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Let $E$ be a Banach Space. Prove that the set of all continuous linear transformations with inverse also continuous is open in $\mathcal{L}(E,E)$ (the set of continuous transformations from $E$ to $E$). Consider the norm $\|T\| = \mbox{sup}\{\|T(x)\|:x\in E, \|x\|=1\}$.

The only fact that i previous know is that if $T$ is linear and $\|T\|<1$, then $I-T$ does have continuous inverse, but i can't really apply this fact.

Any help or hint?

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    $\begingroup$ Have you considered proving that the set of continuous linear transformations without a continuous inverse is closed in $\mathcal{L}(E, E)$? $\endgroup$ – Michael Lee Oct 16 '17 at 21:35
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    $\begingroup$ If $S$ is invertible, and $U$ close to $S$, try writing $U$ as $S(I - T)$. $\endgroup$ – Daniel Fischer Oct 16 '17 at 21:40
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Suppose that $U$ is invertible, consider the ball $B(U,r)$ such that $\|U^{-1}\|r<1$. For every $T\in B(U,r)$, you have $T=U+L$ with $\|L\|<r$. This implies that $T=U(I+U^{-1}L)$ since $\|U^{-1}L\|<1$, $I+U^{-1}L$ is invertible, you deduce that $T$ is invertible.

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