Let $\mathcal{H}$ be a separable Hilbert space with orthonormal base $\{e_i\}\;\;i\in \mathbb{N}$.
Definition: We say a subvector space $W\subset B(\mathcal{H})$ is a Fredholm subspace if there is a constant $M$ such that $Ind(T)\leq M$ for all operators $T \in W$ which are Fredholm. In the other words $W$ is a Fredholm subspace if the Index is a bounded function on $$Fred(W):=\{T\in W \mid\; \text{T is a Fredholm operator\}}$$ According to this definition $W$ is NOT counted as a Fredholm subspace if $W$ contains no Fredholm operator, at all.
Example: Put $\mathcal{H}=\ell^{2}$. $S_1$ is the shift operator on $\ell^{2}.$
We define $W=\{P(S_1)\mid P\;\; \text{is a polynomial of degree at most n\}}$. Then $W$ is a $n+1$ dimensional Fredholm subspace of $B(\ell^2)$. See this post
In this question we would like to ask "Is the space of Fredholm subspaces an open set?"
We try to give a meaning to the latter statement via Grassmanian in $B(\mathcal{H}):$
We define an inner product on $B(\mathcal{H})$ with $<A,B>=\sum \frac{1}{n^2}<Ae_n,Be_n>$. This obviously enable us to define the Grassmanian $G(n, B(\mathcal{H})),$ the space of all $n$ dimensional subvector space of $B(\mathcal{H})$ with a natural topology as follows:
Let $S$ be the unit sphere of $B(\mathcal{H})$ with respect to the norm arising from the above inner product. So $S$ has a natural topology. We define a unique topology on $G(n, B(\mathcal{H}))$ such that the following map be a quotient map $$Span:\{(x_1,x_2,\ldots,x_n) \in \overbrace{S\times S\times \ldots\times S}^{n-times}\mid <x_i,x_j>=0\}\to G(n, B(\mathcal{H}))$$
This map sends $(x_1,x_2,\ldots,x_n)$ to the $n$ dimensional subspace $W$ generated by $x_1,x_2,\ldots,x_n.$
So our question is the following:
Is the space of all Fredhom subspaces of $G(n,B(\mathcal{H})$ an open set? What about if we replace $S$ by unit sphere of $B(\ell^{2})$ with its operator norm and its natural topology?
