Order in which sets are constructed and epsilon-delta proofs

Question

In epsilon-delta proofs, for example, we might do something like this:

Prove that $\lim_{x\to 0}2x =0$

$0 < |x| < \delta \implies |f(x)-0| < \varepsilon$

Of course, we can choose $\delta < \frac{\varepsilon}{2}$ Intuitively, I understand this to mean that if we choose any difference $\varepsilon$, we can choose a corresponding $\delta$ such that this difference is lower than $\varepsilon$. This would seem to imply that the function converges, since it can't be bounded by any difference. At the same time, it seems to say that this function does not converge. Choose any $\delta > 0$. Then choose $2\delta < \varepsilon$. This seems to say the opposite: if you try to look at any region, the error will be greater than 0. Or, you could choose some error bound $\varepsilon > 0$. Then choose $\delta > \frac{\varepsilon}{2}$. This seems like it says that for all $\varepsilon$ you choose the error bound will aways fail in a non-zero region, since $\delta > \frac{\varepsilon}{2} > 0$.

I think this problem comes from the order in which I construct the sets. For example, lets say I want to look at whether $f(x) = x$ is bounded above. If I first say: "choose any finite value $x$" then $f(x)$ is bounded above by $x+1$ Since $x+1$ is finite, the bound is finite. But if I first say: "choose some finite bound $x$" then $f(x+1) > x$, so $f(x)$ is not bounded by any finite value.

Going back to to the episolon-delta definition, it seems that if I construct the error bound $\varepsilon$ first, then I can find a delta where the function is within the error bound. But if I construct the $\delta$ first, then I find that no matter what range of values for x which I choose, the error is always greater than some function of that range of x values.

How should I interpret this? In general, how should I think of the order in which sets in math are constructed?

Answer 1

Keep in mind the following: we say $$\lim_{x\to a}f(x)=b\quad\text{iff}\quad\forall \varepsilon>0\quad\exists\delta>0\quad \lvert x-a\rvert<\delta\implies\lvert f(x)-b\rvert<\varepsilon$$ As you see, the $\delta$ you pick depends on the $\varepsilon$ and not the other way around, and that is crucial.

Answer 2

In your example you’re not actually constructing sets: you’re proving that certain numbers can always be found. Specifically, you’re trying to show that for each $\epsilon>0$ there is a $\delta>0$ such that some statement $S$ involving $\epsilon$ and $\delta$ is true. It should be clear that $\epsilon$ comes first here: you have to show that no matter what $\epsilon>0$ I give you, you can produce a $\delta>0$ that makes $S$ true for that particular $\epsilon$ and $\delta$. The wording makes it clear that the choice of $\delta$ is allowed to depend on what $\epsilon$ you’re given, so $\epsilon$ must come first.

If, on the other hand, you were trying to show that there is some one $\delta>0$ that makes $S$ true for all $\epsilon>0$, matters would be very different. In that case your job would be to find a $\delta>0$ and a proof that no matter what $\epsilon>0$ I might give you, $S$ is true for your $\delta$ and my $\epsilon$. In that case $\delta$ clearly must come first.

In short, what comes first depends on what you’re trying to show.

Your discussion of whether $f(x)=x$ is bounded above suggests that you may not understand what it means to say that a function is bounded above. The fact that $f(x)<x+1$ for any $x\in\Bbb R$ says absolutely nothing about whether $f(x)=x$ is a bounded function: ‘$f$ is bounded above’ does not mean that for each $x$ there is some $b$ such that $f(x)\le b$. Rather, it means that there is some single $b$ such that $f(x)\le b$ for all $x$. In other words, if you want to show that $f$ is bounded above, you have to produce such a $b$ and a proof that $f(x)\le b$ for all $x$; in your terms, $b$ comes first.

In logical notation you seem to be having a bit of trouble with the difference between

$$\forall x\exists y\big(P(x,y)\big)\tag{1}$$

and

$$\exists y\forall x\big(P(x,y)\big)\,,\tag{2}$$

where $P(x,y)$ is some statement involving $x$ and $y$. $(1)$ says that no matter what $x$ I choose, there is a $y$ that makes $P(x,y)$ a true statement. Here the $x$ clearly comes first: the example in the first paragraph above is of this kind. $(2)$ says that there is some single $y$ such that no matter what $x$ I come up with, $P(x,y)$ is going to be a true statement. Here that special $y$ comes first: the example of showing that a function is bounded above is of this kind.

Here is a standard example illustrating the difference:

$$\begin{align*} &\forall x\in\Bbb Z\,\exists y\in\Bbb Z\,(x+y=0)\tag{1'}\\ &\exists y\in\Bbb Z\,\forall x\in\Bbb Z\,(x+y=0)\,.\tag{2'} \end{align*}$$

$(1')$ says that each integer has a negative: if $x$ is any integer, there is an integer $y$ such that $x+y=0$. Of course this is true: we just take $y=-x$. Notice, though, that for each $x$ we need a different $y$. This is why $(2')$ is false: it says that there is one single integer that is the negative of every integer.