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

  1. #1
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    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 A(V) here is an open set of Y_1, could anyone help me out? Thanks!
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  2. #2
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    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?
    Last edited by zzzhhh; October 27th 2009 at 01:38 PM.
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