proof of open mapping theorem

We prove that if $\mathrm{\Lambda }:X\to Y$ is a continuous linear surjective map between Banach spaces, then $\mathrm{\Lambda }$ is an open map. It suffices to show that $\mathrm{\Lambda }$ maps the open unit ball in $X$ to a neighborhood of the origin of $Y$.

Let $U$, $V$ be the open unit balls in $X$, $Y$ respectively. Then $X={\cup }_{k\in \mathbb{N}}kU$, so, since $\mathrm{\Lambda }$ is surjective, $Y=\mathrm{\Lambda }(X)=\mathrm{\Lambda }({\cup }_{k\in \mathbb{N}}kU)={\cup }_{k\in \mathbb{N}}\mathrm{\Lambda }(kU)$. By the Baire category theorem, $Y$ is not the union of countably many nowhere dense sets, so there is some $k\in \mathbb{N}$ and some open set $W\subset Y$ such that $W$ is contained in the closure of $\mathrm{\Lambda }(kU)$.

Let $y_0\in W$, and pick $\eta >0$ so that $y_0+y\in W$ for all $y$ with . Then $y_0$ and $y_0+y$ are limit points of $\mathrm{\Lambda }(kU)$, so there are sequences $x_i^{\prime }$ and $x_i^{\prime \prime }$ in $kU$ with $\mathrm{\Lambda }{x}_i^{\prime }\to y_0$ and $\mathrm{\Lambda }{x}_i^{\prime \prime }\to y_0+y$. Letting $x_i=x_i^{\prime \prime }-x_i^{\prime }$, we have and $\mathrm{\Lambda }{x}_i\to y$. So for any $y\in \eta V$ there is a sequence $x_i$ in $2kU$ with $\mathrm{\Lambda }{x}_i\to y$. Then by the linearity of $\mathrm{\Lambda }$, we have that for any $ϵ>0$ and any $y\in Y$, there is an $x\in X$ with:

and (1)

where $\delta =\eta /2k$.

Now let $y\in \delta V$ and $ϵ>0$. Then there is some $x_1$ with and . Define a sequence $x_n$ inductively as follows. Assume:

(2)

Then by (1) we can pick $x_{n+1}$ so that:

(3)

and , so (2) is satisfied for $x_{n+1}$.

Put $s_n=x_1+x_2+\mathrm{\dots }+x_n$. Then from (3), $s_n$ is a Cauchy sequence, and so, since $X$ is complete, it converges to some $x\in X$. By (2), $\mathrm{\Lambda }{s}_n\to y$, and by the continuity of $\mathrm{\Lambda }$, $\mathrm{\Lambda }{s}_n\to \mathrm{\Lambda }x$, so $\mathrm{\Lambda }x=y$. Also, . Thus $\mathrm{\Lambda }((1+ϵ)U)\supset \eta V$, or $\mathrm{\Lambda }(U)\supset {(1+ϵ)}^{-1}\delta V$. Since this is true for all $ϵ>0$, we have $\mathrm{\Lambda }(U)\supset {\cup }_{ϵ>0}{(1+ϵ)}^{-1}\delta V=\delta V$.