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

Question

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?

Answer 1

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.