Is the image of an open set of a linear transformation still open in its image?

**zzzhhh** · October 26th 2009, 04:46 PM

I met with this problem when reading a proof of rank theorem in Rudin's "Principles of mathematical analysis", P230, as is shown below.

I don't know why $A(V)$ here is an open set of $Y_1$ , could anyone help me out? Thanks!

**zzzhhh** · October 27th 2009, 12:09 AM

The last part of this proof gives me some hints. For $\mathbf y_0\in A(V)$ , find a ${\bf{x}}_0\in V$ such that $A\bf x_0=y_0$ . Define a mapping $B({\bf y )=x}_0+S(\bf{y-y}_0)$ where $S$ is defined as (67) in P229 of this book satisfying $AS=1$ on $Y_1$ . $B(\bf y_0)=x_0$ . It can be seen that $S$ is a linear transformation from $Y_1$ into $\mathbb R^n$ , so it is continuous and in turn makes $B$ continuous. Therefore, the open set $V$ of $\mathbb R^n$ produces by continuity of $B$ a neighborhood $W$ in $Y_1$ containing $\bf y_0$ such that $B(\mathbf y)\in V$ for all $\mathbf y\in W$ . Note that so far $W$ may not lie in $A(V)$ , so next we prove that it really is contained in $A(V)$ . For any $\bf y$ in $W$ , let $\mathbf x=B(\mathbf y)$ , then $\mathbf x\in V$ , so $A\mathbf x=A\bf x_0+y-y_0=y$ , which just shows $\mathbf y\in A(V)$ , or, $W\subseteq A(V)$ , as desired. Note that this means $A(V)$ is only open in $Y_1$ , but not open in $\mathbb R^m$ because $Y_1$ is not necessarily open in $\mathbb R^m$ . However, if we set the codomain to be the rank space assuming subspace topology, any linear transformation is an open map. Is that right?
I don't know if the above proof is what the author originally intended. It seems not and there should be a simpler proof since "it is clear". oops... Rudin's book is going to drive me mad

, should I continue to use his books to study real and complex analysis and functional analysis in future?

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