Comparing different topologies on the Hilbert cube $H = \prod_{n \in \mathbb{N}} [0,\frac 1n]$ This is essentially exercise 8(c) from section 20 (p.128) of Munkres's Topology:

Let $X$ be the set of all the sequences $(x_n)$ of real numbers such that the series $\sum_{n=1}^\infty x_n^2$ converges. The Hilbert cube $H$ is given by 
  $$H \colon= \prod_{n \in \mathbb{N}} \left[ 0, \frac{1}{n} \right].$$
Then $H$ is contained in $X$. 
How are the subspace topologies that $H$ inherits as a subspace of $X$ related when $X$ is given the  (i) product topology? (ii)  uniform topology? (iii) $\ell^2$ topology? (iv) box topology? 

I know that on $X$, the box topology contains the $\ell^2$ topology, which contains the uniform topology, which in turn contains the product topology. 
But which of these chains of inclusions are proper? 
Does any of the above inclusion relations becomes proper when restricted to $H$? 
 A: Let
$\tau_1$ be the product topology on $H$.
$\tau_2$ be the uniform topology on $H$.
$\tau_3$ be the $l^2$ topology on $H$.
$\tau_4$ be the box topology on $H$.
Claim 1: $\tau_1=\tau_2=\tau_3$
Proof of Claim 1: We clearly have that $\tau_1 \subseteq \tau_2 \subseteq \tau_3$, so it suffices to show that $\tau_3 \subseteq \tau_1$. To do this, we will show that if $f \in H$ and $\epsilon>0$, then there exists $U \in \tau_1$ such that $f \in U$ and $||f-g||_2<\epsilon$ whenever $g \in U$.
Thus, let $f\in H$ and let $\epsilon >0$. Choose $N \in \mathbb{N}$ such that $\sum_{k=N+1}^{\infty} \frac{1}{k^2}<\frac{\epsilon^2}{2}$. Next, define $U$ to be the set of all $g \in H$ such that $\max_{1 \leq k \leq N}|g(k)-f(k)|<\frac{\epsilon}{\sqrt{2N}}$. Then $U\in \tau_1$ since $U = \bigcap_{k=1}^N \pi_k^{-1}\big(f(k)-\frac{\epsilon}{\sqrt{2N}}\;,\; f(k)+\frac{\epsilon}{\sqrt{2N}}\big)$, where the maps $\pi_k:H \to [0,1/k]$ are the canonical projections. Then $g \in U$ implies
$$||g-f||_2^2=\sum_{k=1}^{\infty}|g(k)-f(k)|^2 = \sum_{k=1}^N|g(k)-f(k)|^2 + \sum_{k=N+1}^{\infty}|g(k)-f(k)|^2$$ $$<\sum_{k=1}^N \frac{\epsilon^2}{2N} + \sum_{k=N+1}^{\infty} \frac{1}{k^2}<\frac{\epsilon^2}{2}+\frac{\epsilon^2}{2} = \epsilon^2$$ so that $||g-f||_2<\epsilon$, which proves the claim. $\Box$
Claim 2: $\tau_2 \neq \tau_4$
Proof of Claim 2: Let $U=\Pi_n (0,\frac{1}{n})$. Then clearly $U \in \tau_4$. We will show that $U \notin \tau_2$. To prove this, suppose (for contradiction) that $U \in \tau_2$, and fix some $f \in U$. By definition of uniform topology, there exists some $\epsilon>0$ such that $g \in U$ whenever $||f-g||_{\infty}<\epsilon$. Choose $k_1 \in \mathbb{N}$ such that $\frac{1}{k_1}<\epsilon$. Then consider $g \in H$ defined by $g(k_1)=\frac{1}{k_1}$ and $g(k)=f(k)$ for $k \neq k_1$. Then $||f-g||_{\infty}<\epsilon$ but $g \notin U$ (since $g(k_1)=\frac{1}{k_1}$), which is a contradiction. $\Box$
From Claim 1 and Claim 2, we see that $\tau_1 = \tau_2 =\tau_3 \subset \tau_4$
