The function $f:(X,d)\rightarrow(Y,\rho)$ is continuous if and only if $f$ is sequentially continuous (that means $x_n\rightarrow x \Rightarrow f(x_n)\rightarrow f(x)$)
Proof. First I show that if $f$ is continuous then $f$ is sequentially continuous. I consider the sequence $x_n\rightarrow x_0$ so I can find for $\varepsilon =\delta$ a value $n_\delta$ such that $|x_n-x_0|<\delta$. Doing so I can use the hypothesis that $f$ is continuous so $|f(x_n)-f(x_0)|<\varepsilon$.
Now, I show the opposite implication. I know now that
$$
\forall \varepsilon >0 \,\,\,\,\exists n_\varepsilon : \forall n\geq n_\varepsilon \,\,\,\,\,|x_n-x_0|<\varepsilon
$$
So for a certain $\varepsilon$:
$$
|f(x_n)-f(x_0)|<\varepsilon_1
$$
If I call $\varepsilon=\delta$ and $\varepsilon_1=\varepsilon$ we have the definition of continuity.
I'm not really sure, I don't know why. But can this proof be considered acceptable? I mean, it is correct and it is written in a decent way?
Best Answer
I think the forward direction is fine (even then you could probably polish it a bit better), but the backward direction is a little confusing.
No that isn't what you actually know. What you do know is that
$$\lim_{n \to \infty} f(x_n) = f(x)$$
and we want to prove $$\lim_{z \to x} f(z) = f(x).$$
This one says that whenever we are on a neighborhood of $x$, the images are tolerably the same. I think you were too focused on "matching symbols" to make the proof work. user117042 pretty much outlined the proof for you. I'll only comment on why he chose $\delta = 1/n$.
By choosing $\delta = 1/n$, we create a convergent sequence such that its image must also get reasonably close to its target, but that all hinges on continuity, which we assume is not true, hence the contradiction.
Also, I've noticed you aren't using the metrics in $X$ and $Y$ at all and you are simply using absolute values. You need to fix that afterwards too before you submit your answer.