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Let $X,Y$ be normed spaces and $T:X\to Y$ is an open linear map. Show that $T$ is surjective.

In order to show $T$ is surjective let's take $y_0\in Y$ and assume the contrary that $Tx\neq y_0\forall x\in X$.

Now taking $x_0\in X\implies Tx_0\neq y$.

Also $T(B(x_0,r))$ is open. $X=\cup_{n\in \Bbb N}B(x_0,n)\implies T(X)\subset \cup_{n\in \Bbb N} T(B(x_0,n))$.

I am unable to find any contradiction.Can someone kindly help?

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Visually:

$T$ maps $B_1(0)_X$ to an open set containing $0$, because $0=T(0)$. This means the image of $T$ contains some $\epsilon$ ball of $0$: $ B_\epsilon(0)_Y\subseteq T(B_1(0)_X)$. If you blow this ball up you will cover the entire space $Y$. Linearity means that every point in the blown up ball in $Y$ has a pre-image in the blown up ball in $X$.

Writing out the last sentence more concretely, for every $y\in Y$ you have that $\frac\epsilon{2\|y\|} y$ lies in $B_\epsilon(0)_Y$, so must be the image of some $x$ in $B_1(0)$.

It follows: $$T\left(\frac{2\|y\|}\epsilon\, x\right)=\frac{2\|y\|}\epsilon T(x)=\frac{2\|y\|}\epsilon\frac\epsilon{2\|y\|}y=y$$

And $y$ lies in the image of $T$.

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Since $T$ is open and linear we know that $T(X)\subset Y$ must be an open linear subspace. The only such subspace is all of $Y$. To see this last fact let $y_0\not\in T(X)$. Then $y_0\neq 0$. Choose $a_n\in\mathbb{R}$ with $a_n\rightarrow 0$. Since $T(X)$ is a subspace and $y_0\not\in T(X)$ then $a_ny_0\not\in T(X)$ for all $n$. But $a_ny_0\rightarrow 0$ (here we mean the zero vector in $Y$.)

Therefor $Y\setminus T(X)$ is not closed contradicting that $T(X)$ is open.

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    $\begingroup$ Nice answer Owen +1 $\endgroup$ – Learnmore Jan 18 '17 at 2:41

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