I have a question and I have no idea how to solve this: One problem in my Real Analysis text book says:
Show that if $\ell$ is a nonzero linear functional on a normed vector space not necessarily continuous, then $\ell$ maps open sets into open sets.
Isn't it against the definition? I mean saying that $f$ is continuous is equivalent to say that it maps open sets into open sets, right?
Is there any mistake in this problem? Or is just me that I'm misreading something?
Thank you for your help!!
On
A map $f:X\to Y$ between topological spaces is continuous if $f^{-1}(U)$ is open in $X$ whenever $U$ is open in $Y$.
Note that the definition of continuity is different from the statement "continuous is equivalent to say that it maps open sets into open sets". In fact, the constant map $f:\Bbb R\to\Bbb R$ given by $f(x)=0$ is continuous but the image of any open set $U$ under $f$ is $\{0\}$ which is not open in $\Bbb R$.
A map $f:X\to Y$ between topological spaces such that $f(U)$ is open in $Y$ whenever $U$ is open in $X$ is called an open map. A famous example of a map that is open but not continuous is the floor function $\lfloor-\rfloor:\Bbb R\to\Bbb Z$.
In summary, continuous maps are not necessarily open maps and open maps are not necessarily continuous. Your question asks to prove that a certain kind of map is always an open map. Do you have any ideas how to get started?
A continuous map does not necessarily take open sets to open sets: take, e.g. $f(x)=x^2$ on $\mathbb R \rightarrow \mathbb R$ . Then the open set $(-1,1)$ is sent to the non-open set $[0,1)$.