Equivalent of $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ implies an existence of an isomorphism between $[x_n]$ and $[y_n]$

I am reading the book 'Topics in Banach Space Theory'.

Notations: $[x_n]$ denotes the closed linear span of $(x_n)_{n=1}^{\infty}.$ Two bases $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ are equivalent if for all sequence of scalars $(a_n)_{n=1}^{\infty}$, we have $\sum_{n=1}^{\infty} a_n x_n$ converges if and only if $\sum_{n=1}^{\infty} a_n y_n$ converges.

Theorem $1.3.2: \space$Two bases ( or basic sequences) $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ are equivalent if and only if there is an isomorphism $T: [x_n] \rightarrow [y_n]$ such that $Tx_n = y_n$ for each $n$.

Proof: $(\Rightarrow)$ Suppose $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ are equivalent. Let us define $T: [x_n] \rightarrow [y_n]$ by $T(\sum_{n=1}^{\infty}a_nx_n) = \sum_{n=1}^{\infty}a_ny_n$. Then $T$ is one-to-one and onto.

Question: How to prove the bolded sentence? It seems that equivalence of $(x_n)$ and $(y_n)$ ensures well-definedness of $T$.

My attempt: To show $T$ is one-to-one, it suffices to show that $Ker(T) = \{ 0 \}$. Let $x \in [x_n]$ such that $T(x) = 0.$ Then there exists a sequence of unique scalars $(a_n)$ such that $x = \sum_{n=1}^{\infty}a_nx_n$. Hence, we have $\sum_{n=1}^{\infty}a_ny_n = 0$. Since $(y_n)$ is linearly independent, $a_n = 0$ for all $n \in \mathbb{N}$ (I am not sure about this statement). Therefore, $x = 0.$

To prove $T$ is onto, let $y \in [y_n]$. Then there exists a sequence of unique scalars $(a_n)$ such that $y = \sum_{n=1}^{\infty}a_n y_n.$ Let $x = \sum_{n=1}^{\infty}a_n x_n$. $x$ is well-defined because of equivalence. Clearly $T(x) = y.$

• Apr 26, 2017 at 6:11
• I asked the question in your link. The difference between the 2 questions is the converse. Apr 26, 2017 at 10:41

Your solution is very nearly correct. The only suggestion I have is that instead of writing "Since $(y_n)$ is linearly independent" I would write "Since $(y_n)$ is a basic sequence". Linear independence refers strictly to finite linear combinations and as such should not be used in this case. On the other hand, recall that $(v_n)$ is a basic sequence if and only if for every $v\in[v_n]$, there exist unique scalars $(a_n)$ such that $v=\sum_{n=0}^\infty a_nv_n$. So since $\sum_{n=0}^\infty a_ny_n=0=\sum_{n=0}^\infty0y_n$, by uniqueness we conclude $a_n=0$ for all $n\in\mathbb N$.