# Direct sum and intersection of more than two subspaces

In "Linear Algebra Done Right" Axler points out that if you have more then two subspaces, it is not enough to test that each pair intersects only at 0, in order to say that they have direct sum. As an example, he asks to consider the following subspaces:
$$U_1 = \{(x, y, 0)\}$$
$$U_2 = \{(0, 0, z)\}$$
$$U_3 = \{(0, y, y)\}$$

where, he says, $$U_1 \bigcap U_2 = U_1 \bigcap U_3 = U_2 \bigcap U_3 = \{0\}$$ but they do not have direct sum.

I agree that they do not have direct sum, but if I test pair intersection I see that
$$U_1 \bigcap U_3 = \{(0, y, 0)\} \ne \{0\}$$
$$U_2 \bigcap U_3 = \{(0, 0, d)\} \ne \{0\}$$

So, to me it looks like contradiction, and that it is enough to check pair intersection. Where am I wrong?

There is no $$(0,y,0)$$ in $$U_3$$ with $$y\neq 0$$. The elements of $$U_3$$ are of the form $$(0,y,y)$$. So if $$y=0$$, we will have $$(0,0,0)$$. Hence we cannot produce $$(0,y,0)$$ with $$y\neq 0$$.