Unclear about the epsilon-delta definition of continuous mapping

Question

In short, I don't see how the epsilon-delta definition excludes non-injective mappings. For example, I can imagine a modified sine where a single point is excluded, but for which we can still find the finite neighbourhood $\delta$.

Answer 1

Your drawing seems way wrong. You are given $\epsilon$, and you then have to come up with a $\delta$ so that the entire $\delta$-ball around $x_0$ is mapped into the $\epsilon$-ball around $f(x_0)$. If $\epsilon$ is sufficiently small, you won't be able to do that with the graph you show since there will be points arbitrarily close to $x_0$ which are mapped "far" (more than $\epsilon$ away) from $f(x_0)$.

In your graph, a tiny ball around $x_0$ will be mapped to a tiny ball around the y-value of the "missing" point on the graph, except for the single value $f(x_0)$. Except for the single point, the image of the ball will be far from $f(x_0)$.

Answer 2

There are some things you are confusing. First and foremost, injectivity has nothing to do with continuity.

$$f : \mathbb{R} \to \mathbb{R}, x\mapsto x^2$$ is continous but neither injective nor surjective.

And then i suppose there's some confusion about the $\varepsilon, \delta$-criteria.

I'm not sure what you meant by excluding a single point of your function. However, what you certainly can do, roughly speaking, is tearing the graph of a function apart by some piecewise definition such as

$$ f : \mathbb{R} \to \mathbb{R}, x\mapsto \begin{cases} \sin(x) & \ x < 1 \\ 0 & \ x = 1 \\ \sin(x) & \ 1 < x \end{cases}$$

And indeed you can always give me an $\varepsilon$ for which there is a $\delta$ such that the $\varepsilon,\delta$ criteria holds.

But the key is that there is at least one $\varepsilon'$ around $f(1)$ for which it fails and thus the given function is not continous.

Long story short: It's not about finding one $\varepsilon > 0$ for which you can find a $\delta$ such that

$$\vert x-x'\vert < \delta \Rightarrow \vert f(x) - f(x') \vert < \varepsilon$$

You must be able to find such $\delta$ for every $\varepsilon > 0$ for which the implication of the $\varepsilon,\delta$ criteria holds.