# A question about the problem in Functional analysis ( Rudin)

Problem 10( chapter 1, p.39). $X, Y$: topological vector spaces, $dimY<\infty$, $f:X\rightarrow Y$ is linear, and $f(X)=Y$.

(a) Prove that $f$ is an open mapping.

(b) Assume, in addition, that the null space of $f$ is closed, and prove that $f$ is then continuous. Thanks in advance.

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There is a subspace Z of X which is homeomorphic to Y(By Thm 1.21), thus f is an open mapping(any open set of X must contain an open set of Y because Y is given the subspace topology). –  lee Jan 7 '13 at 10:55
For (b), you just look at the Thm 1.18. I think the idea is the same. –  lee Jan 7 '13 at 11:01
Thanks @lee for your idea! :) –  user52523 Jan 8 '13 at 4:53
For (b), you can think another way. Consider the quotient map X->X/N(f), now that the quotient topology is given on X/N(f) which is a finite dimensional topological vector space homeomorphic to Y. Just recall that a map from X to Z if factor through X/Y, then it's continuous if and only if the map from X/Y to Z is continuous. –  lee Jan 8 '13 at 13:50
@ˈjuː.zɚ79365 Ok.. I am in a series of long time exams. I will have time to do that until 7/11, I almost forgot this problem, but I will try to recall the solution and write it down after my huge exams. –  lee Jun 26 '13 at 15:38
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