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$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.

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up vote 2 down vote accepted

Let $y\in B_Y(0,2\varepsilon_1)$. Then $\frac 12 y\in B_Y(0,\varepsilon_1)$, and we can find a sequence $\{x_n\}\subset B_X(0,\varepsilon_2)$ such that $\frac 12 y=\lim_{n\to +\infty}Tx_n$. This gives $y=\lim_{n\to +\infty}T(2x_n)$. Since $\{2x_n\}\subset B_X(0,2\varepsilon_2)$, we have $y\in \overline{T(B_X(0,2\varepsilon_2)}$.

Note that it also works if we replace $2$ by any positive constant.

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