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We take $A$ to be a nonempty subspace of standard Euclidean space, then we have the following statements equivalent:

  1. A is an affine subspace
  2. $$\forall x,y\in A, t\in\mathbb{R}, (1-t)x+ty\in A$$

It looks like just following the definition but here is how to prove it only uses that affine set is a translated linear subspace. How can we show the two statements are the same?

79999
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1 Answers1

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Pick some $p\in A$ and define $A_p=A-p$. Then

  1. Prove that $A_p$ is a vector subspace
  2. Use that $A=p+A_p$ so it is an affine subspace

How to do part 1: Let $(x-p)\in A_p$ with $x\in A$. Then for $\lambda \in\mathbb R$ we have $$ \lambda(x-p)=\underbrace{((1-\lambda)p +\lambda x)}_{\in A}-p\in A_p. $$ Now its left to show that sums are contained

Alex
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  • but you just used the second statement to conclude the last equation in A and therefore in A_p. I thought we are suppose to prove that – 79999 Mar 10 '23 at 02:24
  • Where? I only used that $a-p\in A_p$ if $a\in A$ and thats just the definition of $A_p$ – Alex Mar 10 '23 at 10:08
  • If we assume A is an affine set then the condition we can use is that it preserves any linear transformation for {x} + A. – 79999 Mar 10 '23 at 13:42
  • Whats your defintions of an affine space? I thought it is just a subset of the form $p+V$ for some vector space $V$ – Alex Mar 10 '23 at 14:24
  • yes. but when you use (1-lambda)p+lambda x, this is not the definition for the affine set. the transformation is about one element not the parallel elements like (1-lambda)p – 79999 Mar 10 '23 at 16:03
  • Well I am proving that 2) implies 1) there; so of course I can use 2). (The 2) and 1) from your question) – Alex Mar 11 '23 at 13:32