Confusion with $\epsilon-\delta$ constraints

Question

I'm admittedly a bit embarrassed to ask this, but after completing real analysis I came to the realization that I don't actually understand the constraints of continuity as well actually I thought.

Say we're trying to prove $f(x) $ is continuous at $x = x_0$ using $\epsilon-\delta $. When choosing $\delta$, it is common to express it as a function of $\epsilon$ and $x_0$, and this part I'm okay with.

However, we cannot have it be a function of $x$, and I am struggling to understand why. I can see how it would lead to impossible results if we did, and some part of it is vaguely intuitive. But I have always considered intuition a parlor trick of sorts without having the proper justification behind it, and I'd like some help with clarification.

Answer 1

$\delta$ technically isn't a function, it's a fixed constant. It's a constant that depends on two other constants: $\epsilon$ and $x_0$.

But $x$ is not a constant, it is a variable. And since $\delta$ is (supposed to be) a constant, then $\delta$ can't depend on $x$.

Answer 2

This is just a matter of how the definition is stated. To say that $f$ is continuous at $x_0$ means that:

For all $\epsilon>0$ there exists a $\delta>0$ such that for all $x$, if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<\epsilon$.

So, what's the difference between the role $\epsilon$ and the role of $x$? The difference is that "for all $x$" comes after "there exists a $\delta>0$", whereas "for all $\epsilon>0$" comes before it. This means that we pick $\epsilon$ before we pick $\delta$, but we pick $x$ after we pick $\delta$. So $\delta$ can depend on $\epsilon$, because we know what $\epsilon$ is (it's some fixed number) before we get around to needing to pick $\delta$. On the other hand, we pick $\delta$ before we know what $x$ is and our choice of $\delta$ needs to make the statement "if $|x-x_0|<\delta$ then $|f(x)-f(x_0)|<\epsilon$" be true for all $x$. So we need a single $\delta$ that works for every $x$ at once; we can't use different $\delta$'s for different $x$'s.

The role of $x_0$ is similar to the role of $\epsilon$, except even more strongly so: the number $x_0$ is fixed from the very beginning. In particular, it's fixed before we have to choose a $\delta$, so $\delta$ can depend on what $x_0$ is.

Answer 3

Since you are trying to prove continuity at a point $x_0$, the goal is to, for any fix $\epsilon>0$, find a single $\delta$ so that all values within $\delta$ of $x_0$ map to a point within $\epsilon$ of $f(x_0)$. That $\delta$ cannot be a function of $x$, because it is a single value (positive real number) which is associated with the point $x_0$.

$\delta$ is only a "function" in that for every $x_0$ and $\epsilon$ you choose, this $\delta$ may need to change. If the function is continuous at more than one point (as is usually the case) you may also have different $\delta$'s associated with different $x$ values. When proving continuity at an arbitrary $x$, you could view the expression that you get for $\delta$ (which may depend on $x$ or may not) to be a function of $x$, but still each $\delta$ which is outputted by this function is just one real number, and thus it doesn't really make sense for it to be a "function of $x$".