$(a)$ Show that on $X$, we have the inclusions $ \mbox{ box topology } \ \supset \ \ell^2 \mbox{ topology } \ \supset \ \mbox{ uniform topology}$ Let $X$ be the subset of $\mathbb{R}^\omega$ consisting of all sequences $x$ such that $\sum x_i^2$ converges. Then the formula $$d(x,y) = \left[ \sum_{i=1}^\infty \left( x_i - y_i \right)^2 \right]^{\frac{1}{2}} $$ defines a metric on $X$. On $X$ we have the three topologies it inherits from the box, uniform, and product topologies on $\mathbb{R}^\omega$. We have also the topology given by the metric $d$, which we call the $\ell^2$ topology. 

(a) Show that on $X$, we have the inclusions $$ \mbox{ box topology } \ \supset \ \ell^2 \mbox{ topology } \ \supset \ \mbox{ uniform topology}.$$

as  i got answer  here How can the little l2 topology be finer than the uniform topology?
But  i didn't understand  this...
Pliz  give  me the proper  solution  so that  i can  easily understands....
 A: 
If we have two topologies $\mathcal{T}_1$ and $\mathcal{T}_2$ on $X$ (with local neighbourhood bases $\mathcal{N}_1(x), \mathcal{N}_2(x)$ resp., then $\mathcal{T}_1 \subseteq \mathcal{T}_2$ iff for every $x \in X$ and every $O \in \mathcal{N}_1(x)$ there is some $O' \in \mathcal{N}_2(x)$ with $O' \subseteq O$.  

This is quite clear from the definitions. It might be a lemma in your text, or easily provable yourself.
With the $d$-metric (the $\ell_2$-metric) and the uniform metric $d_u$ we have such a situation: If $x \in X$ and $r>0$ (we can assume $r<1$ wlog, as such balls form a local base anyway) we have $B_d(x,\frac{r^2}{4}) \subseteq B_{d_u}(x,r)$. This implies that $\mathcal{T}_u \subseteq \mathcal{T}_d$, as required.
Also, if we have an open ball $B_d(x,r)$ in the $\ell_2$-metric, we can define the basic box-open set $U(x) = \prod_n (x_n - \frac{r^2}{2^n}, x + \frac{r^2}{2^n})$ and prove that $U(x) \subseteq B_d(x,r)$, showing $\mathcal{T}_d \subseteq \mathcal{T}_{\text{box}}$
E.g. define the sequence in $X$:
$x_1 = (1,0,0,0,\ldots)$
$x_2 = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0, \ldots,)$
$\ldots$
$x_n = (\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}}, \ldots, \frac{1}{\sqrt{n}} (n \text{ terms })\,, 0,0 \ldots,)$
which converges to $(0,0,0,\ldots)$ in the uniform metric (and product topology too), but not in the $d$-metric (where the distance to that point from each $x_n$ is $1$).
On the other hand, it's also easy to find convergent sequences under $d$ that are not box-convergent. Such sequences show the properness of the inclusions.
