Results 1 to 2 of 2

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

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    120

    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.


    I don't know why $\displaystyle A(V)$ here is an open set of $\displaystyle Y_1$, could anyone help me out? Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jul 2009
    Posts
    120
    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?
    Last edited by zzzhhh; Oct 27th 2009 at 12:38 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. linear map open iff image of unit ball contains ball around 0
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 25th 2010, 01:15 AM
  2. Image of a linear transformation matrix...
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Jan 18th 2010, 10:34 AM
  3. Linear Transformation: finding the image.
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Oct 6th 2009, 08:55 PM
  4. A question about rank/image of a linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: Feb 11th 2009, 04:54 PM
  5. Kernal and Image of a Linear Transformation
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Oct 15th 2008, 05:04 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum