Homeomorphism on the Hilbert space We can consider two different topologies on the Hilbert space ; $l^{2}(\mathbb{N})$. 
One is the topology deduced from the norm 
\begin{equation*}
\|f\|=\sqrt{\sum_{n=1}^{\infty} f(n)^{2}}, 
\end{equation*}
and the other one is a subspace topology from $\mathbb{R}^{\omega}$ on $l^{2}(\mathbb{N})$.

I want to prove or disprove that the above two topological spaces are homeomorphic.

The only thing I found is the identity map is not open, so the two topologies are not equal. 

Any help will be appreciated. Thank you.
It is enough to show that $(l^2(\omega), \tau)$ where $\tau$ is the product topology on $\mathbb{R}^{\omega}$ restricted to $l^2(\omega)$ is not completely metrizable. Suppose not. Then $l^2(\omega)$ must be a dense $G_{\delta}$ subset of $\mathbb{R}^{\omega}$. But then $l^2(\omega)$ is comeager in $\mathbb{R}^{\omega}$ which is easily refuted by observing that the sets $A_n = \{x \in l^2(\omega) : \sum_k (x(k))^2 &lt; n\}$ are nowhere dense in $\mathbb{R}^{\omega}$.
Consider the sequence $e_k = \langle 0, 0, \dots, 1, 0, 0, \dots  \rangle$ where $1$ occurs at the $k$-th coordinate. This sequence converges to the constant sequence $\overline{0} = \langle 0, 0, 0, \dots \rangle$ in the (product) topology $\mathbb{R}^{\omega}$ simply because every basic open neighborhood of $\overline{0}$ contains $e_k$ for all sufficiently large $k$. But in the Hilbert space $l_2(\omega)$ this sequence has a constant distance $\sqrt{2}$ from $\overline{0}$. Hence the two topologies are different.