(a) If a function is continuous, and we change the topology of the range to a coarser one, then the function remains continuous. And if a function is not continuous, and we make the topology of the range finer, the function remains noncontinuous. So, for each function we may specify the finest topology out of the three given topologies such that the function is continuous relative to it. For $f$ it is the product topology, so that $f$ is continuous relative to the product topology only, and for $g$ and $h$ it is the uniform topology, so that they are not continuous in the box topology only.

The box topology. $((-1,1),(-\tfrac{1}{4},\tfrac{1}{4}),...,(-\tfrac{1}{k^{2}},\tfrac{1}{k^{2}}),...)$ is open in the box topology, but all three functions have its inverse image equal to $\{0\}$ . So, neither function is continuous relative to the box topology.

The uniform topology. $f^{-1}(B_{\overline{\rho}}(\mathbf{0},1))\subset$ $f^{-1}(\prod_{n\in\mathbb{Z}_{+}}(-1,1))=$ $\{0\}$ . At the same time, if the function $k(t)=(a_{1}t,a_{2}t,\ldots)$ equals $g$ ($a_{n}=1$ ) or $h$ ($a_{n}=\tfrac{1}{n}$ ), and $k(t)\in B_{\overline{\rho}}(\mathbf{x},\epsilon)$ , then for every $n\in\mathbb{Z}_{+}$ , $|x_{n}-a_{n}t|\le$ $\sup_{n\in\mathbb{Z}_{+}}|x_{n}-a_{n}t|\overset{def}{=}\delta<\epsilon$ , and for $|z|<\tfrac{\epsilon-\delta}{2}$ , $|x_{n}-a_{n}(t+z)|\le$ $|x_{n}-a_{n}t|+a_{n}|z|<$ $\delta+\tfrac{\epsilon-\delta}{2}=$ $\tfrac{\epsilon+\delta}{2}<\epsilon$ . Hence, $k((t-\tfrac{\epsilon-\delta}{2},t+\tfrac{\epsilon-\delta}{2}))\subset B_{\overline{\rho}}(\mathbf{x},\epsilon)$ , and $k^{-1}(B_{\overline{\rho}}(\mathbf{x},\epsilon))$ is open. So, $f$ is the only function that is not continuous relative to the uniform topology.

The product topology. It is easy to see that all three functions are continuous relative to the product topology. For $f$ it follows from the fact that if $f(t)\in U$ open in the product topology, then $f(t)\in\prod_{n\in\mathbb{Z}_{+}}U_{n}\subset U$ where for finitely many $n\in\mathbb{Z}_{+}$ , i.e. for $n\in\{n_{1},\ldots,n_{k}\}$ , $U_{n}=(nt-n\epsilon_{n},nt+n\epsilon_{n})$ , and for every other $n\in\mathbb{Z}_{+}$ , $U_{n}=\mathbb{R}$ . Therefore, if $\epsilon=\min_{i=1,\ldots,k}\epsilon_{n_{k}}$ , $f((t-\epsilon,t+\epsilon))\subset U$ , and $f^{-1}(U)$ is open. Therefore, $f$ is continuous relative to the product topology. For $g$ and $h$ we can either use a similar argument, or use the fact that they are already continuous relative to the finer uniform topology.

(b) If a sequence converges to a point, and we change the topology to a coarser one, then the sequence still converges to the point (it may converge to new points as well but not in a Hausdorff space). And if a sequence does not converge to a point (or at all), and we change the topology to a finer one, then the sequence still does not converge to the point (or at all). Therefore, for each sequence we may specify the finest topology out of the three given topologies in which it converges to some point (which is unique because all three topologies are Hausdorff). For $\{\mathbf{w}_{n}\}$ it is the product topology, for $\{\mathbf{x}_{n}\}$ and $\{\mathbf{y}_{n}\}$ it is the uniform topology, and for $\{\mathbf{z}_{n}\}$ it is the box topology (i.e. it converges in all three topologies).

In the product topology (metricized) $\{\mathbf{w}_{n}\}$ converges to $\mathbf{0}$ as $\overline{d}(\mathbf{w}_{k},0)/k=1/k$ , and, therefore, any ball centered at $\mathbf{0}$ has all elements of the sequence starting from some index. Therefore, if it did converge in the uniform topology, it would converge to $\mathbf{0}$ , but in the uniform topology the distance between any $\mathbf{w}_{n}$ and $\mathbf{0}$ is always $1$ ($B_{\overline{\rho}}(\mathbf{0},1)$ does not contain any members of the sequence). So, $\{\mathbf{w}_{n}\}$ converges in the product topology only.

$\{\mathbf{x}_{n}\}$ converges to $\mathbf{0}$ in the uniform topology (and, hence, in the product topology), as for $n>\tfrac{1}{\epsilon}$ , $\mathbf{x}_{n}\in B_{\overline{\rho}}(\mathbf{0},\epsilon)$ , but it does not converge in the box topology, as there are no elements of the sequence contained in the open neighborhood $A=(-1/2,1/2)\times(-1/3,1/3)\times...$ of $\mathbf{0}$ (and $\mathbf{0}$ is the only point it may converge to in the box topology).

$\{\mathbf{y}_{n}\}$ converges to $\mathbf{0}$ in the uniform topology (and, hence, in the product topology), as for $n>\tfrac{1}{\epsilon}$ , $\mathbf{y}_{n}\in B_{\overline{\rho}}(\mathbf{0},\epsilon)$ , but not in the box topology (the same set $A$ shows this).

$\{\mathbf{z}_{n}\}$ converges to $\mathbf{0}$ in all three topologies, as it clearly converges to $\mathbf{0}$ in the finest one — the box topology (for $n>\tfrac{1}{\min\{\epsilon_{1},\epsilon_{2}\}}$ , $\mathbf{z}_{n}\in(-\epsilon_{1},\epsilon_{1})\times(-\epsilon_{2},\epsilon_{2})\times\ldots$ ).