Does the following sequence converge in product, uniform and box topologies?

Consider the product, uniform and box topologies on $\mathbb{R}^\omega$.

In which the topologies does the following sequence converge?

$$z_1 =(1,1,0,0, \ldots)$$

$$z_2 =\left(\frac{1}{2},\frac{1}{2},0,0, \ldots\right)$$

$$z_3 =\left(\frac{1}{3},\frac{1}{3},0,0, \ldots\right)$$

$$\vdots$$

My attempt:

By theorem 19.6 and example 2 page 117 in Munkres' topology book, in the box topology, the sequence $z_n$ will not converge, but in the uniform and product topologies it will converge.

Is this correct?

• This sequence converges in in the box topology. Assume $U=\prod_i\in\omega U_i$ is a neighborhood of $(0,0,...)$ with $U_i\ni 0$ open in $R$. Then for $N$ large enough $1/n\in U_1\cap U_2$. Therefore, $z_n\in U$. Should your example be $z_n=(1/n,1/n,1/n,...)$? Mar 31 '18 at 14:10
• To the proposer: The 10 or 11 people in the world (like me) who don't have a copy of Munkres' book would like to ask what Theorem19.6 and Example 2 Page 117 are. We are curious as to what your mistake was.:) Mar 31 '18 at 19:12
• i lived in Nepal ,,very backward Area.and poor country .. there is no so much higher eduaction system,,like iNDIA,USA ,european counrty...as my english language is also very weak,,,as im newly learning topology..@DanielWainfleet Mar 31 '18 at 20:02

It converges in all three topologies to $(0,0,\ldots)$.

It is known that on $\mathbb{R}^\omega$ we have

$$\text{ product topology } \subseteq \text{ uniform topology } \subseteq \text{ box topology}$$

Therefore, it suffices to show that the sequence converges in the box topology.

Let $\prod_{n\in\mathbb{N}} U_n$ be a basis neighbourhood of $(0,0,\ldots)$. Then $U_1$ and $U_2$ are neighbourhoods of $0$ in $\mathbb{R}$ so there exist $n_1, n_2 \in \mathbb{N}$ such that $n \ge n_1 \implies \frac1n \in U_1$ and $n \ge n_2 \implies \frac1n \in U_2$.

Therefore, $n \ge \max\{n_1, n_2\} \implies \left(\frac1n, \frac1n, 0, 0,\ldots \right) \in \prod_{n\in\mathbb{N}} U_n$.

We conclude $\left(\frac1n, \frac1n, 0, 0,\ldots \right) \xrightarrow{n\to\infty} (0,0,\ldots)$ in the box topology.

• thanks a lots @ mechanodroid Mar 31 '18 at 17:59
• You should also point out that if $(0,0,0,...)\ne x\in \Bbb R^{\omega}$ then in any of the topologies, $x$ has a nbhd that contains only finitely many terms of the series, so $(0,0,0,..)$ is the $only$ limit point of the series. Mar 31 '18 at 19:07
• @DanielWainfleet Doesn't if follow from the fact that all three topologies are Hausdorff? So the only limit point must be precisely the limit $(0,0,\dots)$? Mar 31 '18 at 20:57
• Yes. Obvious. Now that you mentioned it. I am sometimes obsessed with justifying every little detail. If you ask me to prove that $\pi>3.14$ I might start by defining $\Bbb R$. Apr 1 '18 at 16:48