Equivalent characterization of inner direct sum of subspaces

Question

Let $V$ be a vector space and let $A, B$ be subspaces of V. The sum of $A, B$ is the subspace of $V$ given by $$ A + B = \{a+b : a\in A, b\in B\} $$ Moreover if $A \cap B=\{0\}$ then the sum of $A,B$ is called the inner direct sum of $A,B$ and is denoted with $A \oplus B$. Let $A, B, C$ be subspaces of $V$ with $C = A + B$. Then $$ C = A \oplus B \iff \forall c \in C, \exists! a\in A, \exists! b \in B : c = a + b $$

If we replace on the right hand side $\forall c\in C$ with $\exists c\in C$ then does the equivalence still hold?

Answer 1

The equivalence still holds: We have that \begin{align*} C = A \oplus B &\iff \forall c \in C, \exists! a\in A, \exists! b \in B : c = a + b \\ &\,\implies \exists c \in C, \exists! a\in A, \exists! b \in B : c = a + b \end{align*} because we can choose $c = 0$. To show the other implication $$ \exists c \in C, \exists! a\in A, \exists! b \in B : c = a + b \implies \forall c \in C, \exists! a\in A, \exists! b \in B : c = a + b $$ let $c \in C$ such that for $c = a + b$ with $a \in A$, $b \in B$ the summands $a, b$ are unique.

Let $x \in A \cap B$. It then follows from $x \in A$ that $a + x \in A$ and it follows from $x \in B$ that $b - x \in B$. We have that $$ c = a + b = (a + x) + (b - x) $$ so it follows from the uniqueness of $a, b$ that $a+x = a$ and $b-x = b$. Both equations show that $x = 0$. This shows that $A \cap B = \{0\}$ and together with $C = A+B$ that $C = A \oplus B$.

For subspaces $A, B, C \subseteq V$ with $C = A + B$ one can show more generally that the following conditions are all equivalent:

• $C = A \oplus B$, i.e. $A \cap B = 0$
• $\forall c \in C: \exists! a \in A \exists! b \in B: c = a + b$
• $\forall c \in C: \exists! a \in A \exists b \in B: c = a + b$
• $\forall c \in C: \exists a \in A \exists! b \in B: c = a + b$
• $\exists! a \in A \exists! b \in B: 0 = a + b$
• $\exists! a \in A \exists b \in B: 0 = a + b$
• $\exists a \in A \exists! b \in B: 0 = a + b$