Proof of equivalence between limit of a vector field and limit of a scalar field I have a doubt with a proof regarding the following implication.

Consider $F=(f_1,..,f_m): A \subset  \mathbb{R}^n \rightarrow 
> \mathbb{R}^m$ and $\bar{x}$ a limit point for $A$, then
  $$\mathrm{lim}_{x \rightarrow \bar{x}} f_i(x)=l_i  \, \, \, \, \, \,
 \, \, \forall i=1,...,m\implies \mathrm{lim}_{x \rightarrow \bar{x}}
 F(x)=l=(l_1,...,l_m)$$

What I know is that, for the single component $i$,
$$\forall \epsilon>0 \,\, \, \exists \delta_i>0 \,\,\,\,\, | \,\,\,\,\,\,\, x\in A \,\,\,\,\,\,\,\  0<||x-\bar{x}||<\delta_i \implies |f_i(x)-l_i|<\epsilon \tag{A}$$
So here $\delta$ depends on $\epsilon$ choosen but also on $_i$, therefore it's a $\delta_i$.
Defining $\bar{\delta}= min \{d_i \}$ we have
$$ x\in A \,\,\,\,\,\,\,\  0<||x-\bar{x}||<\bar{\delta} \implies |f_i(x)-l_i|<\epsilon \,\,\,\,\,\, \forall i=1,...,m \tag{B}$$
And then also 
$$x\in A \,\,\,\,\,\,\,\  0<||x-\bar{x}||<\bar{\delta} \implies$$
$$||F(x)-l||^2=(f_1(x)-l_1)^2+...+(f_m(x)-l_m)^2<\epsilon^2+...+\epsilon^2=m \epsilon^2 \tag{C}$$
Which means $||F(x)-l||<\sqrt{m} \, \epsilon$. And that proves the statement since $\epsilon$ is arbitrary.

I'm confused about all these $\epsilon$ and $\delta$.
Firstly in $(A)$ I just wrote the definition of limit, which states "$\forall \epsilon$", so there I'm not considering a particular and fixed $\epsilon$ but all of them. 
But in the following statements it looks like I do choose one of the possible $\epsilon$. Here is how I interpreted the reasoning for point $(B)$.


*

*Among all the $\epsilon$ only one is choosen, and the same is used for all the $i$ components

*Each of the components will have a $\delta_i$ corresponding to that particular $\epsilon$ chosen ($\delta$ is always a function of $i$).

*Among all the $\delta_i$ i choose the minimum and, therefore, I can write $(B)$, where $\epsilon$ is still the same for all the $i$ components.


As long as this interpetation is correct, it would not give no problem so far. But in $(C)$ what is proved is that $$||F(x)-l||<\sqrt{m} \, \epsilon$$
Where $\epsilon$ is still the chosen $\epsilon$ in previous point 1.
So, according to me, it does matter that I get a $\sqrt{m} \epsilon$ instead $\epsilon$, since $\sqrt{m} \epsilon >\epsilon$ for sure and this is not exactly in coherence with the definition of limit.
In fact in principle I chose (among all the possible ones) that particular $\epsilon$ but then I get $||F(x)-l||<\sqrt{m} \, \epsilon$.

Is my interpetation wrong? I guess it is since at the end of the proof it is said that $\epsilon$ is arbitrary. 
But I cannot really understand how it is arbitrary. I mean if I go back and choose $\epsilon'=\frac{\epsilon}{\sqrt{m}}$ (where $\epsilon$ is the one choosen firstly), then I do get $||F(x)-l||< \epsilon$ but this still does not prove the limit since $\epsilon > \epsilon'$ so the problem is the same as before.
Probably I'm missing something in the reasoning and I do not see where. Could anyone clarify what is the reasoning behind this proof (like what are the steps followed in terms of reasoning)?
In particular, is $\epsilon$ really choosen? And how to say that $||F(x)-l||<\sqrt{m} \, \epsilon$ proves the statement (the problem is in that $\sqrt{m}$)?

(I'm not looking for an alternative proof, if there is one, but I need to understand this one, as it is and also I know that the reverse implication in the statement holds but I did understand the proof for that revere implication)
 A: Your reasoning towards the end by setting $\epsilon'=\epsilon/\sqrt{m}$ is the correct reasoning to complete the proof formally. Let me explain.
To show that $F(x) \to l$, we need to show that (formally):
$$\forall \eta>0 ~ \exists \nu>0 \,\mid \, \|x-\overline{x}\|<\nu \implies \|F(x)-F(\overline{x})\|<\eta.$$
To resolve your confusion, I have used different symbols. This is logically equivalent to taking arbitrary $\eta$ and finding an appropriate $\nu$. Part of the reason for your confusion is that the author of the proof does not explicitly say 'Given $\epsilon>0$...'
Now, from your statement (A), given $\eta>0$ for the above, we can find $\delta_i$ such that
$$\|x-\overline{x}\|<\delta_i \implies |f_i(x)-l_i|<\epsilon := \eta/\sqrt{m};$$
that is, we apply statement (A) with a relevant choice of $\epsilon$, which is certainly allowed since (A) is true for every $\epsilon$.
Following through the remainder of the proof, we arrive at the desired conclusion, with $\nu = \overline{\delta}$.
However, when proving such results in practice it is difficult to have the foresight to choose $\epsilon$ appropriately. Therefore it is standard practice to follow through with an arbitrary value of $\epsilon$ in the statement (A) and then retrospectively recognize that $\epsilon$ can be made small enough to satisfy the limiting condition for an arbitrarily chosen value $\eta$.
In summary, your confusion arises from the fact that both the assumed statement (A) and the result we are aiming to prove, use an '$\epsilon$' but it is important to be able to distinguish these values, which is why it is often valuable to vary notation as I have above.
A: The translation of "$\forall $" into the English expression "for all" does not capture all the nuances of $\forall$. "$\forall$" can also be translated as "for each", "for each and every", "for every", "for", "whenever we pick", "for each ... that we care to consider", "given an arbitrary ...," etc.
English, being a natural language, is less precise and more lyric than Mathematics. When we cast Mathematics into English we make a choice, but if the choice interferes with your getting through the argument, then make a different choice. After a very short while you end up switching back and for, and come to realized that the meaning is the same: "All men are mortal", "Every man is mortal", "Each man one cares to look at is mortal", "If you choose a man, whether Grek or not, you choose a mortal (being)", "The set of men is contained in the set of mortal( being)s,"...
Similar remarks apply to "$\exists$," which can be translated into English in more than one way. Personally, I have a sense of uneasiness when a statement of the form "$\exists ...$" is proved by contradiction, because it seeems one is making an ontological statement of existance from the fact that assuming othwerwise leads to a contradiction.
Cutting the digression, your translation of "$\forall \epsilon > 0,...$" as "for all $\epsilon > 0$" is only one possibility, "for each $\epsilon > 0$" is another, "for each $\epsilon > 0,\: \text{whether rational or not},... $" is yet another.
