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This question is based on this question.

I was wondering why the author did not end the proof by expressing the $\delta$ in terms of $\epsilon$ like he did in prior examples.

Is this because it is not always possible to express the $\delta$ in terms of $\epsilon$ or is it to do so requires a higher level of mathematics that the author has not covered yet?

Thank you in advance for any help provided.

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    $\begingroup$ What do you mean "express in terms of"? I think you'll find this is the crux of the question. $\endgroup$ – Alex Becker Jul 26 '13 at 21:48
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    $\begingroup$ When you are proving say that if $f$ and $g$ are continuous at $a$, then $fg$ is continuous at $a$, the $\delta$ cannot be absolutely explicit, since it depends on unknown $\delta_f$, $\delta_g$ for the two functions $f$ and $g$. And in general, $\delta$ does not have to be explicit, existence is enough. That said, one can often produce explicit $\delta$. But except in very simple cases, it can be difficult to produce the largest $\delta$ that will work. Luckily, one does not have to in the usual $\epsilon$-$\delta$ proof. $\endgroup$ – André Nicolas Jul 26 '13 at 22:36
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In fact, it turns out that one can choose $\delta$ continuously in terms of $\varepsilon$. More precisely, the following result holds true : If $f:E\to F$ is a continuous map between two metric spaces $E$ and $F$, then there is a continuous function $\delta:E\times (0,\infty)\to (0,\infty)$ such that, for all $(x,y,\varepsilon)\in E\times E\times (0,\infty)$, $$d_E(x,y)<\delta(x,\varepsilon)\implies d_F(f(x),f(y))<\varepsilon\, . $$

I read this several years ago in an American Math. Monthly article, but I can remember neither the title of the article, nor the name(s) of the author(s). It would be nice if someone could give me the reference...

Here is one way of proving this result. Consider the set $$C=\{ (x,y,\varepsilon)\in E\times E\times (0,\infty);\; d(f(x),f(y))\geq \varepsilon\, . $$ Since $f$ is continuous, $C$ is closed in $E\times E\times (0,\infty)$ (endowed with the product topology), and clearly $(x,x,\varepsilon)\not\in C$ for any $(x,\varepsilon)\in E\times (0,\infty)$. It follows that if we fix a metric $\rho$ on $E\times E\times(0,\infty)$ (compatible with the product topology), then ${\rm dist}_\rho \left((x,x,\varepsilon),C\right)>0$.

We take for $\rho$ the metric $$\rho((x,y,\varepsilon), (x',y',\varepsilon'))=\max(d_E(x,x'),d_E(y,y'),\vert\varepsilon-\varepsilon'\vert) $$ and we define $\delta :E\times (0,\infty)\to (0,\infty)$ by $$\delta(x,\varepsilon)={\rm dist}_\rho \left((x,x,\varepsilon),C\right) $$ Then the dunction $\delta$ is continuous, and it is straightforward to check that it has the required property. Indeed, if $x,y\in E$ and $\varepsilon >0$, then $d_E(x,y)=\rho((x,x,\varepsilon), (x,y,\varepsilon))$ by the definition of the metric $\rho$. Hence, if $d_E(x,y)<\delta(x,\varepsilon)$, then $(x,y,\varepsilon)\not\in C$ by the very definition of $\delta(x,\varepsilon)$, i.e. $d_F(f(x),f(y))<\varepsilon$.

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  • $\begingroup$ This is analysis, not calculus, +1. $\endgroup$ – Julien Jul 26 '13 at 22:49
  • $\begingroup$ That's too bad... $\endgroup$ – Etienne Jul 27 '13 at 4:07
  • $\begingroup$ Anyway, all the foundations of real analysis are wrong, as proven here. $\endgroup$ – Julien Jul 27 '13 at 4:13
  • $\begingroup$ A great paper, indeed. $\endgroup$ – Etienne Jul 27 '13 at 11:54
  • $\begingroup$ @julien Does this mean that this answer is not acceptable? Is there another method to come up with a formula that expresses $\delta$ in terms of $\epsilon$? $\endgroup$ – mauna Jul 27 '13 at 19:20
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In these proofs, $\delta$ doesn't always need to be directly written as a function of $\epsilon$ in the usual sense. However, you do need to show that for any $\epsilon$, you can find a satisfactory $\delta$, and the author does exactly that.

Namely, he says that for any $\epsilon>0$, you can follow the following process:

  1. choose an integer $n$ so that $1/n<\epsilon$ (or equivalently, $n>\frac{1}{\epsilon}$).
  2. list the numbers $$\frac{1}{2}; \frac{1}{3}, \frac{2}{3}; \frac{1}{4}, \frac{3}{4}; \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}; \dots ; \frac{1}{n}, \dots , \frac{n-1}{n}$$ and from those, let $\frac{p}q$ be the closest one not equal to $a$.
  3. let $\delta=\frac{p}q$

From there, the author proceeds to show that this choice of $\delta$ works. Can we create some function that takes any $\epsilon$ and does this, spitting out $\delta=\frac{p}q$? Sure. Is that necessary? No; all we needed to show is that for any $\epsilon>0$, we can find this $\delta$.

The reason for writing $\delta$ as some direct function of $\epsilon$ (such as $\epsilon/2$ or $1/\epsilon$) is that in other instances, this is the most concise way to describe the process of finding $\delta$.

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If you are thinking about continuity of a function at a point $a$, you only need to find $\delta$ such that $d(x,a)<\delta$ implies $d(f(x),f(a))<\epsilon$.

But it is possible to obtain $\delta$ which works for every $\epsilon$.

For example, take $f:\mathbb{R}\to \mathbb{R}$ constant, e.g., $f(x)=2$, for all $x\in \mathbb{R}$. Then, given $\epsilon>0$, you can choose any $\delta>0$ and the statement above holds since $d(f(x),f(a))=d(2,2)=0<\epsilon$.

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