# How to prove that $\bigcup_{n=1}^{\infty} K_n = Y$ in the open mapping theorem?

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 $$K_n$$, $$h_n^{-1}(U)\subset K_1$$.
• How do u know that $|| x_p|| \leq n$? You let $x \in K_n$. So there is a sequence $(x_p) \in T(B(0,n))$ such that $x_p \to x$. You don't know that the $x_p$ lie in $B(0,n)$? Only in the image of this ball? – Kamil Dec 11 '19 at 17:28
• @Kamil: I think you and Tsemo are using $x_p$ for different things. If you write $x_p$ as in your comment, then for each $p$ there exists $z_p \in B(0,n)$ such that $T(z_p) = x_p$. This $z_p$ is what Tsemo denotes by $x_p$. – Nate Eldredge Dec 11 '19 at 19:23