In Banach space over $\mathbb{C}$ maximal chain in the collection of all-invariant-subspaces for-compact-operator I saw the page
Maximal chain in the collection of all invariant subspaces for compact operator $K$
And I cannot understand Ali's answer. I'm wondering that where the $X$ is Banach space over "$\mathbb{C}$".
I'm trying to make that if $\mathcal{L}$ is not maximal, then we can make $\mathcal{L} \subset \mathcal{L_2} $ that $\mathcal{L_2}$ is chain of $\mathtt{Lat}K$.
The problem is in below.

Let $X$ be a Banach space over ${\Bbb C}$, and $K\in K(X)$ ($K(X) = $ compact operators space). Show that if ${\cal L}$ is a maximal chain in the $Lat K$ ($Lat K = $ the collection of all invariant subspaces for $K$), then ${\cal L}$ is a maximal chain in the lattice of all subspaces of $X$.

Thank you.
 A: Lemma 1.  Every compact operator on a space of dimension 2 or more has a nontrivial invariant subspace.
Lemma 2.  If $M\subseteq L$ are closed invariant spaces for a compact operator, such that $\text{dim}(L/M)\geq 2$, then there exists a closed invariant space $N$ with $M\subsetneq N\subsetneq L$.
Proof. Apply Lemma (1) to the induced operator on $L/M$.  QED.
Now assume that $\mathcal L$ is a maximal chain in $\text{Lat}(K)$, which is not maximal in the lattice of all subspaces of $X$.
We may then assume that there exists a chain of the form  $\mathcal L\cup\{L\}$, such that $L$ is not invariant under $K$.  Setting
$$L'=[K;L]:= \overline{\text{span}}\{K^nx: n\geq0, x\in L\},$$
we have that $\mathcal L\cup\{L'\}$ is a chain in $\text{Lat}(K)$, so $L'$ must be in $\mathcal L$ by maximality.
Also let $L''$ denote the closed linear span of the union of the members of $\mathcal L$ which are contained in $L$.  It follows that
$$L''\subsetneq L\subsetneq L'.$$
Therefore the codimension of $L''$ in $L'$ is at least 2, and I claim that there can be  no member of $\mathcal L$ in
between $L''$ and $L'$.  This is because every $M$ in $\mathcal L$ obviouly belongs to $\mathcal L\cup\{L\}$, so either
$M\subseteq L$ or $L\subseteq M$.  In the first case $M⊆L''$, and in the second case
$$
  L'=[K;L]⊆[K;M]=M.
  $$
We then get a contradiction by Lemma (2).
