Difference between Skorokhod spaces $D^n$ and $D\times D\times \cdots \times D$ Let $$D=\left\{x:[0,\infty) \rightarrow \mathbb{R}\,\vert\, x\text{ is RCLL}\right\}$$
and let $$D^n=\left\{x:[0,\infty) \rightarrow \mathbb{R}^n\,\vert\, x\text{ is RCLL}\right\}.$$
I would like to understand if there is a difference between the topological spaces $D^n$ and $$\prod_{i=1}^n D:= D\times \cdots\times D.$$
Here, $D^n$ is endowed with the Skorokhod $J_1$ topology (induced by the multidimensional Skorokhod metric). On the other hand, $\prod_{i=1}^n D$ is endowed with the product topology, where each of the $D$'s are given the Skorokhod $J_1$ topology (i.e., each $D$ is given the topology induced by the one-dimensional Skorokhod metric). 
I am particularly interested in convergence in these two spaces. For example, if $\left\{x^m:m\ge 1\right\}$ is a sequence in $D^n$ and $x^m_i\rightarrow x_i$ in $D$ for each $i=1,\dots, n$, then is it true that $x^m\rightarrow x:=(x_1,\dots, x_n)$ in $D^n$? On the other hand, if $x^m\rightarrow x$ in $D^n$, then does $x_i^m\rightarrow x_i$ in $D$ for each $i=1,\dots, n$ (i.e., does $(x^m_1,\dots, x^m_n)\rightarrow (x_1,\dots, x_m)$ in $D\times \cdots \times D$)? 
My guess to the first question is "no", since the set of continuity points of the mapping $t\mapsto (x_1(t),\dots, x_n(t))$ is a subset of the set of continuity points of any individual mapping $t\mapsto x_i(t)$. 
However, my guess to the second question would be "yes" for the opposite reason (i.e., if $T$ is a continuity point of $x\in D^n$, then it must be a continuity point of each of the individual $x_i$'s). 
Insights would be very much appreciated. 
 A: Your guess is correct in both cases. The infinite time Skorokhod metric is kind of annoying to use, so I'll use the following facts.

*

*$x^m \rightarrow x$ in $D[0,\infty)$ if and only if $x^m[0,T] \rightarrow x[0,T]$ for every $T < \infty$ such that $T$ is a continuity point of $x$. Here $x[0,T]$ is the restriction of $x$ to $D[0,T]$ (see Billingsley 1999, Theorem 16.2).


*Suppose $x$ is continuous at the point $t_1$, and $x^m[0,t_2] \rightarrow x[0,t_2]$ for some $t_2 > t_1$. Then $x^m[0,t_1]\rightarrow x[0,t_1]$ (see Billingsley 1999, section 16 lemma 1).
Then I can use the following metric on $D[0,T]$:
$$d_T(x,y) = \inf_{\lambda\in\Lambda_T}\max\left\{\|\lambda - I\|_T,\|x-y\circ\lambda\|_T\right\}.$$
Here, $\Lambda_T$ is the set of continuous, strictly increasing bijections from $[0,T]$ to itself. $\|\cdot\|_T$ is the sup norm up to time $T$ with respect to the $\ell^1$ metric on $\mathbb{R}$ or $\mathbb{R}^n$ (this gives us cleaner calculations). This metric is incomplete, but it generates the J1 topology. I'm abusing notation slightly by using the same notation for the univariate and multivariate Skorokhod metrics.
Now, suppose $x^m \rightarrow x$ (in D^n). Fix any $i \in \{1,\dots,n\}$. Let $T_k$ be a sequence of continuity points of $x$ increasing to infinity. Fix any $k$. Then for each $m$, there must exist a $\lambda_m\in \Lambda_{T_k}$ such that,
$$\max\left\{\|\lambda_m - I\|_{T_k},\|x^m - x\circ\lambda_m\|_{T_k}\right\}< 2d_{T_k}(x^m,x).$$
Then,
$$d_{T_k}(x^m_i,x_i) \leq \max\left\{\|\lambda_m - I\|_{T_k},\|x^m_i - x_i\circ\lambda_m\|_{T_k}\right\}\leq \max\left\{\|\lambda_m - I\|_{T_k},\|x^m - x\circ\lambda_m\|_{T_k}\right\} < 2d_{T_k}(x^m,x).$$
So, $x^m_i[0,T_k] \rightarrow x_i[0,T_k]$ for all $k$. Applying facts 1 and 2 from above, it follows that $x^m_i \rightarrow x_i$ in $D$.
Counterexample:
In the other direction, let $n=2$. Let $\mathbb{I}_A(t)$ be the indicator function (taking the value 1 when $t\in A$ and 0 otherwise). Define,
$$x_1 = x_2 = \mathbb{I}_{t < 1}(t),\quad x^m_1 = \mathbb{I}_{t < 1 - 2^{-m}}(t),\quad x^m_2 = \mathbb{I}_{t < 1 + 2^{-m}}(t).$$
Then it's very clear that $x^m_1 \rightarrow x_1$ and $x^m_2 \rightarrow x_2.$ However, for any $m \in \mathbb{N}$ and $1 + 2^{-m} < T< \infty$ and $\lambda \in \Lambda_T$,
$$\|x^m-x\circ\lambda\|_T\geq 1,$$
so,
$$d_T(x^m,x) \geq 1.$$
Thus, $x^m \nrightarrow x$.
