Does the lack of "for all" in (g) make this statement false?
Exercise 1, part (g):
If $x, y$, and $z$ are vectors in an inner product space such that $\langle x, y\rangle=\langle x, z\rangle$, then $y=z$.
But on page 333:
Theorem 6.1. Let $\mathrm{V}$ be an inner product space. Then for $x, y, z \in \mathrm{V}$ and $c \in F$, the following statements are true.
(a) $\langle x, y+z\rangle=\langle x, y\rangle+\langle x, z\rangle$.
(b) $\langle x, c y\rangle=\bar{c}\langle x, y\rangle$.
(c) $\langle x, 0\rangle=\langle 0, x\rangle=0$.
(d) $\langle x, x\rangle=0$ if and only if $x=0$.
(e) If $\langle x, y\rangle=\langle x, z\rangle$ for all $x \in \mathrm{V}$, then $y=z$.
I have so far:
Proof: $\langle x, y\rangle=\langle x, z\rangle \Leftrightarrow\langle x, y-z\rangle=0, \forall x \in V$ so you can take $x=y-z$ and get $\|y-z\|^2=0 \Rightarrow y=z$
Counter example:
Let $x=(1,0), y=(0,1)$, and $z=(0,-1)$. Then we have $\langle x, y\rangle=$ $\langle x, z\rangle=0$, but $y \neq z$
Am I missing something?
Upon reviewing the questions, I see my mistake, thanks to Sassatelli and Ted for confirming that the for all was indeed the issue:
The theorem says
IF the equality is true FOR ALL $x$, THEN $z=y$
While the (g) says
IF the equality is true FOR ONE $x$, THEN $z=y$