Sufficient condition in proving open mapping theorem for Banach spaces

Question

Statement of the open mapping theorem: Let $X, Y$ be Banach spaces, $T:X\to Y$ be a continuous linear transformation. If T is onto, then T is an open map (i.e. for any open set $O\subset X$, $T(O)\subset Y$ is open.

[EDITED] Claim: It suffices [in proving that T is an open map] to show that $T(B_X(1))$ contains an open ball centered at the origin, where $B_X(1)$ is the open ball of radius 1 centered at the origin in $X$.

I'm not seeing how this condition implies that all open sets map to open sets. Linearity of T seems to be important here, but nothing is coming to mind except the definition $T(af+bg)=aT(f)+bT(g)$ for $a,b\in\mathbb{R}$, $f,g\in X$.

Would greatly appreciate some insight. Thanks!

Answer 1

Yes it is sufficient, because it implies that image of any open ball has nonempty interior.

Answer 2

Suppose the condition satisfied, consider $U$ an open subset of $X$, and $x\in U$, there exists $c\in R$ such that $x+cB(0,1)\subset U$. $T(x+cB(0,1))=T(x)+cT(B(0,1))$ since $T$ is linear, this implies that $T(U)$ is open since $T(B(0,1))$ is open.