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The problem

Define the sum of two vector spaces (often two subspaces of a common vector space) $A$ and $B$ as $A+B=\{a+b: a \in A, b \in B \}.$ Let $U,V,W$ be arbitrary vector spaces. I want to show that

$$U \cap (V+W)=(U \cap V)+(U \cap W)$$

My approach

By starting with $x \in (U \cap V)+(U \cap W)$ and applying the properties of vector spaces (closure under vector addition) to end up with $x \in U \cap (V+W)$, I've shown that $(U \cap V)+(U \cap W) \subset U \cap (V+W)$.

My problem is with the reverse, i.e. showing that $U \cap (V+W) \subset (U \cap V)+(U \cap W)$. Following the usual procedure, I started with:

$x \in U \cap (V+W)$, which implies that $x \in U \land x \in V+W$.

Now, $x \in V+W \implies x=y+z$ for some $y \in V$ and some $z \in W$.

Again, $x=y+z \in U$. If we could show that $y$ and $z$ both comes from $U$, then we will be done. But unfortunately $y+z \in U$ doesn't necessarily imply that $y \in U$ and $z \in U.$ Am I missing some very basic properties of vector spaces that would make this trivial? Any help would be greatly appreciated.

P.S. Just a dumb question. Does the result break down if we replace vector spaces by arbitrary sets in the above problem?

  • 1
    You say "*often* two subspaces of a common vector space", but what does "$a+b$" mean when $a$ and $b$ are not in the same vector space?2017-01-25
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    For this identity to work Either of V or W has to be a subspace of U.2018-10-01

1 Answers 1

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The statement is wrong. If $U$, $V$ and $W$ are all lines in the same plane ($U\neq V$, $V\neq W$ and $W\neq U$) then $U\cap (V+W) = U$ but $(U\cap V)+(U\cap W)$ = $\{0\}+\{0\} =\{0\}$.