# Image of unit ball dense under continuous map between banach spaces

I am assuming that the following problem will require the open mapping theorem, or maybe the closed graph theorem. Any help that can be given will be deeply appreciated. The statement is the following:

Let $T: X \to Y$ a continuous linear operator between Banach spaces $X,Y$. Show that $T$ is open if and only if the image under $T$ of the open unit ball in $X$ is dense in a neighbourhood of the origin in $Y$.

Thank you very much in advance!

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I have the forward direction. But I have no clue about the reverse... – user44069 Feb 21 '13 at 2:39
The reverse direction is contained in the standard proof of the open mapping theorem. Look over the proof on wikipedia or your textbook and see if you can get the gist of it. – JSchlather Feb 21 '13 at 3:55

Suppose that $T: X \to Y$ is an open mapping. Then $T[\mathbb{B}(X)]$ is an open neighborhood of $0_{Y}$, where $\mathbb{B}(X)$ denotes the open unit ball of $X$. We can therefore choose the required neighborhood of $0_{Y}$ to be $T[\mathbb{B}(X)]$ itself, since every topological space is obviously a dense subset of itself.
Suppose that $T[\mathbb{B}(X)]$ is dense in some neighborhood $N$ of $0_{Y}$. Then $\overline{T[\mathbb{B}(X)]}^{\| \cdot \|_{Y}}$ contains $N$. Note that $\displaystyle \bigcup_{k=1}^{\infty} k \cdot N = Y$ because $N$, being a neighborhood of $0_{Y}$, must contain an open ball that is centered at $0_{Y}$. As such, \begin{align} Y &= \bigcup_{k=1}^{\infty} k \cdot N \\ &\subseteq \bigcup_{k=1}^{\infty} k \cdot \overline{T[\mathbb{B}(X)]}^ {\| \cdot \|_{Y}} \\ &= \bigcup_{k=1}^{\infty} \overline{T[k \cdot \mathbb{B}(X)]}^ {\| \cdot \|_{Y}} \\ &\subseteq Y, \end{align} which yields $$Y = \bigcup_{k=1}^{\infty} \overline{T[k \cdot \mathbb{B}(X)]}^{\| \cdot \|_{Y}}.$$ By the Baire Category Theorem, we have $\text{Int} \left( \overline{T[k \cdot \mathbb{B}(X)]}^{\| \cdot \|_{Y}} \right) \neq \varnothing$ for some $k \in \mathbb{N}$. Next, mimic exactly the second half of the proof of the Open Mapping Theorem that is presented in Wikipedia. You will therefore be able to conclude that $T$ is an open mapping.