Can I please receive feedback on my proof/solution below? Thank you.
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Let $\f\colon D\to \R^m$, $D\subseteq\R^n$, and let $\a$ be a cluster point of $D$. Prove $\displaystyle{\lim_{\x\to\a}\f(\x)}$ exists if and only if for every $\epsilon>0$ there exists a $\delta>0$ such that for all
$\x_1,\,\x_2\in D\cap (B(\a,\delta)\setminus\{\a\})$, $\|\f(\x_1) – \f(\x_2)\|<\epsilon$.
$\textbf{Solution:}$ Suppose $\displaystyle{\lim_{\x\to \a}\f(\x)}$ exists, and let $\epsilon >0$. Then, there exists $\delta > 0$ such that $||\f(\x)-\f(\a)|| <\frac{\epsilon}{2}$ for all $\x \in (B(\a,\delta)\setminus\{\a\}) \cap D$. Let $\x_1,\,\x_2\in (B(\a,\delta)\setminus\{\a\})\cap D$. Then, $||\f(\x_1)-\f(\a)|| <\frac{\epsilon}{2}$ and $||\f(\x_2)-\f(\a)|| <\frac{\epsilon}{2}$. Therefore, by triangle inequality, $$||\f(\x_1)-\f(\x_2)|| \le ||\f(\x_1)-\f(\a)|| + ||\f(\x_2)-\f(\a)|| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$
Conversely, we must show there do not exist two sequences $\x_n \to \a$ and $\y_n \to \a$, with $\f(\x_n)\to L_1$ and $\f(\y_n) \to L_2$ and $L_1 \neq L_2$, because if they do exist, then the limit can not exist.
Let $L_2>L_1$. Then, we can choose $\epsilon>0$ such that for any $x\in B(L_2;\epsilon)=(L_2-\epsilon,L_2+\epsilon)$ and for any $y\in B(L_1;\epsilon)=(L_1-\epsilon,L_1+\epsilon)$, we have $|x-y|>\epsilon$. This follows since $|\x-\y|\geq \x-\y>(L_2-\epsilon)-(L_1+\epsilon)=(L_2-L_1)-2\epsilon$, which is greater than $\epsilon$ if $\epsilon<\frac{L_2-L_1}{3}$.
Choose such an $\epsilon$. By hypothesis, there exists $\delta_1>0$ such that if $\x,\y \in B(\a;\delta_1)\backslash\{\a\}$, then $|\f(\x)-\f(\y)|<\epsilon$. Since $\x_n \to \a$ and $\f(\x_n)\to L_1$, there exists $N_{\x}$ such that for all $n\geq N_{\x}$, we have $\x_n \in B(\a;\delta)\backslash\{\a\}$ and $\f(\x_n)\in B(L_1;\epsilon)$. Similarly, there exists $N_{\y}$ such that for all $n\geq N_{\y}$, we have $\y_n \in B(\a;\delta)\backslash\{\a\}$ and $\f(\y_n) \in B(L_2;\epsilon)$. Take any $n_0 \geq \max\{N_x,N_{\y}\}$. Then $\x_{n_0},\y_{n_0}\in B(\a;\delta_1)\backslash\{\a\}$, but
$$
|\f(\x_{n_0})-\f(\y_{n_0})|\geq \f(\x_{n_0})-\f(\y_{n_0})>(L_2-L_1)-2\epsilon>\epsilon,
$$
a contradiction.
In the higher dimension case, the argument follows that $\epsilon>0$ can be chosen small so that $d(B(L_1;\epsilon),\, B(L_2;\epsilon))>\epsilon$.We wish to have $\epsilon$ sufficiently small such that $d(B(L_1, \epsilon), B(L_2,\epsilon)) > \epsilon$. So, let $F(\epsilon) = d(B(L_1, \epsilon), B(L_2,\epsilon)) – \epsilon$. Then $F(0) = ||L_2-L_1|| > 0$. Hence, for $\epsilon$ sufficiently small, $F(\epsilon) > 0$.
Best Answer
Your proof for the first part is correct. For the second part, you basically showed that if $x_n \to a$, then $f(x_n)\to L$ for some $L$. The problem though is that, a priori, this $L$ that arose depends on the sequence you picked. You have to show there aren't two sequences $x_n \to a$ and $y_n \to a$, with $f(x_n)\to L_1$ and $f(y_n) \to L_2$ and $L_1 \neq L_2$.
I'll sketch the idea assuming we are in $\mathbb{R}$. WLOG, suppose $L_2>L_1$. Then I claim we can choose $\epsilon>0$ so that for any $x\in B(L_2;\epsilon)=(L_2-\epsilon,L_2+\epsilon)$ and for any $y\in B(L_1;\epsilon)=(L_1-\epsilon,L_1+\epsilon)$, we have $|x-y|>\epsilon$. This follows easily since $|x-y|\geq x-y>(L_2-\epsilon)-(L_1+\epsilon)=(L_2-L_1)-2\epsilon$, which is greater than $\epsilon$ if $\epsilon<\frac{L_2-L_1}{3}$.
Choose such an $\epsilon$. By hypothesis, there exists $\delta_1>0$ such that if $x,y \in B(a;\delta_1)\backslash\{a\}$, then $|f(x)-f(y)|<\epsilon$. Since $x_n \to a$ and $f(x_n)\to L_1$, there exists $N_x$ such that for all $n\geq N_x$, we have $x_n \in B(a;\delta)\backslash\{a\}$ and $f(x_n)\in B(L_1;\epsilon)$. Similarly, there exists $N_y$ such that for all $n\geq N_y$, we have $y_n \in B(a;\delta)\backslash\{a\}$ and $f(y_n) \in B(L_2;\epsilon)$. Take any $n_0 \geq \max\{N_x,N_y\}$. Then $x_{n_0},y_{n_0}\in B(a;\delta_1)\backslash\{a\}$, but $$ |f(x_{n_0})-f(y_{n_0})|\geq f(x_{n_0})-f(y_{n_0})>(L_2-L_1)-2\epsilon>\epsilon, $$ a contradiction.
In the higher dimension case, it's an easy argument that $\epsilon>0$ can be chosen small so that $d(B(L_1;\epsilon),\, B(L_2;\epsilon))>\epsilon$.