difference between uniform topology and product topology can anyone make an example that implies the difference  between uniform topology and product topology on $\Bbb{R}^\infty$ ($\Bbb{R}\times \Bbb{R} \times \Bbb{R} \times \Bbb{R} \times ....$)?
 A: The set $U = \{(x_n): \exists M: \forall n |x_n| < M  \}$, the set of all bounded sequences in $\mathbb{R}^{\infty}$ is open in the uniform topology, but not in the product topology.
To see this, let $(x_n)$ be any sequence in $U$, with corresponding bound $M$. If $d_u((x_n), (y_n)) < 1$, where $d_u$ is the uniform distance on $\mathbb{R}^{\infty}$, this means that all $|x_n - y_n| < 1$, and so $|y_n| < |x_n| + 1 < M+1$ for all $n$, so $(y_n)$ is also bounded. So the whole open ball around $(x_n)$ with radius 1 is contained in $U$, so every $(x_n)$ in $U$ is an interior point, meaning $U$ is open.
If however $U_1 \times U_2 ...\times U_n \times \mathbb{R} \times \mathbb{R} \ldots $is any basic open set of $\mathbb{R}^{\infty}$ in the product topology, where the $U_i$ are non-empty open subsets of $\mathbb{R}$, then we can choose points $n, n+1,\ldots$ for the higher coordinates, so every basic open subset of $\mathbb{R}^{\infty}$ in the product topology contains unbounded sequences. So $U$ has no interior points at all in the product topology.
This shows that they are quite different.
Another difference is that the former is not separable, while the product topology is.
