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I came across the following statements while reading:

Consider two affine subspaces $L = x_0 + U$ and $L = x'_0 + U'$ of a vector space $V$.

Then, $L \subseteq L'$ iff $U \subseteq U'$ and $x_0 −x'_0 \in U'$.

My original question: why is the condition $x_0 −x'_0 \in U'$ required for $L$ to be a subspace of $L'$?

I found an answer here, but I'm not sure I understand this part:

"Then, let us consider an element $z\in U_1$. We know by definition that $x_1+z$ belongs to $L_1\subset L_2=x_2+U_2$. It follows that $z\in (x_2-x_1)+U_2=U_2$ because we already know that $x_1-x_2 \in U_2$."

how does it follow that $z\in (x_2-x_1)+U_2$ (and how is this equal to $U_2$)?

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    $\begingroup$ Think about two parallel planes in 3-space. $\endgroup$ – JonathanZ supports MonicaC Oct 18 '20 at 16:18
  • $\begingroup$ The explanation was getting pretty long for a comment, so I broke it out into an answer below. $\endgroup$ – JonathanZ supports MonicaC Oct 18 '20 at 18:59
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For the necessity of $x_0 −x'_0 \in U'$, think about two parallel (but not coincident) planes in $\mathbb R ^ 3$.

For the second part it's mostly notational shortcuts that might be easier to follow if we write everything out explicitly (using the variables used in the linked question, not in your version):

$x_1 + z \in x_2+U_2$ means there is a $u_2 \in U_2$ such that $x_1 + z = x_2 +u_2$.

So $z = (x_2 - x_1) +u_2$, which is exactly what you need for $z \in (x_2 - x_1) + U_2$.

As for "and how is this equal to $U_2$", recall that $(x_2 - x_1) \in U_2$, so we're using the fact/lemma that:

If $U \subset V$ is a linear subspace of $V$, then for any $u \in U$, $u + U = U$.

You could write out of proof of this for yourself, but you could also just think about some examples in $\mathbb R ^3$. If you take a line (through the origin) and and "slide" it in its own direction (as opposed to "to the side") then you get the same line. Or for a plane (again through the origin), pick a vector in that plane and "translate" the plane in that direction. Again you get the same plane back.

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