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Let $A_1=P_1+V_1,A_2=P_2+V_2$ be affine spaces. My teacher uses $\dim$ on affine spaces and the embedded vector spaces interchangeably, which is correct by definition for $\dim A_1=\dim V_1$, but isn't necessarily correct for the results of operations like $+$ and $\cap$, which have different semantics for affine and vector spaces.


Does $\dim (A_1+A_2) = \dim (V_1+V_2)$?

Direct proof

$$\dim(A_1+A_2=\langle A_1\cup A_2\rangle=P_1+\langle V_1\cup V_2\cup\{\overset{\rightarrow}{P_1P_2}\}\rangle)=\dim(\langle V_1\cup V_2\cup\{\overset{\rightarrow}{P_1P_2}\}\rangle)\geq\dim(\langle V_1\cup V_2\rangle=V_1+V_2)$$

Counterexample

Let there be two parallel planes $p=A+V,q=B+W$. Clearly $$V=W\implies V+W=V=W\implies \dim V=\dim W=\dim (V+W)=2$$ On the other hand:$$p+q=A+\mathbb{R}^3\implies\dim p+q=\dim \mathbb{R}^3=3\geq\dim(V+W)$$


Does $\dim (A_1\cap A_2) = \dim (V_1\cap V_2)$?

Direct proof

$$\dim (A_1\cap A_2) = \dim(\{\overset{\rightarrow}{P_1P_2}\,\left|\,P_{1,2}\in A_1\cap A_2\})\right.\leq\dim (V_1\cap V_2)$$

Counterexample

Let there be two parallel planes $p=A+V,q=B+W$. Clearly $$V=W\implies V\cap W=V=W\implies \dim V=\dim W=\dim (V\cap W)=2$$ On the other hand:$$p\cap q=\emptyset\implies p\cap q\text{ is not a subspace of }\mathbb{R}^3\implies\dim (p\cap q)\text{ is undefined }\not=\dim(V\cap W)$$


Is my reasoning sound?

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