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Let $\mathbb{R}^\omega$ denote the set of all the (infinite) sequences of real numbers. Then which of the following sequences in $\mathbb{R}^\omega$ are convergent (and if so, then to which points(s)) in $\mathbb{R}^\omega$ in the product, uniform, and box topologies? $$ \begin{align} w_1 &= \left( 1, 1, 1, 1, \ldots\right), \\ w_2 &= \left( 0, 2, 2, 2, \ldots \right), \\ w_3 &= \left( 0, 0, 3, 3, 3, \ldots \right), \\ & \ldots; \end{align} $$ $$ \begin{align} x_1 &= \left( 1, 1, 1, 1, \ldots\right), \\ x_2 &= \left( 0, \frac 1 2, \frac 1 2, \frac 1 2, \ldots \right), \\ x_3 &= \left( 0, 0, \frac 1 3, \frac 1 3, \frac 1 3, \ldots \right), \\ & \ldots; \end{align} $$ $$ \begin{align} y_1 &= \left( 1, 0, 0, 0, \ldots\right), \\ y_2 &= \left( \frac 1 2, \frac 1 2, 0, 0, 0, \ldots \right), \\ y_3 &= \left( \frac 1 3, \frac 1 3, \frac 1 3, 0, 0, 0, \ldots \right), \\ & \ldots; \end{align} $$ $$ \begin{align} z_1 &= \left( 1, 1, 0, 0, 0, \ldots\right), \\ z_2 &= \left( \frac 1 2, \frac 1 2, 0, 0, 0, \ldots \right), \\ z_3 &= \left( \frac 1 3, \frac 1 3, 0, 0, 0, \ldots \right), \\ & \ldots. \end{align} $$

My effort:

We can prove the following result.

Let $X$ be a non-empty set with topologies $T_1$ and $T_2$ such that $T_1 \subset T_2$. Let $p_n$ be a sequence of points of $X$, and let $p \in X$. Then we have the following:

If $p_n $ converges to $p$ in the topology $T_2$, then $p_n$ converges to $p$ in the topology $T_1$ also. In other words, if $p_n$ fails to converge to $p$ in $T_1$, then $p_n$ fails to converge to $p$ in $T_2$ also.

Am I right?

Also, the product and uniform topologies on $\mathbb{R}^\omega$ are the metric topologies induced by the metrics $D$ and $\tilde{\rho}$, respectively, on $\mathbb{R}^\omega$ defined as follows: $$D\left( x, y \right) = \sup \left\{ \ \frac{ \min \left\{ \left\vert x_n - y_n \right\vert, 1 \right\} }{n} \ \colon \ n \in \mathbb{N} \ \right\}$$ and $$\tilde{\rho}\left( x, y \right) = \sup \left\{ \ \min \left\{ \left\vert x_n - y_n \right\vert, 1 \right\} \ \colon \ n \in \mathbb{N} \ \right\}$$ for all $x = \left(x_n \right)_{n \in \mathbb{N}}$ and $y = \left( y_n \right)_{n \in \mathbb{N}}$ in $\mathbb{R}^\omega$. Am I right?

Moreover, in a metric space, every convergent sequence is Cauchy; so every non-Cauchy sequence in a metric space fails to converge.

We can also state and prove the following.

Let $p_n = \left( \xi_{n1}, \xi_{n2}, \xi_{n3}, \ldots \right)$ be a sequence of points in $\mathbb{R}^\omega$, and let $p = \left( \xi_1, \xi_2, \xi_3, \ldots \right)$ be a point of $\mathbb{R}^\omega$. Suppose that $\mathbb{R}^\omega$ is given the product topology. Then the sequence $p_n$ converges to $p$ in $\mathbb{R}^\omega$ if and only if, for each $j \in \mathbb{N}$, the sequence $\left( \xi_{nj} \right)_{n \in \mathbb{N}}$ converges to $\xi_j$ in $\mathbb{R}$.

Am I right? If so, can we extend the above result (or part of it) to the uniform and the box topologies on $\mathbb{R}^\omega$?

If the last result I've stated is correct, then in the product topology on $\mathbb{R}^\omega$, all of the given sequences can only converge to the point $\mathbf{0} = (0, 0, 0, \ldots)$. So none of these sequences can converge to a point distinct from $\mathbf{0}$ in either the uniform or the product topology. Am I right?

Now for all $n \in \mathbb{N}$, we have $$D\left( w_n , \mathbf{0} \right) = \frac 1 n \to 0 \ \mbox{ as } \ n \to \infty, $$ showing that $w_n$ does converge to $\mathbf{0}$ in the product topology. Am I right?

