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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.

  • 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?

  1. The sequence $( \mathbf{w}_n )_{n=1}^\infty$ given by $$\begin{align} \mathbf{w}_1 &= (1, 1, 1, 1, 1, \ldots) \\ \mathbf{w}_2 &= (0, 2, 2, 2, 2, \ldots) \\ \mathbf{w}_3 &= (0, 0, 3, 3, 3, \ldots) \\ \mathbf{w}_4 &= (0, 0, 0, 4, 4, \ldots) \\ &\vdots \\ \mathbf{w}_n &= ( \overbrace{0 , \ldots , 0}^{n-1\text{ times}}, n, n, \ldots ) \\ &\vdots \end{align}$$

  2. The sequence $( \mathbf{x}_n )_{n=1}^\infty$ given by $$\begin{align} \mathbf{x}_1 &= \left(1, 1, 1, 1, 1, \ldots\right) \\ \mathbf{x}_2 &= \left(0, \tfrac 12, \tfrac 12, \tfrac 12, \tfrac 12, \ldots\right) \\ \mathbf{x}_3 &= \left(0, 0, \tfrac 13, \tfrac 13, \tfrac 13, \ldots\right) \\ \mathbf{x}_4 &= \left(0, 0, 0, \tfrac 14, \tfrac 14, \ldots\right) \\ &\vdots \\ \mathbf{x}_n &= ( \overbrace{0 , \ldots , 0}^{n-1\text{ times}}, \tfrac 1n, \tfrac 1n, \ldots ) \\ &\vdots \end{align}$$

  3. The sequence $( \mathbf{y}_n )_{n=1}^\infty$ given by $$\begin{align} \mathbf{y}_1 &= \left(1, 0, 0, 0, 0, \ldots\right) \\ \mathbf{y}_2 &= \left(\tfrac 12, \tfrac 12, 0, 0, \ldots\right) \\ \mathbf{y}_3 &= \left(\tfrac 13, \tfrac 13, \tfrac 13, 0, 0, \ldots\right) \\ \mathbf{y}_4 &= \left(\tfrac 14, \tfrac 14, \tfrac 14, \tfrac 14, 0, \ldots\right) \\ &\vdots \\ \mathbf{y}_n &= ( \overbrace{\tfrac 1n , \ldots , \tfrac 1n}^{n\text{ times}}, 0, 0, \ldots ) \\ &\vdots \end{align}$$

  4. 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?

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    $\begingroup$ My answer to this earlier instance of the question may help a bit, though most of the hints in it are already present in G. Sassatelli’s answer here. $\endgroup$ – Brian M. Scott Apr 8 '15 at 16:44
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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.

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  • $\begingroup$ you've stated that if a topology is coarser than a Hausdorff topology, then it is Hausdorff as well. Are you sure? Or, do you want to replace the word "coarser" with the word "finer"? $\endgroup$ – Saaqib Mahmood Apr 19 '15 at 16:18
  • $\begingroup$ Yes, @SaaqibMahmuud, you're right. I'll edit it right away. Thank you! $\endgroup$ – user228113 Apr 19 '15 at 18:34
  • $\begingroup$ we can also state the following: a sequence $x_n \to x $ in the product topology on $\Pi_{\alpha \in J} X_\alpha$, where $J$ is finite or infinite, if and only if for each $\beta \in J$ the sequence $\pi_\beta (x_n) \to \pi_\beta(x)$. But we we can only state the only-if part in the box topology in case $J$ is infinite. Am I right? $\endgroup$ – Saaqib Mahmood Apr 20 '15 at 2:55
  • $\begingroup$ is the following correct? If $x_n \to x$ in $\Tau_1$ and if $\Tau_1$ is finer than $\Tau_2$, then $x_n \to x$ in $\Tau_2$ as well; in other words, if $\Tau_1$ is finer than $\Tau_2$ and if $x_n$ fails to converge to $x$ in $\Tau_2$, then $x_n$ fails to converge to $x$ in $\Tau_1$ either. $\endgroup$ – Saaqib Mahmood Apr 20 '15 at 3:36
  • $\begingroup$ @SaaqibMahmuud The first statement is correct in the part regarding the product topology, but it is incorrect in the part regarding the box topology: what holds in the general case is the "only if" part ($x_n\rightarrow x$ in the box topology only if $\forall\alpha\pi_\alpha(x_n)\rightarrow\pi_\alpha(x)$), and the "if" part does not (it's not true that $x_n\rightarrow x$ in the box topology if $x_n\rightarrow x$ pointwise). $$\ $$ Of course, if $I$ is finite those three topologies are coincident.$$\ $$ Your second statement is true.$$\ $$ $\endgroup$ – user228113 Apr 20 '15 at 5:52

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