I just have a few question about some things in Banach spaces. Let $X$ be a separable, reflexive Banach space with basis $\{e_{i}\}$.
Let $X_{n} = \text{span}\{e_{1},...,e_{n}\}$, then consider the closed ball $\bar{B_{R_{o}}(0)} \subset X_{n}$ for some $R_{o} > 0$. Would it follow immediately that $\bar{B_{R_{o}}(0)}$ is convex and compact in $X$?
Would it follow that $\bar{B_{R_{o}}(0)}$ is homeomorphic to the closed unit ball in $\mathbb{R}^{n}$?
Also if we construct such subspaces $X_{n}$ for each $n \in \mathbb{N}$. Why does it follow that $\cup_{n=1}^{\infty}X_{n}$ is dense in $X$? It seems that $\cup_{n=1}^{\infty}X_{n}$ should immediately be equivalent to $X$.
Thanks for any help.
Yes. The space $X_n$ with the norm induced from $X$ is a finite-dimensional normed space. The closed ball is bounded and closed, therefore compact.
Yes. Every convex compact set with nonempty interior in $n$-dimensional space is homeomorphic to the Euclidean closed unit ball. One can follow the proof here.
The [Schauder] basis assumption implies that $\{e_i\}$ has dense linear span in $X$. This linear span is exactly the union of all spaces $X_n$. So the union is dense. To see that it need not be all of $X$, take, for example, $\ell^2$, and observe that $\bigcup_{n=1}^\infty X_n$ does not contain the element $a=(1,1/2,1/3,1/4,\dots)$. Indeed, if $a$ was an element of the union $\bigcup_{n=1}^\infty X_n$, it would have to belong to some $X_n$, which is clearly false.