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)
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.