converse of open mapping theorem in functional analysis. 
Let $X,Y$ be two Banach spaces over $\mathbb K$ and let $T:X\to Y$ be $\mathbb K-$linear and continuous . Then how to show that 'if T is open then it is surjective?'

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Thank you..
 A: The converse is true.
We have from hypothesis of the converse statement that $T(x)$ is an  open subspace of $Y$ $(1)$
Thus $T(X)$ ha a nonempty interior(actualy it is equal to its interior).
Thus exists a point $z \in T(X)$ and $r>0$ such that $B(z,r) \subseteq T(X)$,and because $T(X)$ is a subspace and $z \in T(X)$ we have that $$B(0,r)=z-B(z,r)\subseteq T(X)$$
Now we will prove that $Y \subseteq T(X)$ 
Let $y \in Y$,then $\frac{r}{2}\frac{y}{||y||} \in B(0,r)$ because $$||\frac{r}{2}\frac{y}{||y||}||=\frac{r}{2}<r \Rightarrow \frac{r}{2}\frac{y}{||y||} \in B(0,r) \subseteq T(X) \Rightarrow y=\frac{2||y||}{r}\frac{r}{2}\frac{y}{||y||} \in B(0,r) \subseteq T(X)$$ because $T(X)$ is a subspace.
Thus $y \in T(X)\Rightarrow Y \subseteq T(X)$ $(2)$
Combining $(1),(2)$ we have that $T(X)=Y$ thus $T$ is onto.
A: If $T$ is an open mapping, then 
1) the image of the unit ball, $T(B(0,1))$ is open,
2) the image of the unit ball contains an open ball centered at 0 with radius $r>0$, and
3) if a point $y\in Y$ has norm $\alpha$, then $y \in T( B(0,1+ \alpha/r))$.
This "proof" works for if $K$ is the real number field.  I'm not sure if it works for finite fields.  
