Homework problem on continuity Let $U =\{A \in M_{n} : A \text{ is invertible}\}$ (where $M_{n}$ is the space of all $n\times n$ matrices).
$U$ is an open subset of $M_{n}$. Define $\alpha: U \rightarrow M_{n}$ by $\alpha(A)=A^{-1}$. Prove that $\alpha$ is continuous.
I suppose to use the  identity $A^{-1}-B^{-1} = A^{-1}(B-A)B^{-1}$ and take norm on both side and using the inequality about the norm of a product of a bounded linear operator. The hint say but it leaves open the problem of finding an upper bound for $\|B^{-1}\|$. One strategy to follow is to use the triangle inequality: $\|B^{-1}\|=\|(B^{-1}-A^{-1}+A^{-1})\|\leq\|B^{-1}-A^{-1}\|+\|A^{-1}\|$. If insert this into inequality, we can use ordinary algebra to bring to the left-side of the inequality all of the terms involving $\|B^{-1}-A^{-1}\|$. Then we can make appropriate estimates to control the "coefficient" of$\|B^{-1}-A^{-1}\|$, so that we can divide by it. I do not really see how this works and I try to work on it. 
Do I still use definition of continuity to do this? With $\epsilon, \delta$ ?
make $\|B-A\| \leq \delta$. 
$\|\alpha(A)-\alpha(B)\| = \|A^{-1}-B^{-1}\|=\| A^{-1}(B-A)B^{-1}\| \\ 
\leq \|A^{-1}\| \|B-A\| \|B^{-1}\| \\ 
\leq \|A^{-1}\| \|B-A\| (\|B^{-1}-A^{-1}\|+\|A^{-1}\|) $ 
Not sure if this is what I should do and how this works out to bring to left side.
 A: You are quite close to a proof. If you solve for $\|A^{-1}-B^{-1}\|$ you get
$$\| A^{-1} - B^{-1}\| \leq \frac{\|A^{-1}\|^2\|A-B\|}{1-\|A^{-1}\|\|A-B\|} \, .$$
Being sloppy here we can just take the limit as $B\to A$ and so $\|A^{-1}-B^{-1}\| \to 0$, which proves the statement. However, if we want to be somewhat careful it might be good to think a moment or two about the denominator. I do believe there is a result stating that if $A,B \in U$, then $\|A^{-1}\|\|A-B\| < 1$. You might want to take a closer look at that.
A: Let $\alpha(X)=X^{-1}$, also $X^{-1}X=I$ then by taking parial derivatives:
$\partial X^{-1}X+X^{-1}\partial X=0$
Therefore: $\partial X^{-1}X=-X^{-1}\partial X X^{-1}$, and finally $\partial X^{-1}=-X^{-1}\partial X X^{-1}$,
Now you have:
$\alpha(X+\partial X)-\alpha(X)\approx-X^{-1}\partial X X^{-1}$ (linearization w.r.t $\partial X$ or first order Taylor approximation w.r.t $\partial X$)
Now for $\|\partial X\|\leq \delta$, by Cauchy-Schwarz:
$\|\alpha(X+\partial X)-\alpha(X)\|\leq \|X^{-1}\|\|\partial X\|\| X^{-1}\|$
Let $K\geq \|X^{-1}\|^2$, then
$\|\alpha(X+\partial X)-\alpha(X)\|\leq K \delta$
which implies Lipschitz continuity.
