# $X,Y$ be Banach , $T \in \mathcal B(X,Y)$ be onto ; then , for every sequence $y_n \to y \in Y$ , $\exists x_n \to x\in X$ s.t. $T(x_n)=y_n , T(x)=y$?

Let $X,Y$ be Banach spaces , $T:X \to Y$ be a surjective continuous linear transformation , then is it true that for every convergent sequence $\{y_n\}$ in $Y$ , converging to $y \in Y$ , there exist a sequence $\{x_n\}$ in $X$ , converging to $x \in X$ , such that $T(x_n)=y_n , \forall n \in \mathbb N$ and $T(x)=y$ ?

## 1 Answer

If $T$ is onto, this is the open mapping theorem:

You first choose $x$ with $Tx=y$. Then you look at the images of small open balls around $x$. These images are open and contain $y$ and thus contain all but finitely many $y_n$. Hence, you can choose $x_n$ in the balls as needed.

• Could you please expand a bit on this? Unfortunately I am not able to fill in all the details. I see that for any $B_r (x) \subset X$, the image $T (B_r(x)$ is open and $y \in T (B_r(x))$. Therefore, there exists an $\epsilon > 0$ such that $B_{\epsilon} (y) \subset T (B_r(x))$, but I do not see how to get the result from here. – jvnv May 28 '16 at 11:08
• Sure. You consider a sequence of such $r$. E.g., $r=\frac{1}{k}$. Once a $y_n$ is no longer in the image, you can allocate a $x_n$ from the last ball. If a $y_n$ is in all such images, you just take a $x_n$ from the ball with radius $\frac{1}{n}$. If it is in none of the images, you just allocate any $x_n$ from the preimage. Overall, for every $k$, all but finitely many $x_n$ are in the ball with radius $\frac{1}{k}$. Hence, $x_n \to x$. – Julian Braun May 30 '16 at 11:55