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?



$T$ 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 $T$.

  • $\begingroup$ Question 1: How to show that $T(B_1(0)_X)\subset B_\epsilon (0)_Y$ Question 2:*Every point in the blown... up ball in $X$* --How ,we only have subset relation not equal to realtion $\endgroup$ – Learnmore Jan 17 '17 at 18:31
  • $\begingroup$ The map $T$ is an open map, meaning that the images of open sets are also open sets. $B_1(0)_X$ is open and contains $0$, so its image must be open an contain $0$. This means the image must contain some ball $B_\epsilon(0)_Y$ around $0$. Here I see that I have written the wrong inclusion, it should be $T(B_1(0)_X)\supset B_\epsilon(0)_Y$ not $\subset$. I think this also addresses your question 2? $\endgroup$ – s.harp Jan 17 '17 at 18:39
  • $\begingroup$ Yes right ,it's perfect now +1 $\endgroup$ – Learnmore Jan 17 '17 at 18:50

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.