Non-surjective linear operator that is open?

Question

Is there any non-surjective linear operator that is open?

By the open mapping theorem, $T:X \to Y$ is open iff $T$ is surjective. Is it possible then to have a non-surjective linear operator that is open?

I have been thinking of $T:\mathbb{R}^2 \to \mathbb{R}^2,\ (x, y) \to (x, 0)$

Here T is not surjective but the set $A = {(x, 0) \in \mathbb{R}^2} \subset \mathbb{R}^2$ is closed: Let $(z^{(n)})_n \subset A,\ z^{(n)} = (x_n, y_n)$ be some sequence with $\lim x_n = x$. Then the sequence converges in $A$, and therefore $A$ is closed. So this wouldnt work as an example.

Answer 1

No such $T$ exists. $X$, of course, is open in $X$, so if $T$ is open then $T(X)$ must be open in $Y$. But $T(X)$ is a linear subspace of $Y$. If $T$ is not surjective, then $T(X)$ is a proper linear subspace, and one may see that a proper linear subspace can never be open.

Answer 2

If the range of $T$ is open it contains an open ball centered at $0.$ By the homogeneity the range is equal to the whole space.

The operator does not need to be additive neither bounded, just homogeneous. Observe that homogeneity implies $T0=0$ and that has been used in the first sentence.