Hilbert Spaces and linear subspaces Let X and Y be linear subspaces of a Hilbert space $\mathcal{H}$. $\\$
Recall that $\\$
$X + Y$ = {$x+y: x \in X, y \in Y$} $\\$
Show that $\\$
$(X+Y)^\bot$ = $X^\bot\cap Y^\bot$ $\\$
I tried solving the problem as follows;$\\$
Let $z\in(X+Y)^\bot$, if$\;$ $\forall$ $x+y\in(X+Y)$ we have $<z,x+y> = 0$. And if $x+y\neq 0$, this implies $z=0$.Hence $(X+Y)^\bot=z=0$. $\\$
Again from $<z,x+y> = 0$, we have $<z,x+y> = <z,x>+<z,y>=0$. Which implies $z\in X^\bot$ and $z\in Y^\bot$. Thus $X^\bot\cap Y^\bot=z=0$. $\\$
Therefore, $(X+Y)^\bot$ = $X^\bot\cap Y^\bot$. $\\$
I don't know if my working is logically correct. Your comments and hints, would be greatly appreciated. Thanks.
 A: Let $z \in (X+Y)^{\perp}$. This means $\langle z,u \rangle =0$ for all $u \in X+Y$. In particular, take $u=x+0$, where $x \in X$ and $0 \in Y$ (since $Y$ is a subspace so $0 \in Y$). Thus $\langle z,x \rangle =0$. Consequently $z \in X^{\perp}$.
Now you can proceed similarly to show that $z \in Y^{\perp}$. This would prove that $(X+Y)^{\perp} \subseteq X^{\perp} \cap Y^{\perp}$. You will now have to show the other containment.
A: With
$z \in (X + Y)^\bot \tag 1$
we have by definition
$\langle z, x + y \rangle = 0, \; \forall x + y \in X + Y; \tag 2$
so in particular if we take
$y = 0, \tag 3$
then
$\forall x \in X, \; \langle z, x \rangle = 0 \Longrightarrow z \in X^\bot; \tag 4$
likewise with
$x = 0 \tag 5$
we obtain
$\forall y \in  Y, \; \langle z, y \rangle = 0 \Longrightarrow z \in Y^\bot; \tag 6$
combining (4) and (6) yields
$z \in X^\bot \cap Y^\bot, \tag 7$
and thus we have
$(X + Y)^\bot \subset X^\bot \cap Y^\bot. \tag 8$
Going the other way,
$z \in X^\bot \cap Y^\bot \Longrightarrow z \in X^\bot, \; z \in Y^\bot$
$\Longrightarrow \forall (x, y) \in X \times Y, \; \langle z, x \rangle = \langle z, y \rangle = 0, \tag 9$
whence
$\forall x + y \in X + Y, \; \langle z, x + y \rangle = \langle z, x \rangle + \langle z, y \rangle = 0 + 0 = 0, \tag{10}$
whence
$X^\bot \cap Y^\bot \subset (X + Y)^\bot. \tag{11}$
Now by virtue of (8) and (11) we conclude that
$(X + Y)^\bot = X^\bot \cap Y^\bot. \tag{12}$
$OE\Delta.$
