Please, answer me that how is the set of all bounded invertible operators (for example on a Hilbert space) clopen (closed and open) in the set of all bounded surjective operators? In fact, which topology do imply this?

  • $\begingroup$ Did you manage to show that the set of invertible operators is open? $\endgroup$ – Davide Giraudo Sep 25 '12 at 19:02
  • $\begingroup$ (I meant, for classical operator norm) $\endgroup$ – Davide Giraudo Sep 25 '12 at 19:13

In the set $\mathcal{Sur}(H)$ of bounded surjective operators you should take norm topology norm induces by $\mathcal{B}(H)$, where $H$ is a Hilbert space. By $\mathcal{Inv}(H)$ we denote the set of invertible operators.

Recall that a bounded surjective linear operator between Banach spaces $A\in\mathcal{B}(X,Y)$ is invertible iff there exist $c>0$ such that for all $x\in X$ we have $\Vert A(x)\Vert\geq c\Vert x\Vert$. This is direct corollary of open mapping theorem.

Take arbitrary $A\in\mathcal{Inv}(H)$, which is, of course, surjective. Then we have $c>0$ such that for all $x\in H$ we have $\Vert A(x)\Vert\geq c\Vert x\Vert$. Then for all $B\in\mathcal{Sur}(H)$ with $\Vert B-A\Vert<c$ and all $x\in H$ we get $$ \Vert B(x)\Vert\geq\Vert A(x)\Vert-\Vert(A-B)(x)\Vert\geq c\Vert x\Vert-\Vert A-B\Vert\Vert x\Vert=(c-\Vert A-B\Vert)\Vert x\Vert $$ By previous result this means that $B\in\mathcal{Inv}(H)$. Since for all $A\in\mathcal{Inv}(H)$ we found $c>0$ such that $\Vert A-B\Vert<c$ and $B\in\mathcal{Sur}(H)$ implies $B\in\mathcal{Inv}(H)$, then $\mathcal{Inv}(H)$ is open in $\mathcal{Sur}(H)$.

As for the closedness I'll refer to the article Closure of invertible operators on a Hilbert space. The main result of this article is the following: $$ A\in\mathrm{Closure}_{\mathcal{B(H)}}(\mathcal{Inv}(H))\Longleftrightarrow\mathrm{dim}\;\mathrm{Ker}(A)=\mathrm{dim}\;\mathrm{Ker}(A^*). $$ Since we are working in the set of surjective operators we always have $\mathrm{Ker}(A^*)={0}$. Now take arbitrary $A\in\mathrm{Closure}_{\mathcal{Sur}(H)}(\mathcal{Inv}(H))$, since $\mathrm{Ker}(A^*)={0}$, by previous result we get $\mathrm{Ker}(A)={0}$. Thus $A$ is surjective and injective, hence invertible by corollary of open mapping theorem.

In fact $\mathcal{Inv}(H)$ is open not even in $\mathcal{Sur}(H)$ but also in $\mathcal{B}(H)$. One can prove it using Neumann series. There you can find a proof of this fact.

Moreover $\mathcal{Inv}(H)$ is path connected - see theorem 5.29 and corollary 5.30 in Douglas's Banach Algebra Techniques.

As for the set $\mathcal{Sur}(H)$ it is also open. See MSE question Why is the space of surjective operators open?.

This small survey shows us the following structure. The set of surjective operators is open in $\mathcal{B}(H)$ but not connected. It have at least two components. The first one is $\mathcal{Inv}(H)$, it is open, connected and consist of invertible operators. The second one is $\mathcal{Sur}(H)\setminus\mathcal{Inv}(H)$, it is consist of surjective not invertible operators. By open mapping theorem this means that operators of $\mathcal{Sur}(H)\setminus\mathcal{Inv}(H)$ have non trivial kernel.

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    $\begingroup$ By the way, the proof of openness within the set of surjective operators is easier (no need for Neumann series). A surjective operator $A$ is invertible iff it has a lower bound: $\|Ax\|\ge c\|x\|$ for some $c>0$. Any operator $B$ with $\|A-B\|<c$ has a lower bound as well, by the triangle inequality. $\endgroup$ – user31373 Sep 25 '12 at 23:14

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