Convergent sequence in product space on $\mathbb{R}^{\omega}$ I am confused about the concept of convergent sequence in product space when learning Munkres's Topology, especially when I am comparing two related exercises of it.
The exercise 6 of section 19 titled "The Product Topology" considers the convergent sequence in product space (its solution can be found at Convergence in product topology):

Let $x_1, x_2, \ldots$ be a sequence of the points of the product space $\Pi X_{\alpha}$. (show that) This sequence converges to the point $x$ if and only if the sequence $\pi_{\alpha}(x_1), \pi_{\alpha}(x_2), \ldots$ converges to $\pi_{\alpha}(x)$ for each $\alpha$. 

The exercise 4(b) of section 20 titled "The Metric Topology" considers convergent sequence in product space on $R^{\omega}$ (the hint of its solution can be found at Convergence of sequence in uniform and box topologies). Consider, for instance, 

the sequence $w_1 = (1,1,1,1,\ldots), w_2 = (0,2,2,2,\ldots), w_3 = (0,0,3,3,\ldots), \ldots$

The $w$ sequence converges to the "zero" point, i.e., $0 = (0,0,0,0, \ldots)$ in product space on $R^{\omega}$ because $D(0, w_k) = 1 / k$, and therefore, any ball $B_{D}(0, \epsilon)$ contains all elements of the sequence starting from some large enough index.
Moreover, we can check that the sequence $\pi_{\alpha}(w_1), \pi_{\alpha}(w_2), \ldots$ converges to $\pi_{\alpha}(0)$ for each $\alpha$.
In other words, for the $w$ sequence, the statement of exercise 6 is satisfied. However, is it generally true for the sequence in product space on $\mathbb{R}^{\omega}$? Formally,

Problem: Let $x_1, x_2, \ldots$ be a sequence of the points of the product space on $\mathbb{R}^{\omega}$. Is it true that this sequence converges to the point $x$ if and only if the sequence $\pi_{\alpha}(x_1), \pi_{\alpha}(x_2), \ldots$ converges to $\pi_{\alpha}(x)$ for each $\alpha$? 

 A: The product, uniform and box topologies on $\mathbb{R}^\omega$ are distinctly different topologies.  It is therefore unsurprising that the details of convergence of a sequence is different in these as well.  When speaking about the product space $\mathbb{R}^\omega$, we are talking about this set endowed with the usual product topology.  As exercise 6 of section 19 indicates, for any product space $X = \prod_{\alpha} X_\alpha$ a sequence $\langle \mathbf{x}_n \rangle_n$ in $X$ converges to $\mathbf{x}$ iff for each $\alpha$ the sequence $\langle \pi_\alpha ( \mathbf{x}_n ) \rangle_n$ converges to $\pi_\alpha ( \mathbf{x} )$ in $X_\alpha$.  However if you endow $X$ with a different topology the same result will not hold.
There are two things you can take form this result:  Let $X = \prod_{\alpha} X_\alpha$ (as a set), let $\langle \mathbf{x}_n \rangle_n$ be a sequence in $X$ and $\mathbf{x} \in X$.  Then 


*

*if $X$ is given a topology coarser than the product topology, then if for each $\alpha$ the sequence $\langle \pi_\alpha ( \mathbf{x}_n ) \rangle_n$ converges to $\pi_\alpha ( \mathbf{x} )$ in $X_\alpha$, it follows that $\langle \mathbf{x}_n \rangle_n$ converges to $\mathbf{x}$.

*if $X$ is given a topology finer than the product topology, then if $\langle \mathbf{x}_n \rangle_n$ converges to $\mathbf{x}$, it follows that for each $\alpha$ the sequence $\langle \pi_\alpha ( \mathbf{x}_n ) \rangle_n$ converges to $\pi_\alpha ( \mathbf{x} )$ in $X_\alpha$.

But this is true in general: the convergence of a given sequence is dependent on the topology, and the same sequence may converge in one topology, and not in another (or even to a different limit).
As perhaps a more basic example, consider $\mathbb{R}$ with the usual metric topology, and also with the lower-limit (or Sorgenfrey) topology: let's denote that space by $\mathbb{R}_{\text{S}}$.  As we know, in $\mathbb{R}$ a sequence $\langle x_n \rangle_n$ converges to $x$ iff for each $\epsilon > 0$ there is an $N $ such that $| x_n - x | < \epsilon$ for all $n \geq N$.  However in $\mathbb{R}_{\text{S}}$ convergence of sequences is different:

a sequence $\langle x_n \rangle_n$ converges in $\mathbb{R}_{\text{S}}$ to $x$ iff for each $\epsilon > 0$ there is an $N $ such that $x \leq x_n < \epsilon$ for all $n \geq N$.

In particular, the sequence $\langle 1 - \frac{1}{n} \rangle_n$ converges (to $1$) in $\mathbb{R}$, but is not convergent in $\mathbb{R}_{\text{S}}$.  ($[1,2)$ is an open neighbourhood of $1$ in $\mathbb{R}_{\text{S}}$ which contains no members of this sequence.)
A: Yes, because the metric topology on $\mathbb{R}^w$ (as defined in Munkress theorem 9.5, chapter II) agrees with the product topology, as it's proved in the aforementioned theorem.
