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Let T:X-->Y be a linear map.

--> this direction follows by definition of an open set. For the other direction, this is how I've started attempting it:

Let U be open in X. take an element in T(U), say T(u) where u is in U. We want to show T(u) is contained in an open ball in T(U). Suppose B(0,a) is the ball contained in the image of the unit ball under T. Then T(ua/2||T(u)||) has norm a/2, and so is in B(0,a), and so we can find an open ball around ua/2||T(u)|| which is contained in the image of the unit ball under T. I'm not sure where to go from here (or even if this is the right direction!). I would appreciate if someone could explain how to prove this harder direction.

lkjhgfdsa
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    Why does $T(\frac{ua}{2||u||})$ have norm 2? All we know is that $\frac{ua}{2||u||}$ has norm $2$. – mathworker21 Apr 13 '17 at 13:46
  • sorry - I should have written T(ua/2||T(u)||), which has norm a/2 by linearity of T - I will edit to correct this in my question. – lkjhgfdsa Apr 13 '17 at 13:55

1 Answers1

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Let $U$ be open. Wlog. we may assume $0\in U$ and $B_1(0)\subset U$ (by scaling and translating). Let $y\in T(U)$, then there is $x\in U$ s.t. $Tx=y$. As $U$ is open, there is some $r\leq 1$ s.t. $B_r(x)\subset U$. Now $T(x+B_r(0))=y+T(B_r(0))$ is a neighborhood of $y$, since $T(B_r(0))$ contains a neighborhood of $0$ and translations are homeomorphisms.

Sebastian Bechtel
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