I have a question. In my introductory course on functional analysis, we are proving the open mapping theorem. It states:
Theorem: Let $X$ and $Y$ be Banach spaces and $T: X \rightarrow Y$ a surjective bounded operator. Then $T$ is open.
The proof in my notes follows some steps. I write $cl$ for the closure, and $int$ for the interior. For $n \geq 1$, we write $K_n= cl \ T(B(0,n))$, where $B(0,n)$ is the open ball centered at $0$.
Question 1: I have to prove that $$\bigcup_{n=1}^{\infty} K_n = Y. $$ I let $y \in Y$. Then by surjectivity there exists a $x \in X$ such that $T(x) = y$. Now I have to show there exists a $N \geq 1$ such that $T(x) \in cl \ T(B(0,N))$. How to pick this $N$?
I know that I can write $$ B(x, N) = x + B(0, N)$$ by linearity. But not sure how to proceed.
Question 2: We use a corollary of the Baire category theorem to deduce the existence of a $n \in \mathbb{N}$ such that $int \ K_n$ is non-empty. This I understand. But then my course notes say that from this we can deduce that $cl \ T(B(0,1))$ has non-empty interior? Why is this true??
Thanks in advance for any help.
For question $1$, take $N$ be an integer superior to $\|x\|$ where $T(x)=y$.
For 2, if $x$ is an element of $K_n$, $x=lim_pT(x_p)$ where $\|x_p\|\leq n$. Consider $y_p={1\over n}x_p$, $lim_p(T(y_p))={1\over n}lim_pT(x_p)={1\over n}x$. This implies that the image of $K_1$ by $h_n(x)=nx$ contains $K_n$. Let $U$ be a non empty subset of $K_n$, $h_n^{-1}(U)\subset K_1$.