Prove that a Schauder basis consists of isolated points.

Question

Let $X$ be a Banach space. Suppose $(x_n) \in X$ is a sequence such that every $x \in X$ has a unique representation in the form $x = \sum_{n = 1}^{\infty} \lambda_ n x_n$. Prove that the set $\{x_n\}$ consists of isolated points.

Details

• Such a sequence is a Schauder basis of the space
• The goal is to prove that each point in $\{x_n\}$ is an isolated point of the set $\{x_n\}$
• Tried to prove by contradiction, no success

Answer 1

For a fixed $n$, the linear functional $f(x) = \lambda_n$ is continuous (discussed here). Since $f_n(x_n) = 1$, there is a neighborhood of $x_n$ where $f_n>0$. This neighborhood contains no other basis elements.