If $T:\mathbb{R}^n \to \mathbb{R}^m$ is linear and injective, then $T^{-1}(B)$ is Borel for Borel $B$.

Asked Oct 25 '14 at 21:53

Active Oct 25 '14 at 21:53

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If $T:\mathbb{R}^n \to \mathbb{R}^m$ is linear and injective, then $T^{-1}(B)$ is Borel for Borel $B$.

Is it possible to prove this theorem?

asked Oct 25 '14 at 21:53

agha

http://math.stackexchange.com/questions/112985/every-linear-mapping-on-a-finite-dimensional-space-is-continuous – aram Oct 25 '14 at 22:14
Yes, it is possible to prove this theorem. – Oct 25 '14 at 23:29

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