Talk:Uniform continuity

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This page doesn't need math mode, for anyone with a Unicode-compatible browser.

We should consider making the math renderer produce HTML for simple expressions like this that can trivially be rendered as HTML + Unicode.

Yes, please, in my browser the equations are about twice the font size of the rest of the article and it's quite jarring. I don't see any way to overcome this using PNG equations, since there's no way to know how any given browser will render the text. But as another option, how about a perference setting that causes equations to be displayed in MathML instead? If a browser knows how to handle MathML, then it probably knows how to make it look good as well. Bryan
We're not XML-clean yet, so we can't embed MathML in our output stream and expect anything to be able to use it. --Brion 08:45 Jan 7, 2003 (UTC)

The "In other words, the Slope is Bounded" comment is false. sqrt(x) is uniformily continuous on [0, infinity) but the slope can be made arbitrarily large by looking at points near 0.

\frac{\sqrt{x}-\sqrt{0}}{x - 0} = \frac{1}{\sqrt{x}}

so i'm taking down the "slope is bounded" comment. AlfredR 05:00, 6 December 2006 (UTC)

[edit] Continuous functions on a closed real interval

The introduction stated that continuous functions on a closed interval are uniformly continuous. We need compactness here, as the example f:\mathbb{R}\rightarrow\mathbb{R}, f(x) = x2 shows: Choose ε = 1, let δ > 0 and set y = x − δ / 2. Then | x − y | = δ / 2 < δ, but |x^2-y^2| = |x-y| |x+y| = \delta/2 |2x-\delta/2| \rightarrow \infty as x \rightarrow \infty. --Lacce (talk) 13:45, 22 December 2007 (UTC)

[edit] You gave a wrong definition of uniformly continous

When the author writes

If X and Y are subsets of the real numbers, d1 and d2 can be the standard Euclidian norm, |\cdot|, yielding the definition: if there exists a δ > 0 such that, for all ε > 0 and x,y \in X, | x − y | < δ implies | f(x) − f(y) | < ε.

s/he is trivially wrong. The right definition over the reals is

for every ε > 0, there exists a δ > 0 (depending on ε!!!) such that, for all x,y \in X, | x − y | < δ implies | f(x) − f(y) | < ε.

It's clearly a typo, since the author gave the right definition in the case of general metric spaces. If you use the definition I'm claiming to be wrong, even the identity function f(x)=x is not uniformly continous: let suppose that such a δ exists, and call it D. Let's take x and x+D/2, wich are closer than D. You will have |f(x)-f(x+D/2)| = D/2, wich is far to be smaller of any possible ε > 0... With the right definition, you will just take δ(ε) = ε and everything works fine. G2h2 (talk) 08:23, 7 April 2008 (UTC)

The comment is correct, the sentence
if there exists a δ > 0 such that, for all ε > 0 and x,y \in X, | x − y | < δ implies | f(x) − f(y) | < ε.
should become
if for all ε > 0 there exists a δ > 0 such that, for all x,y \in X, | x − y | < δ implies | f(x) − f(y) | < ε.
Stoppato (talk) 10:04, 7 April 2008 (UTC)

[edit] misleading lead?

The lead says "furthermore the size of the changes in f(x) depends only on the size of the changes in x but not on x itself". I know this is meant to be an informal statement, so maybe I should not be too picky, but I find this somewhat misleading. The changes in f(x) do depend on x (as well as on the size of the changes to x) but they can be bounded in a way that only depends on the size of the changes. Would saying that make the lead too technical? (I think anybody wanting to learn about uniform continuity will have to be prepared to encounter some technicality anyway.) Marc van Leeuwen (talk) 16:34, 17 May 2008 (UTC)