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 Fredholm operators $T \in W$. According  to this definition $W$ is  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](http://mathoverflow.net/questions/215501/an-equivalent-relation-on-the-space-of-polynomials-in-one-complex-variable)

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](https://en.wikipedia.org/wiki/Quotient_space_(topology)#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?