Hilbert spaces and orthogonality sets

I need to prove if $X$ is a Hilbert space and $M$ and $N$ it's closed: $$(M+N)^\perp=M^\perp\cap N^\perp$$ thanks

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Do you mean $X$ is a Hilbert space, and $M,N\subseteq X$ are closed? It's a bit unclear. Also, what have you tried? Where are you stuck? – Stahl Mar 27 '13 at 23:16

This is true for every sets $M$ and $N$, in any inner-product space $X$. They don't need to be subspaces, and they don't need to be closed. Note that the orthogonal of any set is always a closed subspace.
In words: a vector $x$ is orthogonal to $M+N$ if and only if it is orthogonal to $M$ and $N$.
Sufficient: assume $(x,m)=0$ for all $m\in M$ and $(x,n)=0$ for all $n\in N$. Then $(x,m+n)=(x,m)+(x,n)=0$ for all $m\in M$ and all $n\in N$. So $(x,k)=0$ for all $k=m+n\in M+N$.
Necessary: assume $(x,k)=0$ for all $k\in M+N$. Since $M\subseteq M+N$,, this implies $(x,m)=0$ for all $m\in M$. Likewise, $(x,n)=0$ for all $n\in N$.
Now a more instructive way for the "necessary" direction: first observe the fundamental related fact $$S\subset T\quad\Rightarrow\quad T^\perp\subseteq S^\perp.$$ Since $M$ and $N$ are both contained in $M+N$, it folows that $(M+N)^\perp \subseteq M^\perp$ and $(M+N)^\perp \subseteq N^\perp$. Hence $(M+N)^\perp \subseteq M^\perp\cap N^\perp$.