I was reading this question, and the related answers, which popped out in merit to my previous question about continuity while searching over here: $\lim_{x\to 0} f(x)$ where $0$ is isolated in the domain of $f$?
Now as fare as I understood there are two main ways to prove that function is continuous at $x = 0$: the topological one and the analytical one ($\epsilon$-$\delta$), and in the latter case it turns out to be that the definition is vacuously satisfied.
I was wondering: is there some other way to prove that $f: \{0\} \cup (2, 3)$ with $f(x) = x^2$ is continuous at $x = 0$?
I apologise for the questions, hoping it's not meaningless. I am just curious: I thought to put myself into some non mathematics student, someone rather far from the mathematical world, hence like some amateur or biology students (and don't get this wrong, I'm not denigrating biologists, I am just saying that they don't have to know and study those rigorous definition for continuity), with the will to understand continuity in some easy way. Is there something or are those the only two ways?
Thank you!!
EDIT: what do I mean?
The "idea" (which I believe will turn out into something trivial or wrong) came from the approach to continuity in $\mathbb{R}^2$, where for what I understood reading over here certain answers and questions is something like: "$f(x, y)$ is continuous at a point $(x_0, y_0)$ in the domain, if I can find a function of a distance (whether it be the Euclidean distance or another one), such that$$|f(x, y) - f(x_0, y_0)| \leq h(d(x, y))$$
Where $h(d(x, y))$ is a distance function, for example I could have $h = x^2+y^2$ or $h = \sqrt{x^2+y^2}$ or $h = |x|+|y|$ and so on.
This is good because we know that distances have the property to go to zero when the argument is zero, that is for example when $(x_0, y_0) \to (0, 0)$ we have $h(d(0, 0)) \to 0$ as it happens with the above distances.
Example:
$$f(x, y) = \begin{cases} \dfrac{x^3}{x^2+y^2} & (x, y) \neq (0, 0) \\\\ 0 & (x, y) = (0, 0) \end{cases}$$
Continuity at zero:
$$\vert|f(x, y) - f(0, 0)\vert| = \bigg|\frac{x^3}{x^2+y^2}\bigg| = x \frac{x^2}{x^2+y^2} \leq x \to 0$$
Hence $f$ is continuous at zero.
So?
What I mean hence is: can we do the same in $\mathbb{R}$? For example: say we have $f(x) = x^2$ with domain the whole real axis. Then the continuity, say, at $x = 4$ would read something like
$$|f(x) - f(4)| = |x^2-4| \leq x^2$$
And $x^2$ is a distance in $\mathbb{R}$ so we are done.
- Is this nonsense? Is this the actual $\epsilon-\delta$"revised"?
I mean in this case it would be trivial to show $x^2$ is continuous at $x = 0$, when the domain is $\{0\} \cup (2, 3)$.