Let $X,Y$ be normed spaces and $T:X\to Y$ is an open linear map. Show that $T$ is surjective.
In order to show $T$ is surjective let's take $y_0\in Y$ and assume the contrary that $Tx\neq y_0\forall x\in X$.
Now taking $x_0\in X\implies Tx_0\neq y$.
Also $T(B(x_0,r))$ is open. $X=\cup_{n\in \Bbb N}B(x_0,n)\implies T(X)\subset \cup_{n\in \Bbb N} T(B(x_0,n))$.
I am unable to find any contradiction.Can someone kindly help?
Best Answer
Visually:
$T$ maps $B_1(0)_X$ to an open set containing $0$, because $0=T(0)$. This means the image of $T$ contains some $\epsilon$ ball of $0$: $ B_\epsilon(0)_Y\subseteq T(B_1(0)_X)$. If you blow this ball up you will cover the entire space $Y$. Linearity means that every point in the blown up ball in $Y$ has a pre-image in the blown up ball in $X$.
Writing out the last sentence more concretely, for every $y\in Y$ you have that $\frac\epsilon{2\|y\|} y$ lies in $B_\epsilon(0)_Y$, so must be the image of some $x$ in $B_1(0)$.
It follows: $$T\left(\frac{2\|y\|}\epsilon\, x\right)=\frac{2\|y\|}\epsilon T(x)=\frac{2\|y\|}\epsilon\frac\epsilon{2\|y\|}y=y$$
And $y$ lies in the image of $T$.