The Cartesian Product of two metric spaces and sequences that converge Prove that if $(X,d)$ is the Cartesian product of the two metric spaces $(X_1,d_1)$ and $(X_2,d_2)$, then a sequence ${(x_n^1,x_n^2)}$ in $X$ converges to $(x^1,x^2)$ if and only if $x_n^1 \rightarrow x^1$ and $x_n^2 \rightarrow x^2$
my proof:$(\rightarrow)$ Suppose $(x_n^1,x_n^2)\rightarrow (x^1,x^2)$. Choose $\epsilon >0$ so that $\exists N\in\mathbb{N} $ such that $d((x_n^1,x_n^2),(x^1,x^2))<\epsilon\space,\space \forall n\geq N $. Then it follows by definition $d_1(x_n^1,x^1)+d_2(x_n^2,x^2)<\epsilon$. Then $d_1(x_n^1,x^1)<\epsilon_1 \space$and $d_2(x_n^2,x^2)<\epsilon_2$ where $\epsilon=\epsilon_1+\epsilon_2$ and $\epsilon_1>0,\epsilon_2>0$ are both chosen arbitrary for $n\geq N$. Then $x_n^1\rightarrow x^1$ and $x_n^2\rightarrow x^2$.
$(\leftarrow)$ Suppose $x_n^1\rightarrow x^1$ and $x_n^2\rightarrow x^2$. Choose $\epsilon_1>0$ and $\epsilon_2>0$ so $\exists N_1,N_2\in\mathbb{N}$ such that $d_1(x_n^1,x^1)<\epsilon_1 \space$and $d_2(x_n^2,x^2)<\epsilon_2\space \forall n\geq max\{N_1,N_2\}$. Take $\epsilon=\epsilon_1+\epsilon_2$ so that $d_1(x_n^1,x^1)+d_2(x_n^2,x^2)<\epsilon\space, \forall n\geq max\{N_1,N_2\}$. Then $d((x_n^1,x_n^2),(x^1,x^2))<\epsilon\space,\space \forall n\geq max\{N_1,N_2\}$. Therefore, $(x_n^1,x_n^2)\rightarrow (x^1,x^2)$
Let me know if my proof is wrong and where it needs correcting
 A: 
$(\rightarrow)$ Suppose $(x_n^1,x_n^2)\rightarrow (x^1,x^2)$. Choose
$\epsilon >0$ so that $\exists N\in\mathbb{N} $ such that
$d((x_n^1,x_n^2),(x^1,x^2))<\epsilon\space,\space \forall n\geq N $.
Then it follows by definition
$d_1(x_n^1,x^1)+d_2(x_n^2,x^2)<\epsilon$. Then
$d_1(x_n^1,x^1)<\epsilon_1 \space$and $d_2(x_n^2,x^2)<\epsilon_2$
where $\epsilon=\epsilon_1+\epsilon_2$ and $\epsilon_1>0,\epsilon_2>0$
are both chosen arbitrary for $n\geq N$. Then $x_n^1\rightarrow x^1$
and $x_n^2\rightarrow x^2$.

The problem is that $\epsilon_1$ and $\epsilon_2$ aren't chosen arbitrarily! They need to satisfy $d_i(x_n^i,x^i)<\epsilon_i<\epsilon$. But you don't actually need those $\epsilon_i$. Here's my suggestion:

$(\rightarrow)$ Suppose $(x_n^1,x_n^2)\rightarrow (x^1,x^2)$. Choose $\epsilon >0$ so that $\exists N\in\mathbb{N} $ such that $d((x_n^1,x_n^2),(x^1,x^2) \epsilon\space,\space \forall n\geq N $.
Then it follows by definition
$d_1(x_n^1,x^1)+d_2(x_n^2,x^2)<\epsilon$. Then
$d_1(x_n^1,x^1)<\epsilon$ and $d_2(x_n^2,x^2)<\epsilon$ (since distances are non-negative). Then $x_n^1\rightarrow x^1$ and $x_n^2\rightarrow x^2$.

Now for the second part:

$(\leftarrow)$ Suppose $x_n^1\rightarrow x^1$ and $x_n^2\rightarrow x^2$. Choose $\epsilon_1>0$ and $\epsilon_2>0$ so $\exists
> N_1,N_2\in\mathbb{N}$ such that $d_1(x_n^1,x^1)<\epsilon_1 \space$and
$d_2(x_n^2,x^2)<\epsilon_2\space \forall n\geq max\{N_1,N_2\}$. Take
$\epsilon=\epsilon_1+\epsilon_2$ so that
$d_1(x_n^1,x^1)+d_2(x_n^2,x^2)<\epsilon\space, \forall n\geq
> max\{N_1,N_2\}$. Then
$d((x_n^1,x_n^2),(x^1,x^2))<\epsilon\space,\space \forall n\geq
> max\{N_1,N_2\}$. Therefore, $(x_n^1,x_n^2)\rightarrow (x^1,x^2)$

The way you choose $\epsilon$ is again not arbitrary! I would suggest not choosing $\epsilon_1$ and $\epsilon_2$. Instead:

$(\leftarrow)$ Suppose $x_n^1\rightarrow x^1$ and $x_n^2\rightarrow x^2$. Choose $\epsilon>0$. So $\exists N_1,N_2\in\mathbb{N}$ such that $d_1(x_n^1,x^1)<\frac{\epsilon}{2} \space$and $d_2(x_n^2,x^2)<\frac{\epsilon}{2},\space \forall n\geq max\{N_1,N_2\}$.Then $d_1(x_n^1,x^1)+d_2(x_n^2,x^2)<\epsilon\space, \forall n\geq max\{N_1,N_2\}$. Then $d((x_n^1,x_n^2),(x^1,x^2))<\epsilon\space,\space \forall n\geq max\{N_1,N_2\}$. Therefore, $(x_n^1,x_n^2)\rightarrow (x^1,x^2).$

