This is Exercise 4(b) in section 20 of Munkres' Topology (2nd edition).
Let $\mathbb{R}^\omega$ denote the set of all infinite sequences of real numbers.
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Let the uniform metric $\tilde{\rho}$ on $\mathbb{R}^\omega$ be defined as follows: for any $x = (x_i)_{i \in \mathbb{N}}$, $y=(y_i)_{i \in \mathbb{N}} \in \mathbb{R}^\omega$, $$\tilde{\rho}(x,y) \colon= \sup \left\{ \ \min \left( \ \vert x_i – y_i \vert \ , 1 \ \right) \ \colon \ i = 1, 2, 3, \ldots \ \right\}.$$
Then the topology induced by this metric is called the uniform topology on $\mathbb{R}^\omega$. -
The product topology on $\mathbb{R}^\omega$ has as a basis all sets of the form
$$U_1 \times U_2 \times U_3 \times \cdots, $$
where each $U_i$ is open in $\mathbb{R}$ and the $U_i$ are distinct from $\mathbb{R}$ for only finitely many $i$. -
The box topology on $\mathbb{R}^\omega$ has as a basis all sets of the form
$$ (a_1, b_1) \times (a_2, b_2) \times (a_3, b_3) \times \cdots, $$
where $(a_i)_{i \in \mathbb{N}}, (b_i)_{i \in \mathbb{N}} \in \mathbb{R}^\omega$ such that $a_i < b_i$ for each $i$, and $(a_i, b_i)$ denotes the segment (i.e. open interval) with $a_i$ as the left endpoint and $b_i$ as the right endpoint.
In which of the above three topologies are the following sequences convergent?
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The sequence $( \mathbf{w}_n )_{n=1}^\infty$ given by
$$\begin{align}
\mathbf{w}_1 &= (1, 1, 1, 1, 1, \ldots) \\
\mathbf{w}_2 &= (0, 2, 2, 2, 2, \ldots) \\
\mathbf{w}_3 &= (0, 0, 3, 3, 3, \ldots) \\
\mathbf{w}_4 &= (0, 0, 0, 4, 4, \ldots) \\
&\vdots \\
\mathbf{w}_n &= ( \overbrace{0 , \ldots , 0}^{n-1\text{ times}}, n, n, \ldots ) \\
&\vdots
\end{align}$$ -
The sequence $( \mathbf{x}_n )_{n=1}^\infty$ given by
$$\begin{align}
\mathbf{x}_1 &= \left(1, 1, 1, 1, 1, \ldots\right) \\
\mathbf{x}_2 &= \left(0, \tfrac 12, \tfrac 12, \tfrac 12, \tfrac 12, \ldots\right) \\
\mathbf{x}_3 &= \left(0, 0, \tfrac 13, \tfrac 13, \tfrac 13, \ldots\right) \\
\mathbf{x}_4 &= \left(0, 0, 0, \tfrac 14, \tfrac 14, \ldots\right) \\
&\vdots \\
\mathbf{x}_n &= ( \overbrace{0 , \ldots , 0}^{n-1\text{ times}}, \tfrac 1n, \tfrac 1n, \ldots ) \\
&\vdots
\end{align}$$ -
The sequence $( \mathbf{y}_n )_{n=1}^\infty$ given by
$$\begin{align}
\mathbf{y}_1 &= \left(1, 0, 0, 0, 0, \ldots\right) \\
\mathbf{y}_2 &= \left(\tfrac 12, \tfrac 12, 0, 0, \ldots\right) \\
\mathbf{y}_3 &= \left(\tfrac 13, \tfrac 13, \tfrac 13, 0, 0, \ldots\right) \\
\mathbf{y}_4 &= \left(\tfrac 14, \tfrac 14, \tfrac 14, \tfrac 14, 0, \ldots\right) \\
&\vdots \\
\mathbf{y}_n &= ( \overbrace{\tfrac 1n , \ldots , \tfrac 1n}^{n\text{ times}}, 0, 0, \ldots ) \\
&\vdots
\end{align}$$ -
The sequence $( \mathbf{z}_n )_{n=1}^\infty$ given by
$$\begin{align}
\mathbf{z}_1 &= \left(1, 1, 0, 0, 0, \ldots\right) \\
\mathbf{z}_2 &= \left(\tfrac 12, \tfrac 12, 0, 0, 0, \ldots\right) \\
\mathbf{z}_3 &= \left(\tfrac 13, \tfrac 13, 0, 0, 0, \ldots\right) \\
\mathbf{z}_4 &= \left(\tfrac 14, \tfrac 14, 0, 0, 0, \ldots\right) \\
&\vdots \\
\mathbf{z}_n &= \left(\tfrac 1n, \tfrac 1n, 0, 0, 0, \ldots\right) \\
&\vdots
\end{align}$$
My effort:
I claim that the sequence $( \mathbf{w}_n )_{n=1}^\infty$ converges in the product topology to the point $\mathbf{w} \colon= (0, 0, 0, \ldots) \in \mathbb{R}^\omega$. Let $U \colon= \Pi_{i \in \mathbb{N}} U_i$ be a product topology basis element containing $\mathbf{w}$. Let $i_1, \ldots, i_m$ be the finitely many indices for which $U_i$ is distinct from $\mathbb{R}$. Then $\mathbf{w}_n \in U$ for all $n > \max\{ i_1, \ldots, i_m\}$ because the $i$-th coordinate of $\mathbf{w}_n$ is zero for all $i \leq \max\{ i_1, \ldots, i_m\}$ if $n > \max\{ i_1, \ldots, i_m\}$.
