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 $K_n$, $h_n^{-1}(U)\subset K_1$.

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  • $\begingroup$ 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? $\endgroup$ – Kamil Dec 11 '19 at 17:28
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    $\begingroup$ @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 $x_p$. $\endgroup$ – Nate Eldredge Dec 11 '19 at 19:23

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