Syntax of an epsilon delta proof/why is this version incorrect

Question

So we have the regular $\delta$-$\epsilon$ definition of continuity as:

(1) For all $\epsilon>0$, there exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.

My question is why is the following definition incorrect?

(2) There exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$ for all $\epsilon>0$.

The obvious response is that $\forall x$ $\exists y$ $\neq$ $\exists y$ $\forall x$ (or rather, they are not always equal), but look harder at the grammar: is that necessarily what is going on?

Let us define $p:=$ "There exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$"

So we now have the statement:

"$p$ is true for all $\epsilon>0$"

Isn't that sentence identical to the English sentence: "for all $\epsilon>0$, $p$ is true"?

In which case you would have (1):

For all $\epsilon>0$, there exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.

The counterargument that $\forall x$ $\exists y$ $\neq$ $\exists y$ $\forall x$ makes sense if you look at it immediately from a logical view, but because of the way English sentences work (and their vagueness, in a case like this), the "for all $\epsilon>0$" clause can be placed anywhere without changing the meaning of the statement in English.

To illustrate better what I'm talking about, let us imagine that we write our definition as follows:

There exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$ (for all $\epsilon>0$).

Or, maybe:

There exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$ (N.B. we're talking about all $\epsilon>0$ here!).

Aren't these parentheticals sort of "overlying" the whole statement? Is the fault in my reasoning that $\epsilon$ is being "bound" to an if-then statement before it has been defined? Or am I just blatantly incorrect and this is a case of $\forall x$ $\exists y$ $\neq$ $\exists y$ $\forall x$.

The essential point of all of this is that, if you have the statement: $$ \forall A \exists B : {B \subseteq A} $$ Then in English, this is equivalent to saying, both "For all A, there exists a B such that if B then A", and, "There exists a B such that if B then A, for all A".

Answer 1

The English sentence

There exists $A$ such that $P(A,B)$ for all $B$.

is ambiguous -- it can either mean $\exists A\forall B\,P(A,B)$ or $\forall B\exists A\,P(A,B)$, and there is no generally observed convention about which of them it ought to be understood as.

Therefore, for the sake of clarity sentences of this shape should be avoided in mathematical writing.

In symbolic logic, there is no ambiguity because the quantifier should always come first. Some beginning students (and, sadly, some teachers) write things like "$P(x) \,\forall x$", as a kind of shorthand for the English wording "$P$ of $x$ holds for all $x$", but this is not proper syntax for predicate logic. If you use symbols, you should stick to the grammar that goes with them.

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On the other hand, symbolic logic regrettably has a different ambiguity, namely whether $\forall x\;P(x)\Rightarrow Q$ means $(\forall x\;P(x))\Rightarrow Q$ or $\forall x\;(P(x)\Rightarrow Q)$. There are conventions for which one to choose here, but unfortunately they give different results! Explicit parentheses are recommended for both of these meanings, unless you're really really sure that your readers know and follow the same convention that you do.

Answer 2

The definition is incorrect because the order in which the values are picked matters. I'm glad you asked this because this is not nearly hammered as much as it should be to get people to understand and appreciate what is going on the first time around.

First you pick $\epsilon$. THEN you pick $\delta$. The $\delta$ depends on the $\epsilon$. This is hugely important because if you were to pick another $\epsilon$ you would get a different $\delta$. (Caveat: if you're talking about the case of uniform continuity, then no matter what $\epsilon$ you pick, the same $\delta$ works for any of them).

For me, the easiest way to grasp the intuition behind this is to draw it out. Pick a continuous function and draw it. Then draw 2 lines parallel to the y-axis a certain distance apart (say half an inch, or whatever distance you want). At the points they intersect the curve (i.e. function) mark the corresponding values on the y-xis. So now you have your first ($\epsilon_1$, $\delta_1$) pair. Repeat this process but this time around change the distance between the 2 parallel lines, so that you get another ($\epsilon_2$, $\delta_2$) pair.

Remember that $\epsilon_1$ =/= $\epsilon_2$ by design. Now look at $\delta_1$ and $\delta_2$. If you picked a continuous function (but not uniformly continuous) $\delta_1$ won't be the same as $\delta_2$. If you picked a uniformly continuous function, they will be the same.

For these kinds of long-winded explanations that articulate the intuition you're supposed to grasp, I highly recommend the book "Introduction to Real Analysis" by Michael J. Schramm. It is not nearly as popular as the terse Rudin, but I found it invaluable in helping me to understand how to do analysis. (I don't mean to bash Rudin. His book is great for exercises once you've had some hand-holding from Schramm.)