TITLE: Show that an open linear map between normed spaces is surjective.
QUESTION [15 upvotes]: 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?

REPLY [18 votes]: Since $T$ is open and linear we know that $T(X)\subset Y$ must be an open linear subspace. The only such subspace is all of $Y$. To see this last fact let $y_0\not\in T(X)$. Then $y_0\neq 0$. Choose $a_n\in\mathbb{R}$ with $a_n\rightarrow 0$. Since $T(X)$ is a subspace and $y_0\not\in T(X)$ then $a_ny_0\not\in T(X)$ for all $n$. But $a_ny_0\rightarrow 0$ (here we mean the zero vector in $Y$.)  
Therefor $Y\setminus T(X)$ is not closed contradicting that $T(X)$ is open.