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

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• Oct 26th 2009, 04:46 PM
zzzhhh
Is the image of an open set of a linear transformation still open in its image?
I met with this problem when reading a proof of rank theorem in Rudin's "Principles of mathematical analysis", P230, as is shown below.
http://www.tuchuan.com\d/2009102708423112091.jpg

I don't know why $\displaystyle A(V)$ here is an open set of $\displaystyle Y_1$, could anyone help me out? Thanks!
• Oct 27th 2009, 12:09 AM
zzzhhh
The last part of this proof gives me some hints. For $\displaystyle \mathbf y_0\in A(V)$, find a $\displaystyle {\bf{x}}_0\in V$ such that $\displaystyle A\bf x_0=y_0$. Define a mapping $\displaystyle B({\bf y )=x}_0+S(\bf{y-y}_0)$ where $\displaystyle S$ is defined as (67) in P229 of this book satisfying $\displaystyle AS=1$ on $\displaystyle Y_1$. $\displaystyle B(\bf y_0)=x_0$. It can be seen that $\displaystyle S$ is a linear transformation from $\displaystyle Y_1$ into $\displaystyle \mathbb R^n$, so it is continuous and in turn makes $\displaystyle B$ continuous. Therefore, the open set $\displaystyle V$ of $\displaystyle \mathbb R^n$ produces by continuity of $\displaystyle B$ a neighborhood $\displaystyle W$ in $\displaystyle Y_1$ containing $\displaystyle \bf y_0$ such that $\displaystyle B(\mathbf y)\in V$ for all $\displaystyle \mathbf y\in W$. Note that so far $\displaystyle W$ may not lie in $\displaystyle A(V)$, so next we prove that it really is contained in $\displaystyle A(V)$. For any $\displaystyle \bf y$ in $\displaystyle W$, let $\displaystyle \mathbf x=B(\mathbf y)$, then $\displaystyle \mathbf x\in V$, so $\displaystyle A\mathbf x=A\bf x_0+y-y_0=y$, which just shows $\displaystyle \mathbf y\in A(V)$, or, $\displaystyle W\subseteq A(V)$, as desired. Note that this means $\displaystyle A(V)$ is only open in $\displaystyle Y_1$, but not open in $\displaystyle \mathbb R^m$ because $\displaystyle Y_1$ is not necessarily open in $\displaystyle \mathbb R^m$. 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?