Let $V$ be a vector space and $U$ and $W$ subspaces. Recall that
$$
U^0+W^0 β(U β© W)^0 .
$$
When $V$ is an inner product space we have the similar looking inclusion
$$
U^{β}+W^{β} β(U β© W)^{β} .
$$
This latter inclusion may be strict outside of finite dimension.
See MSE
So what about the former inclusion?
Let's try and prove the reverse inclusion $(U β© W)^0 β U^0+W^0$ directly in the finite dimensional case (rather than appealing to a dimension argument).
So let $f β(U β© W)^0$, that is, $f: V β π½$ with $\left.f\right|_{U β© W}=0$. We need to find $g β U^0$ and $h β W^0$ with $f=g+h$. First we find a subspace $X β U$ with $$ (U β© W) β X=U $$ We can do this by taking a basis for $U β© W$ and extending it to one for $U$. We call this a direct complement for $U β© W$ in $U$. Likewise we find $Y β W$ with $$ (U β© W) β Y=W $$ One checks that $$ (U β© W) β X β Y=U+W $$ So finally we find a subspace $Z β V$ with $$ ((U β© W) β X β Y) β Z=V $$ Now we define $g, h: V β π½$ by giving them on each summand in this direct sum decomposition: \begin{array}c & U β© W & X & Y & Z \\ \hline g & f / 2 & 0 & f & f / 2 \\ h & f / 2 & f & 0 & f / 2 \end{array} Then indeed $g+h=f$ and $\left.g\right|_U=0$ and $\left.h\right|_W=0$ (note if 2 is not invertible in $π½$ this "symmetric" construction can be easily modified).
Our proof does not mention dimensions. But we do use finite dimensionality, extending bases for a subspace to the whole space (to show every subspace has a direct complement). Can this be done outside of finite dimension too? Well yes, if we assume something called Zornβs Lemma: this is an axiom in mathematics which is (probably) not necessary for most of, for example, my own subject number theory (and one which many mathematicians try to avoid). But it seems to be unavoidable in certain contexts.
So what about the former inclusion?
Let's try and prove the reverse inclusion $(U β© W)^0 β U^0+W^0$ directly in the finite dimensional case (rather than appealing to a dimension argument).
So let $f β(U β© W)^0$, that is, $f: V β π½$ with $\left.f\right|_{U β© W}=0$. We need to find $g β U^0$ and $h β W^0$ with $f=g+h$. First we find a subspace $X β U$ with $$ (U β© W) β X=U $$ We can do this by taking a basis for $U β© W$ and extending it to one for $U$. We call this a direct complement for $U β© W$ in $U$. Likewise we find $Y β W$ with $$ (U β© W) β Y=W $$ One checks that $$ (U β© W) β X β Y=U+W $$ So finally we find a subspace $Z β V$ with $$ ((U β© W) β X β Y) β Z=V $$ Now we define $g, h: V β π½$ by giving them on each summand in this direct sum decomposition: \begin{array}c & U β© W & X & Y & Z \\ \hline g & f / 2 & 0 & f & f / 2 \\ h & f / 2 & f & 0 & f / 2 \end{array} Then indeed $g+h=f$ and $\left.g\right|_U=0$ and $\left.h\right|_W=0$ (note if 2 is not invertible in $π½$ this "symmetric" construction can be easily modified).
Our proof does not mention dimensions. But we do use finite dimensionality, extending bases for a subspace to the whole space (to show every subspace has a direct complement). Can this be done outside of finite dimension too? Well yes, if we assume something called Zornβs Lemma: this is an axiom in mathematics which is (probably) not necessary for most of, for example, my own subject number theory (and one which many mathematicians try to avoid). But it seems to be unavoidable in certain contexts.