How to prove that $U=\{f\in X^G: f(e)=b_k\}$ is open in $X^G=\{f\,\,|\,\,f:G\to X\}.$

Question

Let $G$ be a Boolean group of the cardinality $\mathfrak{c}=2^\mathbb{N}$ and $X=\{a\}∪\{b_n: n \in \mathbb{N}\}$ a convergent sequence with the limit point $a$. If $e$ is the neutral element of $G$ and $h:G\to X$, then $h(e) = b_k$, for some $k\in \mathbb{N}$.

How to prove that $U=\{f\in X^G: f(e)=b_k\}$ is open in $X^G=\{f\,\,|\,\,f:G\to X\}.$

Thanks for you help!