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Let $A$ be an affine space over a vector space $V$. Let $B$ be a nonempty subset of $A$ such that $W:=\{a-b\in V:a,b\in B\}$ is a subspace of $V$. In this case, how do I prove that for each $b\in B$ and $y\in W$, $b+y\in B$?

Thank you in advance.


This is wrong, hence you can't prove it :-)

As an example choose $A=V=\mathbb{R}^2$ and $B:=\{(x,y) \ | \ x=0 \lor y=0\}$ as the union of the two axes. You can easily see, that $W=\mathbb{R}^2$ holds (for $v=(x,y)\in V$ take $a=(x,0)$ and $b=(0,-y)$). But obviously $B$ is not an affine space. E.g. $(0,0)+(1,1) =(1,1)\notin B$ but $(0,0) \in B$ and $(1,1) \in W$.

  • $\begingroup$ Then, wikipedia gives a wrong definition of affine subspace. What would be a correct definition of affine subspace? Should it be a nonempty subset $B\subset A$ such that$ that $B$ is an affine space over $W$? $\endgroup$ – Rubertos Sep 18 '16 at 0:48
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    $\begingroup$ @Rubertos Yes. It is a (some say nonempty, this varies depending on your source) subset, which is itself an affine space. $\endgroup$ – ctst Sep 18 '16 at 0:51
  • $\begingroup$ It's the first time I see wikipedia is totally wrong.. Really surprising. Thank you. $\endgroup$ – Rubertos Sep 18 '16 at 0:59
  • $\begingroup$ @Rubertos No problem. I mentioned it in the wiki discussion page (since I am not actually active there, I refrain from changing it directly). You might think of the wikidefinition the right way, if you keep track of all the differences (starting and ending point) and see if this is a vector space (much bigger than the translation space) but that would be kindof horrible. The wikis definition is at least strongly misleading. You should stick with the is itself an affine space (which is the correct definition). $\endgroup$ – ctst Sep 18 '16 at 1:08

If $B$ is itself an affine space of $V$ and a subset of $A$, then we get the desired conclusion.

Since $A$ is an affine space of $V$, there exists a subspace $U$ of $V$ and a vector $v$ in $V$ such that $A = v + U = \{v + u:u \in U\}.$ Since $B$ is also affine and a subset of $A$, there exists a subspace $U'$ of $U$ such that $B = v + U' = \{v + u':u \in U'\}$. Let $a, b \in B$. Then, there exist $x_a,x_b \in U'$ such that $$ a = v + x_a \\ b = v + x_b. $$ If $y \in W$, there exist $a,b \in B$ such that $y = a - b = (v + x_a) - (v + x_b) = x_a - x_b \in U'$. Let $b = v + x_b' \in B$. Then, $$ b + y = v + x_b' + (x_a - x_b) = v + (x_b' + x_a - x_b). $$ $U'$ is a subspace of $V$, so $U'$ is closed under addition and scalar multiplication. Furthermore, $x_b',x_a,x_b\in U'$. Hence, $(x_b'+x_a-x_b) \in U'$, so $v + (x_b' + x_a - x_b)\in B$.

  • $\begingroup$ Why should $v + (x_b' + x_a - x_b)\in B$ hold? $B$ is just a subset of $A$. It just hold $v + (x_b' + x_a - x_b)\in A$! $\endgroup$ – ctst Sep 18 '16 at 0:45
  • $\begingroup$ @ctst I edited my post to fix this. It's true that a subset alone won't work. Instead, $B$ has to be an affine space as well. $\endgroup$ – Alex Ortiz Sep 18 '16 at 0:59
  • $\begingroup$ @Rubertos I edited my answer to show how you can prove the statement in the case that $B$ is also an affine space. $\endgroup$ – Alex Ortiz Sep 18 '16 at 1:00
  • $\begingroup$ @AOrtiz I'm confused now. So is an affine subspace required to be an affine space over $V$ or $W$? Do you mean $W$ instead of $V$? $\endgroup$ – Rubertos Sep 18 '16 at 1:00
  • $\begingroup$ @Rubertos The statement is that the affine spaces we are considering are all affine spaces over $V$. $\endgroup$ – Alex Ortiz Sep 18 '16 at 1:01

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