Is the linear operator $(Tx)_n = \frac{x_n}{n}$ a homeomorphism?

Question

Let $X$ be an Hilbert subspace of $\ell^2$ over $\mathbb{C}$

$\forall v \in \ell^2$ let indicate with $v_n$ the nth element of the sequence $v$

Let $Y = \{y \in \ell^2 \backslash \exists x \in X: y_n = \frac{x_n}{n} \}$

Let $T: X \to Y$ be the linear operator such that $(Tx)_n = \frac{x_n}{n}$

My question is: Is $T$ a homeomorphism?

Answer 1

It need not be for an infinite-dimensional subspace. Suppose the zero sequence and at least a countable subset of the countable family of sequences $\mathcal{F} = \{s_k\}_{k=1}^\infty, \text{where } (s_k)_n = \delta_{kn}$ is in X. ($s_k$ is the sequence that has zeros as entries except for the kth entry, where it is 1.)

Note that $Ts_k \rightarrow 0 \in \ell^2$, but $s_k \nrightarrow 0$. This shows that $T^{-1}$ is not continuous. Hence T is not a homeomorphism.