How to prove that $\dim(U_1 \cap U_2 \cap U_3) \geq 1$? I need to prove that if $V$ is a vector space and $U_1, U_2, U_3$ are sub-spaces of $V$, and $\dim V = 7,$ $\dim(U_1) = \dim(U_2)=\dim(U_3)=5$ then $\dim(U_1\cap U_2 \cap U_3) \geq 1$.
I tried to prove it, but it didn't work.

By modular law,
$$\mathrm{dim}(U_1\cap U_2) = \mathrm{dim}(U_1)+\mathrm{dim}(U_2)-\mathrm{dim}(U_1+U_2)\ge3$$
(using $\mathrm{dim}(U_1+U_2)\le \mathrm{dim}(V)=7$)
$$\implies \mathrm{dim}(U_1\cap U_2\cap U_3) = \mathrm{dim}(U_1\cap U_2)+5-\mathrm{dim}((U_1\cap U_2)+U_3)\ge 1$$
(using the previous result and $\mathrm{dim}((U_1\cap U_2)+U_3)\le \mathrm{dim}(V)=7$)
