Proof of Functions from R to R I need hints in how to prove this.

Given $X,Y\subset \Bbb R$, and let $F=${$f:\Bbb R \rightarrow \Bbb R| f(X)=0$} and $G=${$g:\Bbb R \rightarrow \Bbb R| g(Y)=0$} prove that:

*

*$Hom(\Bbb R)=F+G\iff X\cap Y=\emptyset$


*$F\cap G=${$0$} if and only if $X\cup Y=\Bbb R$


*$Hom(\Bbb R)=F\oplus G \iff Y=\Bbb R -X$
Where  $Hom(\Bbb R)$ stands for the set of all the functions from  $\Bbb R$ to  $\Bbb R$

My thinking: For the first one we know that $Hom(\Bbb R)=F+G$, we assume that $X\cap Y\neq \emptyset$, we pick a $v \in X\cap Y$ and the by doing $(F+G)(v)$ we get a contradiction. How do I tackle the other part? and the other ones where should I start from?
Thanks for reading.

For one direction of the second point, if $p\in R$  \  $(X\cup Y),$ then $\exists h:R\to R\;(\; h(F\cup G)\subset\{0\}\;\land h(p)=1\;).$ Then $h\in F\cap G$ but $h$ is not the constant $0$-function.
For one direction of the third point, if $Y=R$ \ $X$ and $h:R\to R$, let $f(z)=0$ for $z\in X$ and $f(z)=h(z)$ for $z\in Y,$ while $g(z)=0$ for $z\in Y$ and $g(z)=h(z)$ for $z\in X.$
Note: Using the notation $f''X=\{f(z):z\in X\}$ (where $f''$ is read $f$-double-tick, or something like it), it would be better to write $F=\{\;f:R\to R\; |f''X\subset \{0\}\;\},$ and similarly for $G.$ It is not necessary to assume that $X$ (or $Y$) is not empty.
The $f''$ notation has the advantage that,in general, a member of dom$(f)$ may  also be a subset of dom$(f)$, so $f(X)$ may be ambiguous when $X\subset $dom$(f)$.
