(a) To show that a topology $\mathcal{T}$ is finer than a metric topology $d$ , it is enough to show that for any $B_{d}(x,\epsilon)$ there is $U\in\mathcal{T}$ such that $x\in U\subset B_{d}(x,\epsilon)$ (i.e. we need to show there is a neighborhood contained in the ball of the center point only, as for any other point in the ball we can first find another ball centered at it; the argument is similar to the second part of Lemma 20.2). Let $M(\mathbf{x})=\{\mathbf{y}||x_{i}-y_{i}|^{2}<\epsilon^{2}/2^{i}\}$ . Then it is open in the box topology, and for every $\mathbf{y}\in M(\mathbf{x})$ , $\sum_{i=1}^{\infty}y_{i}^{2}<$ $\sum_{i=1}^{\infty}(|x_{i}|+\tfrac{\epsilon^{2}}{2^{i}})^{2}$ converges (because $|x_{i}|$ ’s are bounded) and $\sum_{i=1}^{\infty}|x_{i}-y_{i}|^{2}<\epsilon^{2}$ , so that $M(\mathbf{x})\subset B_{l^{2}}(\mathbf{x},\epsilon)\subset B_{\overline{\rho}}(\mathbf{x},\epsilon)$ (the second inclusion follows from $\overline{\rho}(\mathbf{x},\mathbf{y})\le d(\mathbf{x},\mathbf{y})$ ).

(b) The finer the topology of a space, the finer the topology of any its subspace. So, given (a) and the fact that the uniform topology is finer then the product topology, to show they are all different as subspace topologies, we only need to find subsets of $\mathbb{R}^{\infty}$ open in a finer topology but not in a coarser one. The set $A$ we used in 4(b) is open in the box topology, and its intersection $A'$ with $\mathbb{R}^{\infty}$ is open in the subspace. $\mathbf{0}\in A'$ , but for every $\epsilon>0$ , $B_{l^{2}}(\mathbf{0},\epsilon)\cap\mathbb{R}^{\infty}$ has a point not in $A'$ (just make a sufficiently large coordinate $\epsilon/2$ and all others $0$ ). Further, for every $\epsilon>0$ and $\delta>0$ , for a sufficiently large $n$ , $(\underbrace{\delta/2,...,\delta/2}_{n},0,0,...)\in$ $(B_{\overline{\rho}}(\mathbf{0},\delta)-B_{l^{2}}(\mathbf{0},\epsilon))\cap\mathbb{R}^{\infty}$ . Finally, for every $0<\epsilon<1$ and $\delta>0$ , for a sufficiently large $k$ , $(\tfrac{\delta}{2},\delta,\tfrac{3\delta}{2},...,\tfrac{k\delta}{2},0,0,...)\in$ $(B_{D}(\mathbf{0},\delta)-B_{\overline{\rho}}(\mathbf{0},\epsilon))\cap\mathbb{R}^{\infty}$ .

(c) The box topology is strictly finer than the $l^{2}$ -topology on $H$ : $A\cap H=\prod_{n=1}^{\infty}[0,\tfrac{1}{n+1})$ is an open neighborhood of $\mathbf{0}$ in $H_{b}$ , but the intersection of any $B_{l^{2}}(\mathbf{0},\epsilon)$ with $H$ has a point $(\underbrace{0,\ldots,0}_{n-1},\tfrac{1}{n+1},0,0,\ldots)$ for a sufficiently large $n$ which is not in $\prod_{n=1}^{\infty}[0,\tfrac{1}{n+1})$ .

The other three topologies are the same on $H$ . To show this we show that the product topology is finer than the $l^{2}$ -topology on $H$ . Intuitively, this is true because unbounded spaces in the product of open sets (a basis element for the product topology) becomes bounded by the sequence $\{\tfrac{1}{n}\}$ which converges in squares. For every $\epsilon>0$ , consider the ball $B=B_{l^{2}}(\mathbf{x},\epsilon)$ . We only need to show that $\mathbf{x}$ has an open neighborhood in $H_{p}$ that lies within the ball (see the comment for (a)). Consider the following set $B'=\prod_{i\le n}((x_{i}-\tfrac{\epsilon^{2}}{2n},x_{i}+\tfrac{\epsilon^{2}}{2n})\cap[0,\tfrac{1}{i}])\times$ $\prod_{i>n}[0,\tfrac{1}{i}]$ where $n$ is sufficiently large such that $\sum_{i>n}\tfrac{1}{i^{2}}<\tfrac{\epsilon^{2}}{2}$ . $B'$ is open in $H_{p}$ . Further, for $\mathbf{y}\in B'$ , $d^{2}(\mathbf{x},\mathbf{y})<$ $\sum_{i\le n}\tfrac{\epsilon^{2}}{2n}+\sum_{i>n}\tfrac{1}{i^{2}}<$ $\epsilon^{2}$ , hence, $\mathbf{x}\in B'\subset B$ .