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Let $\mathcal{C}$ be a chain of subspaces of a Banach space $\mathcal{X}$. For each $\mathcal{Y}\in\mathcal{C}$, define its immediate predecessor \begin{equation*}\mathcal{Y}_{-}=\bigvee\{\mathcal{Z}\in\mathcal{C}:\mathcal{Z}\subset \mathcal{Y},\mathcal{Z}\neq\mathcal{Y}\}.\end{equation*}

I encounter the following statement in one paper:

The condition $\mathcal{Y}_{-}=\mathcal{Y}$ implies that $\mathcal{C}$ contains elements having infinite codimension in $\mathcal{X}$.

I tried to prove this statement but do not actually know where to start. Can someone give some hint?

Thanks!

1 Answers 1

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I don't know much about operator algebras but I'll give it a shot!

Let $\mathcal{Z}\subsetneq \mathcal{Y} = \mathcal{Y}_-$. If we suppose for now that $\dim(\mathcal{Y})-\dim(\mathcal{Z})$ is finite, then $\mathcal{Y}/\mathcal{Z}$ is finite dimensional.

Now $\mathcal{Y} = \mathcal{Y}_-$ implies that there is an ascending chain of subspaces from $\mathcal{Z}$ up to $\mathcal{Y}$, but by finite dimensionality this chain must terminate after finitely many steps. But this would mean that there is a subspace properly contained in $\mathcal{Y}$ which is equal to $\mathcal{Y}$, which is an absurdity. Thus we can conclude that $\dim(\mathcal{Y})-\dim(\mathcal{Z})$ is not finite.

So it is impossible for $\mathcal{Y}/\mathcal{Z}$ to be finite dimensional, whence it is impossible for $\mathcal{X}/\mathcal{Z}$ to be finite dimensional.

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    What you said is essentially true! Thanks!2012-05-30