Question about $\delta$'s in the $\epsilon$-$\delta$ definition of continuity

Question

If a function $f:M\to\mathbb{R}$ is continuous at point $x_0$ we know that for an arbitrary $\epsilon>0$ there exists a $\delta>0$ such that for all $x\in M$ and $|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$. Let's call those $\epsilon$ and $\delta(\epsilon)$ a pair, $(\epsilon,\delta)$.

What happens if I shrink the $\delta$? Does this imply that $|f(x)-f(x_0)|$ also shrinks?

My intuition says that we can't make any claim on the behaviour of $|f(x)-f(x_0)|$. Sure if $\delta$ attains a value which is very small then it will be smaller than another $\delta'$ which belongs to a pair $(\epsilon',\delta')$ where the $\epsilon'<\epsilon$. But if I shrink $\delta$ only a bit what happens then? How do I argue in a formal way?

Answer 1

Nothing happens.

Continuity says that if you are given $\epsilon > 0$ then you can find $\delta > 0$ such that the following statement is true:

$(*)$ if $|x-x_0| < \delta$ then $|f(x)-f(x_0)| < \epsilon$.

So, let's suppose that the $\epsilon$ has been chosen, and let's fix its value. And having done that, let's choose a value of $\delta$ which makes statement $(*)$ true. Then any smaller value will also make statement $(*)$ true. In other words, if $0 < \delta' < \delta$ and if $(*)$ is true then the following statement (obtained by substituting $\delta'$ in place of $\delta$) is also true:

$(*')$ if $|x-x_0| < \delta'$ then $|f(x)-f(x_0)| < \epsilon$.

The reason for this is that $|x-x_0| < \delta'$ and $\delta' < \delta$ together imply $|x-x_0| < \delta$.

Answer 2

If you've not yet played around with what pairs delta and epsilon work, you might want to try something like this: https://www.desmos.com/calculator/iejhw8zhqd which can give you a feel for what happens as you shrink delta. The question you are asking is basically "if I narrow my x-window, does that necessarily shrink the y-window?" the answer could be yes or no, but depends on the behaviour of your function. It might be instructive to try it for a function like f(x) = sin(1/x) (with f(0) = 0) to test your intuition.