Consider $\displaystyle \mathbb{R}$ in the standard topology and $\displaystyle \mathbb{R}^{\omega}$ in the uniform topology.

Consider the function $\displaystyle g:\mathbb{R} \Rightarrow \mathbb{R}^{\omega}$ given by $\displaystyle g(t)=(t,t,t, \ldots)$.

QUESTION: Is $\displaystyle g$ continuous??

MY ATTEMPT:

Let $\displaystyle \bold{x} \in \mathbb{R}^{\omega}$ where $\displaystyle \bold{x}=(x_1,x_2,\ldots)$.

Let $\displaystyle 1> \epsilon >0$.

To see whether $\displaystyle g^{-1}(B_{\overline{\rho}}(\bold{x},\epsilon))$ is open in $\displaystyle \mathbb{R}$ or not.

I have worked out that $\displaystyle B_{\overline{\rho}}(\bold{x},\epsilon)=\bigcup_{ \delta < \epsilon}U(\bold{x},\delta)$, where

$\displaystyle U(\bold{x},\delta)=(x_1-\delta ,x_1+\delta )\times (x_2-\delta ,x_2+\delta )\times \ldots$

Using this i proved that $\displaystyle g^{-1}(B_{\overline{\rho}}(\bold{x},\epsilon))=\bigcup _{ \delta < \epsilon}\left(\bigcap_{i \in \mathbb{Z}^+}(x_i- \delta, x_i + \delta) \right)$

I am not able to see whether $\displaystyle \bigcup_{ \delta < \epsilon}\left(\bigcap_{i \in \mathbb{Z}^+}(x_i- \delta, x_i + \delta) \right)$ is open in $\displaystyle \mathbb{R}$ or not. Can someone please help me on this??