$X, Y $ : Banach space, $T : X \to Y$ : linear bounded operator, onto. I'm studying open mapping theorem, but how can I prove this?
If $B_Y (0, \epsilon_1 ) \subset \overline{T(B_X (0, \epsilon_2 ))}$ then $B_Y ( 0, 2 \epsilon_1 ) \subset \overline{T(B_X (0, 2 \epsilon_2 ))} $
Here $B_X (0, a) := \{ x \in X \;|\; \| x \|_X < a \}$ and the overline(bar) means its closure.