Am I right? If so, is there any point other than $\mathbf{w}$ to which this sequence can converge in the product topology?
The sequence $( \mathbf{x}_n )_{n=1}^\infty$ converges to the point $\mathbf{x} \colon= (0, 0, 0, \ldots) \in \mathbb{R}^\omega$ in the product topology. Let $U \colon= \Pi_{i \in \mathbb{N}} U_i$ be again a product topology basis element containing $\mathbf{x}$. Let $i_1, \ldots, i_m$ be the finitely many indices for which $U_i$ is distinct from $\mathbb{R}$. Then there exist positive real numbers $\delta_1, \ldots, \delta_m$ such that the open interval $(-\delta_k, +\delta_k) \subset U_{i_k}$ for each $k = 1, \ldots, m$. Let $\delta \colon= \min\{\delta_1, \ldots, \delta_m\}$. Then the open interval $(-\delta, +\delta) \subset U_{i_k}$ for each $k = 1, \ldots, m$. Let $N \in \mathbb{N}$ such that $N > 1/\delta$. Then $\mathbf{x}_n \in U$ for all $n > N$ because each co-ordinate of $\mathbf{x}_n$ is either $0$ or $1/n$ which both belong to $(-\delta, +\delta)$.
And a similar kind of reasoning holds for $( \mathbf{y}_n )_{n=1}^\infty$ and $( \mathbf{z}_n )_{n=1}^\infty$, both of which also converge to the point $(0, 0, 0, \ldots) \in \mathbb{R}^\omega$ in the product topology.
Am I right? And if so, then is there any other point to which the sequence $( \mathbf{x}_n )_{n=1}^\infty$ can converge in the product topology?
What is the situation for each of these sequences in $\mathbb{R}^\omega$ in the box topology?
What is the situation for each of these sequences in $\mathbb{R}^\omega$ in the uniform topology?
Best Answer
You may preliminarly observe that these topologies are related in the following way:
the product topology (from now on $Top_1$) is coarser than the uniform topology
the uniform topology (from now on $Top_2$) is coarser that the box topology (from now on $Top_3$)
It's a flashy way to state $Top_1\subseteq Top_2\subseteq Top_3$.
This means that every limit point in $Top_3$ is a limit point in $Top_2$, and every limit point in $Top_2$ is a limit point in $Top_1$.
First question: you're right.
Second question: there isn't, since the product of Hausdorff topologies is Hausdorff (therefore uniqueness of limit holds). The same holds for $Top_2$ and $Top_3$ (if a topology is finer than a Hausdorff topology, it is Hausdorff as well).
If I may suggest a viable strategy, the last fact tells you that you need only prove convergence in the coarsest possible.
An example of what I mean is: $x_n\rightarrow (0,0,\ldots)$ in $Top_2$ (easy proof), and therefore in $Top_1$, but it's not convergent in $Top_3$.
Indeed, if $x_n\rightarrow x$ in $Top_3$, then $x_n\rightarrow x$ in $Top_2$, therefore $x=(0,0,\ldots)$.
But let $(0,0,\ldots)\in U=\Pi_{i\in\mathbb{N}}(-\frac{1}{i},\frac{1}{i})$. You can easily show that $\forall n\ x_n\notin U$.
Anyway, your proof that $x_n\rightarrow (0,0,\ldots)$ in $Top_1$ works as well.
You may try $y_n$ and $z_n$ by yourself with this approach.