However, in the uniform metric, we have $$\tilde{\rho}\left(w_n, \mathbf{0} \right) = 1$$ for all $n \in \mathbb{N}$, showing that $w_n$ cannot converge to $\mathbf{0}$ and hence $w_n $ cannot converge to any other point of $\mathbb{R}^\omega$ either. Therefore $w_n$ cannot converge to any point of $\mathbb{R}^\omega$ in the box topology either. Am I right?

For each $n \in \mathbb{N}$, we have $$D\left( x_n , \mathbf{0} \right) = \frac{1}{n^2} \to 0 \ \mbox{ as } \ n \to \infty,$$ showing that $x_n$ converges to $\mathbf{0}$ in the product topology. Am I right?

For each $n \in \mathbb{N}$, we have $$ \tilde{\rho}\left( x_n, \mathbf{0} \right) = \frac 1 n \to 0 \ \mbox{ as } \ n \to \infty,$$ which shows that $x_n$ converges to $\mathbf{0}$ in the uniform topology also. Am I right?

Now there is no $x_n$ in the box topology basis element $$ \left(- \frac{1}{1^3}, \frac{1}{1^3} \right) \times \left( - \frac{1}{2^3}, \frac{1}{2^3} \right) \times \left( - \frac{1}{3^3}, \frac{1}{3^3} \right) \times \left( \frac{1}{4^3}, \frac{1}{4^3} \right) \times \cdots $$ containing $\mathbf{0}$. So $x_n$ cannot converge in the box topology. Am I right?

For each $n \in \mathbb{N}$, we have $$\tilde{\rho}\left( y_n, \mathbf{0} \right) = \frac 1 n \to 0 \ \mbox{ as } \ n \to \infty,$$ which implies that $y_n$ converges to $\mathbf{0}$ in the uniform topology and hence also in the product topology. Am I right?

However, there is no $y_n$ in the box topology basis element $$\left(- \frac 1 2, \frac 1 2 \right) \times \left(- \frac 1 3, \frac 1 3 \right) \times \left(- \frac 1 4, \frac 1 4 \right) \times \cdots$$ containing $\mathbf{0}$, thus showing that $y_n$ cannot converge in the box topology. Am I right?

Let $$ \left( a_1, b_1 \right) \times \left( a_2, b_2 \right) \times \left( a_3, b_3 \right) \times \cdots$$ be a box topology basis element containing $\mathbf{0}$. Then $a_j < 0 < b_j$ for each $j \in \mathbb{N}$. By the Archimedian property of real numbers, there are natural numbers $N_1$ and $N_2$ such that $b_1 N_1 > 1$ and $b_2 N_2 > 1$. Let $N$ be a natural number such that $N \geq \max \left\{ N_1, N_2 \right\}$. Then $$z_n \in \left( a_1, b_1 \right) \times \left( a_2, b_2 \right) \times \left( a_3, b_3 \right) \times \cdots$$ for all $n \in \mathbb{N}$ such that $n > N$, from which it follows that $z_n$ converges to $\mathbf{0}$ in the box topology. Hence $z_n$ converges to $\mathbf{0}$ in both the uniform and the product topology also. Am I right?

Have I been able to get the correct conclusions? If not, where have I erred? If yes, then can we determine the convergence of these sequences using only the machinery developed by Munkres up to this point in the book?

An afterthought:

If that if and only if result about the product topology is correct, then we don't need to bother doing the calculations involving the metric $D$. Am I right?

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  • $\begingroup$ @HennoBrandsma what is it that you've editted in my post? $\endgroup$ – Saaqib Mahmood Jun 12 '16 at 5:22
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    $\begingroup$ Added proof-verification tag. $\endgroup$ – Henno Brandsma Jun 12 '16 at 5:57
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Yes, $D$ is not needed if you have shown the coordinatewise characterisation.

All the reasoning seems correct. BTW, note that $z_n$ is essentially a sequence that lives on the first 2 coordinates, so all topologies coincide there (it's essentially $\mathbb{R}^2$ in different metrics).

I discuss this problem and a related one on continuous functions here, e.g.

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  • $\begingroup$ can we obtain these conclusions using only what Munkres has discussed up to sec. 20 in the book Topology, 2nd edition? $\endgroup$ – Saaqib Mahmood Jun 12 '16 at 5:21
  • $\begingroup$ @SaaqibMahmuud I don't know that book by heart, so I'm not sure. But it's entirely elementary. $\endgroup$ – Henno Brandsma Jun 12 '16 at 5:57

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