Let $(a_n)_{n\geq 1}$ be a sequence of complex numbers, and let $a\in \mathbb{C}$. It is well-known that, if the following statement $$\forall \epsilon>0\exists N\in \mathbb{N}:(n\geq N\implies |a_n-a|<\epsilon)$$ is satisfied, then $\lim_{n\to \infty}a_n=a$, as a definition. I have seen a lot of proofs that contains something with limits. My experience says that the following statements are equivalent, which I'd like to confirm, $$\forall \epsilon>0\exists N\in \mathbb{N}:(n\geq N\implies |a_n-a|<\epsilon)\tag{a}$$ $$\forall K>0\forall \epsilon>0\exists N\in \mathbb{N}:(n\geq N\implies |a_n-a|<K\epsilon)\tag{b}$$ $$\exists\alpha>0\forall \epsilon\in (0,\alpha)\exists N\in \mathbb{N}:(n\geq N\implies |a_n-a|<\epsilon)\tag{c}$$
I have seen a lot of proofs that indirecly use $(b)\implies (a)$, and $(c)+(b)$ mixed implies $(a)$.
My short proof of equivalences is:
$(a)\implies (b)$: Assume $K>0$, then putting $\epsilon=K\epsilon'>0$, it implies $\epsilon'>0$.
$(b)\implies (a)$: If $K,\epsilon>0$, then put $\epsilon'=K\epsilon>0$.
$(a)\iff (c)$: It is obvious. (The $\impliedby$ can be seen in this LINK).
You are correct that all of these statements are equivalent. For $(a)\iff (c)$, you may want to put your reasoning into notation, although it is fairly obvious. To show complete equivalence, we can say something like $|a_n|-a<\epsilon<\alpha \leq E\in [\alpha,\infty)$, and we already have that this is true $\forall\epsilon \in (0,\alpha)$ and finally $\forall\epsilon' \in(0,\alpha) \cup [\alpha,\infty) \iff \forall \epsilon'>0$.
EDIT: I wrote this before Hopeless added the link. For all interested, the link, as it addresses the specific question on $(a) \iff (c)$, has more succinct arguments.