So, I have an Calculus 1 exam at uni coming up and am currently working through the topic "Continuity". I've been trying to do some proofs and a few weeks back in tutorial we had this exercise:
$$ f : (0,1] \cup \{ 2 \} \to \Bbb R $$
$$ x \mapsto \begin{cases} 1 & \text{if } x \in (0,1] \\ 5 & \text{if } x = 2\end{cases} $$
Now, by definition, $f$ is continous in point $x_0 \iff$ $\forall$ $x_n \in D: \lim_{n\to \infty} x_n = x\implies \lim_{n\to \infty} f(x_n)= f(x) = f(\lim_{n\to \infty} x_n) $ .
We did it in two parts, but I'll just do the first, because the second is analogous.
What we did: Let $x_0 \in (0,1] $ and $ x_n \in D $ so that $ x_n \to x_0 $. (That's clear, of course.)
But the next step wasn't clear to me, even after I asked my tutor. " For $ \epsilon := \frac {1}{2} \exists n_0 \in \Bbb N : \forall n \ge n_0 : |x_n-x| \le \frac {1}{2} = \epsilon$. "
Why can we use have a fixed value for $\epsilon $?
My tutor said something about how the interval looks so it doesn't reach 2. (I think, it lies back a while) But I thought $\epsilon$ should be general and only $\gt 0$. I tried looking online, but I haven't seen this method we used in the tutorial anywhere. It works, but why?
Hopefully someone can enlighten me.
(Note: The steps that followed I understand; we adapt the intervals for the values $x$ can be in the chosen interval and finally, we come to the conclusion that $ 0 \lt x_n \le \frac {3}{2} $ and that the limit is 1 because $f$ is defined as 1 for the chosen interval. So $f$ is continuous in the interval.)
I think the question is proving the function is continuous on $D = (0,1] \cup \{2\}$.
Now, by definition you provide, $f$ is continuous on $D$ if and only if for each $x_0 \in D$, for a sequence $x_n$ in $D$ such that $\lim_{n \to \infty} x_n = x_0$, we have $\lim_{n \to \infty} f(x_n) = f(x_0)$. It is easy to obtain when $x_0 \in (0,1]$ as you proved.
To see why $f$ is continuous at $x=2$, take $\varepsilon > 0$ arbitrarily. Choose $\delta = \frac{1}{2}$. Then, for any $x \in D$, $|x-2| < \delta$, we have $x = 2$, and $|f(x)-f(2)| = 0 < \varepsilon$. Hence, $f$ is continous at $x=2$.
In fact, you can do the same argument to prove the function is continuous on the isolated points.
Update: Now we can use sequence criterion to prove $f$ is continous at $x_0=2$. Suppose $x_n$ converges to $x_0 = 2$.
By definition of convergent sequence, $x_n$ converges to $x_0$ if and only if for all $\delta > 0$, there is some $n_0 \in \mathbb{N}$ such that $|x_n - x_0| < \delta$, for all $n \ge n_0$. Note that it holds for all $\delta$. Because we have had $x_n$ converges to 2, so we can pick one $\delta$. Here we shall pick $\delta = 1/2$ (Why? It will become clearly in the next argument).
Now, let $\varepsilon > 0$. Choose $n \ge n_0$. Then for all $x_n \in D$, $|x_n - x_0| < \delta$, we have $x_n = 2$. Here you can do same argument in the first proof to show $f_n$ converges to $f(2)$.