition of (2) being true is false. Therefore f is not topsyturvy continuous at x_{0}.
Last edited by Ivar Sand (20130204 22:24:02)
I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway.
Re: Topsyturvy continuity
Well, anonimnystefy, let's see. We go to the Wikipedia page named "Continuous function", the paragraph "Weierstrass definition (epsilondelta) of continuous functions" and the sentence "For any number ε > 0,
". This sentence expresses something like: given an ε > 0, a δ > 0 exists
, where ε is part of a condition in the codomain and δ is part of a condition in the domain. Topsyturvy continuity, on the other hand, looks like: given a δ > 0, a k > 0 exists
, where δ is part of a condition in the domain and k is part of a condition in the codomain.
To summarise: The continuity of a function f at x_{0} is defined: Given an ε > 0, a δ > 0 exists so that ∣f(x)  f(x_{0})∣ < ε whenever ∣x  x_{0}∣ < δ. The topsyturvy continuity of a function f at x_{0} is defined: Given a δ > 0 a k > 0 exists so that f(x)  f(x_{0}) ≤ k x  x_{0} whenever x  x_{0} < δ.
I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway.
 bob bundy
 Moderator
Re: Topsyturvy continuity
hi Ivar Sand
I need to think about your definition some more. Most sources seem to regard functions where the derivative goes to infinity at a point, as not differentiable there. eg This site specifically excludes y = x^(1/3) at x = 0.
http://wwwmath.mit.edu/~djk/calculus_b … ion02.html
But does that invalidate your proof? Hmmmm. Thinking ........................
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Re: Topsyturvy continuity
I see, Bob. I agree that f is not differentiable at x_{0}. I'm sorry about that. I propose to replace in post #4: "Suppose f is differentiable at a point x_{0} and that its derivative at x_{0} is infinite" by: "Suppose f is not differentiable at a point x_{0} because its derivative is infinite there"
Last edited by Ivar Sand (20130201 20:22:16)
I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway.
Re: Topsyturvy continuity
Now I have updated post #4. I have removed the word differentiable and also made other changes in order to make the mathematics more precise.
I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway.
