Let $X$ and $Y$ Banach spaces wrt $\|.\|_X$ and $\|.\|_Y$ norms respectively and $S \subseteq X$.
$T:X \to Y$ is a linear and bounded (continuous) map
Show that $\theta_X \in int(S) \Rightarrow \theta_Y \in int(T(S)) $ (where “int” is interior of a set and $\theta$ s are zeros of spaces)
I have written if $\theta_X \in int(S) \Rightarrow \exists r \gt 0 , B(\theta_X,r) \subseteq S$ ($B(\theta_X ,r)$ is open ball which has $\theta_X$ as center and $r$ as radius)
I have been trying to show that :
$ \exists \varepsilon \gt 0 , B(\theta_Y,\varepsilon) \subseteq T(S)$ i.e. $\theta_Y $ is an interior point of $T(S)$ but I stuck. I know it is very basic but I cannot guess anything about it. Is there any mistake in my writtens? How can I use being Banach space in here? I need a proof or guidance as simple as possible. Thanks in advance for helps
You need that $T$ is onto. Otherwise, just take $T=0$ as a counterexample. So lets suppose it is onto. The trick is to consider the union of $T(nB)$ where $B$ is the standard unit ball in $X$ and $n\in \mathbb{N}$.
Since $T(nB) = nT(B)$ (just scaling), you can now feel that if $T(B)$ is "thin", then the union might be "skin" as well (contradicting to the surjectivity of $T$). To make the argument rigorous you need Baire's category theorem, or you could consult the proof section of https://en.wikipedia.org/wiki/Open_mapping_theorem_(functional_analysis).