I am looking at a problem in Conway's functional analysis text. It is problem II.5.7, which states: "If $T$ is compact and $\mathscr{M}$ is an invariant subspace for $T$, show that $T|_{\mathscr{M}}$ is compact." I take this to mean that for every bounded sequence ${x_m}_{m\in\mathbb{N}}$ in $\mathscr{M}$, we need to show that there is a subsequence ${x_{m_k}}_{k\in\mathbb{N}}$ such that $T|_{\mathscr{M}}x_{m_k} = Tx_{m_k}$ converges to some element $y$, and $y$ must be in $\mathscr{M}$. If $\mathscr{M}$ is closed, this is pretty much trivial. However, I am confused as to how to handle it when $\mathscr{M}$ is not closed.
Now there are seemingly various answers to this problem on StackExchange, but they all seem to ignore this fact that $y$ must be in $\mathscr{M}$. My question is: am I misunderstanding the problem, or is it stated incorrectly? Thanks.
Indeed, the claim is false in general if the invariant subspace is not assumed to be closed.
Consider $T : \ell^2 \to \ell^2$ given by $T(x_n)_n = \left(\frac{x_n}n\right)_n$. Then $T$ is compact and $c_{00}$, the space of all finitely-supported sequences, is $T$-invariant.
Consider the sequence given by $y_n = \left(\frac12, \frac14, \ldots, \frac1{2^n}, 0, 0, \ldots\right) \in c_{00}$. Then $(y_n)_n$ converges to the vector $\left(\frac1{2^n}\right)_n \in \ell^2$ (in particular it is bounded) so $(Ty_n)_n$ converges to the vector $\left(\frac1{n2^n}\right)_n \in \ell^2 \setminus c_{00}$.
Hence there isn't a subsequence of $(Ty_n)_n$ which converges to a vector in $c_{00}$.