Showing a subspace is closed

50 Views Asked by At

Let $X = (\textbf{c}_{0},\|\cdot\|_{\infty})$ and $Y$ be defined as $$ Y := \bigg\{ \{x_{i}\} \in \textbf{c}_{0} : \sum_{i=1}^{\infty} \frac{x_{i}}{2^{i}} = 0 \bigg\}. $$

(1) Show that $Y$ is a complete subspace of $X$.

Since $X$ is Banach, it is complete. So, it suffices to show that $Y$ is closed in $X$. Let $y \in \overline{Y}$, then there exists $\{y_{n}\} \subset Y$ such that $y_{n} \to y$ as $n \to \infty$. I don't know where to go from here...

In addition, how may these be shown?

(2) For $x = (2,0,0,\ldots) \in \textbf{c}_{0}$, show that $d_{\infty}(x,Y) = 1$.

(3) Show that there exists no $y \in Y$ satisfying $\|x-y\|_{\infty} = d_{\infty}(x,Y)$.