Bounding $\delta$ without the loss of generality while proving non uniform continuity

Question

To show $f(x)$ isn't uniformly continuous on an interval $I$, I show that there is an $\epsilon$ such that for every $\delta$ there exist $x, y \in I$ such that $|x-y|<\delta$ and $|f(x)-f(y)|\ge\epsilon$.

In most of these proofs I have to put some boundary on $\delta$ to get to a solution. For example - $\frac 1 x$ isn't uniformly continuous on $(0,\infty)$, since I can choose $\epsilon=1$ and $x=\delta, y=\frac\delta2$.

Now I have $|f(x)-f(y)|=|\frac 1 \delta - \frac 2 \delta |=\frac 1 \delta $

To say this is bigger then $\epsilon$, I should say that without the loss of generality - $0<\delta<1$.

Only then I get the desired result - $\frac 1 \delta > 1 \ge \epsilon$.

But how is it legal to bound $\delta$, if the definition is "for all $\delta$"?

Answer 1

As you wrote:

for every $\delta$ there exist $x, y \in I$ such that $|x-y|<\delta$ and $|f(x)-f(y)|\ge\epsilon$.

This means, that you have to pick an arbitrary $\delta$ and find the pair $x,y$ (with distance among them less than $\delta$) for which the inequality holds.

You, however, picked the pairs first, saying:

$x=\delta, y=\frac\delta2$

This naturally bounded $\delta$.

The definition tells you, that "for every $\delta$ there is a pair..." and yet you treat it as "for every pair there's a $\delta$...".

Once again: If you manually choose some $x,y$ pair (which are, in fact, close enough, so that $|x-y|<\delta$), you lose generality. You have to find the fitting pair, not choose them. That is, what bounds $\delta$.

The correct way to go is:

1. Suppose, you have $\delta$;
2. Try, finding $\{x,y,\epsilon\}$ such that, these two conditions hold for any $\delta$:

(a) $|x-y|<\delta$;

(b) $|f(x)-f(y)|\ge\epsilon$

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As for your example: we've got $f(x)=1/x$

I'll first eliminate $y$ by defining $y=x+\delta$. That way, $d(x,y)<\delta$ (I know, here needs to be a $\le$ sign, but it doesn't really matter. If you substitute $x+\delta$ by $x+\delta/2$, you'll see, the end result is the same).

Now, all there's left to do is find out, if there is a pair $x,\epsilon$, such that $|f(x)-f(x+\delta)|\ge\epsilon$.

$$|f(x)-f(x+\delta)|\ge\epsilon$$

$$|1/x-1/(x+\delta)|\ge\epsilon$$

$$|\frac{\delta}{x(x+\delta)}|\ge\epsilon$$

$\delta$ was known to be positive, $x$ is also positive, because we are interested in $(0,\infty)$, so we get rid of module signs:

$$\frac{\delta}{x(x+\delta)}\ge\epsilon$$

$$\frac{\delta}{\epsilon}\ge x(x+\delta)$$

$$x^2+x\delta-\delta/\epsilon \le 0$$

The question reduces to: "is there a positive $\epsilon$, so that the last inequality holds?"

And actually - there is. The inequality holds if the quadratic equation $x^2+x\delta-\delta/\epsilon = 0$ has two solutions, that is, when its discriminant is any real nonzero number.

$$D = \delta^2 + 4 \delta/\epsilon > 0$$

$$ \delta^2 > -4\delta/\epsilon $$

$$ \delta > -4 / \epsilon $$

Actually, $\delta$ being positive, this inequality holds for any nonzero $\epsilon$.

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To sum up: The question is, can we find a single $\epsilon > 0$, which leaves the possibility for any $\delta > 0$ to find $x$ (or, the pair $x,y$), holding aforesaid inequalities.

I hope, I've written everything as clearly, as it gets.

Answer 2

Suppose you could prove your statement for any $\delta$ below some bound. That is, suppose you find some $\delta_0>0$ so that you can find $x,y\in I$ with $|x-y|<\delta$ and $|f(x)-f(y)|\geq\epsilon$ for any $\delta\leq\delta_0$.

You can extend this to the rest of the values of $\delta$ easily: Suppose $\delta>\delta_0$. Then take the pair $(x,y)$ corresponding to $\delta=\delta_0$. Then $|x-y|<\delta_0\leq\delta$ and $|f(x)-f(y)|\geq\epsilon$ as desired.

This is why you can restrict the values of $\delta$ from above. (But mind you that you cannot restrict them from below, since you want to show the thing for arbitrarily small $\delta$.)