Topsy-turvy continuity

Ivar Sand · 2013-01-30 00:09:39

Background. As you would now, the continuity(1)(2) 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 proceeding with a condition in f's domain ("then there exists a δ > 0 so that ∣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 topsy-turvy 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 topsy-turvy 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 topsy-turvy 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 topsy-turvy continuous function should be continuous. However, the topsy-turvy continuity concept may be weaker than standard continuity so that a continuous function may not be topsy-turvy continuous at all points. Therefore, topsy-turvy continuity should imply continuity. Indeed it does, as is stated in the following theorem.

Theorem. A topsy-turvy continuous function f is continuous.
Proof: Let ε > 0 be given. Choose some point x_0 in the domain of f. Since f is topsy-turvy 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 topsy-turvy 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, One-Variable Calculus, with an Introduction to Linear Algebra, Second edition, 1967, p. 130-131.
(2) The wikipedia page for Continuous function.

Last edited by Ivar Sand (2023-09-13 02:49:57)

Bob · 2013-01-30 01:35:51

hi Ivar,

Have you considered the converse theorem?

Bob

anonimnystefy · 2013-01-30 02:18:41

Where did you get the definition in the first paragraph from, because I see a different one on Wikipedia...

Ivar Sand · 2013-01-30 22:07:24

No, Bob, I haven't. I suppose you are thinking about whether continuity implies topsy-turvy continuity. I don't think this is true so a counter-proof 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, topsy-turvy 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 topsy-turvy 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 topsy-turvy continuous at x_0.

Last edited by Ivar Sand (2023-09-13 20:30:57)

Ivar Sand · 2013-01-30 22:44:14

Well, anonimnystefy, let's see. We go to the Wikipedia page named "Continuous function", the paragraph "Weierstrass and Jordan definitions (epsilon–delta) of continuous functions" and the sentence "For any positive real number ..." 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. Topsy-turvy 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 topsy-turvy 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.

Last edited by Ivar Sand (2023-09-13 20:51:48)

Bob · 2013-01-30 23:33:35

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://www-math.mit.edu/~djk/calculus_b … ion02.html

But does that invalidate your proof? Hmmmm. Thinking ........................

Bob

Ivar Sand · 2013-01-31 21:16:55

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 (2023-09-13 20:54:21)

Ivar Sand · 2013-02-03 23:26:59

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.

Math Is Fun Forum

#1 2013-01-30 00:09:39

Topsy-turvy continuity

#2 2013-01-30 01:35:51

Re: Topsy-turvy continuity

#3 2013-01-30 02:18:41

Re: Topsy-turvy continuity

#4 2013-01-30 22:07:24

Re: Topsy-turvy continuity

#5 2013-01-30 22:44:14

Re: Topsy-turvy continuity

#6 2013-01-30 23:33:35

Re: Topsy-turvy continuity

#7 2013-01-31 21:16:55

Re: Topsy-turvy continuity

#8 2013-02-03 23:26:59

Re: Topsy-turvy continuity

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