Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Ivar Sand****Member**- Registered: 2013-01-22
- Posts: 14

Background. As you would now, the continuity[sup]12[/sup] of a function f at x[sub]0[/sub], a point in the domain of f, is defined using a condition in f's codomain first ("given ε > 0 ∣f(x) - f(x[sub]0[/sub])∣ < ε") and then proceeds with a condition in f's domain (" then there exists a δ > 0 ∣x - x[sub]0[/sub]∣ < 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[sup]2[/sup].)

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[sub]0[/sub] if for a given δ > 0 a k > 0 exists such that whenever |x - x[sub]0[/sub]| < δ, it is true that |f(x) - f(x[sub]0[/sub])| ≤ k |x - x[sub]0[/sub]|.

Comments:

- This is just a simple definition of topsy-turvy continuity.

- Note that k may depend on x[sub]0[/sub] and δ.

- Note that ≤ is used instead of <. This is to cover the case x = x[sub]0[/sub] (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[sub]0[/sub] in the domain of f. Since f is topsy-turvy continuous at x[sub]0[/sub], a positive k exists so that when |x - x[sub]0[/sub]| < ε it is true that |f(x) - f(x[sub]0[/sub])| ≤ k |x - x[sub]0[/sub]|.

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[sub]0[/sub])| ≤ k |x - x[sub]0[/sub]| still holds with the same constant k when |x - x[sub]0[/sub]| < δ.

When |x - x[sub]0[/sub]| < δ it is also true that |x - x[sub]0[/sub]| < ε/k since δ ≤ ε/k, and multiplying both sides by k yields k |x - x[sub]0[/sub]| < k*ε/k = ε. Using this in |f(x) - f(x[sub]0[/sub])| ≤ k |x - x[sub]0[/sub]| gives |f(x) - f(x[sub]0[/sub])| < ε still maintaining |x - x[sub]0[/sub]| < δ.

We conclude that for an ε > 0 a δ > 0 exists so that |f(x) - f(x[sub]0[/sub])| < ε whenever |x - x[sub]0[/sub]| < δ, and this means that f is continuous at x[sub]0[/sub]. 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.

I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,376

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

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,518

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

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

Offline

**Ivar Sand****Member**- Registered: 2013-01-22
- Posts: 14

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[sub]0[/sub]. This means that (f(x) - f(x[sub]0[/sub])) / (x - x[sub]0[/sub]) approaches infinity as x approaches x[sub]0[/sub], or, in other words, that for every M > 0 a δ > 0 exists so that (f(x) - f(x[sub]0[/sub])) / (x - x[sub]0[/sub]) > M whenever |x - x[sub]0[/sub]| < δ and x ≠ x[sub]0[/sub]. (We need the condition x ≠ x[sub]0[/sub] 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[sub]0[/sub]|, and we get

|f(x) - f(x[sub]0[/sub])| > M |x - x[sub]0[/sub]| whenever |x - x[sub]0[/sub]| < δ and x ≠ x[sub]0[/sub] (1)

Now, topsy-turvy continuity at x[sub]0[/sub] means: given a δ' > 0, a k > 0 exists so that

|f(x) - f(x[sub]0[/sub])| ≤ k |x - x[sub]0[/sub]| whenever |x - x[sub]0[/sub]| < δ' (2)

Let us try to prove that f is not topsy-turvy continuous at x[sub]0[/sub]. 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[sub]1[/sub] = x[sub]0[/sub] + min(δ',δ)/2. Since |x[sub]1[/sub] - x[sub]0[/sub]| < δ and |x[sub]1[/sub] - x[sub]0[/sub]| < δ' x[sub]1[/sub] fits into both (1) and (2). From (1) we get:

|f(x[sub]1[/sub]) - f(x[sub]0[/sub])| > (k + 1) |x[sub]1[/sub] - x[sub]0[/sub]|,

and from (2) we get:

|f(x[sub]1[/sub]) - f(x[sub]0[/sub])| ≤ k |x[sub]1[/sub] - x[sub]0[/sub]|.

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[sub]0[/sub].

*Last edited by Ivar Sand (2013-02-03 23:24:02)*

I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway.

Offline

**Ivar Sand****Member**- Registered: 2013-01-22
- Posts: 14

Well, anonimnystefy, let's see. We go to the Wikipedia page named "Continuous function", the paragraph "Weierstrass definition (epsilon-delta) 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. 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[sub]0[/sub] is defined:

Given an ε > 0, a δ > 0 exists so that ∣f(x) - f(x[sub]0[/sub])∣ < ε whenever ∣x - x[sub]0[/sub]∣ < δ.

The topsy-turvy continuity of a function f at x[sub]0[/sub] is defined:

Given a δ > 0 a k > 0 exists so that |f(x) - f(x[sub]0[/sub])| ≤ k |x - x[sub]0[/sub]| whenever |x - x[sub]0[/sub]| < δ.

I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,376

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Ivar Sand****Member**- Registered: 2013-01-22
- Posts: 14

I see, Bob. I agree that f is not differentiable at x[sub]0[/sub]. I'm sorry about that. I propose to replace in post #4:

"Suppose f is differentiable at a point x[sub]0[/sub] and that its derivative at x[sub]0[/sub] is infinite"

by:

"Suppose f is not differentiable at a point x[sub]0[/sub] because its derivative is infinite there"

*Last edited by Ivar Sand (2013-01-31 21:22:16)*

Offline

**Ivar Sand****Member**- Registered: 2013-01-22
- Posts: 14

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.

Offline

Pages: **1**