An apparent inconsistency in the form of Spin-z operator for spin 1/2 system Problem:
This is problem 7 from the first chapter of Modern Quantum Mechanics by Sakurai (page 59).

Consider a ket space spanned by the eigenkets $\{ \mid a'\rangle \}$ of some Hermitian operator $A$. There is no degeneracy.
(a) Prove that $\prod\limits_{a'} (A - a')$ is the null operator.
(b) Explain the significance of $\prod\limits_{a'' \neq a'} \frac {(A - a'')}{(a' - a'')}$.
(c) Illustrate (a) and (b) using $A$ set equal to $S_z$ of a spin $\frac{1}{2}$ system.


My Work

*

*Construction of null operator and identity operator from eigenbasis of Hermitian operator:

(a) To show that it's a null operator it's sufficient to take some arbitrary $\mid \gamma \rangle$ belonging to the linear span of the eigenbasis $\{ \mid a' \rangle\}$, and show that
$$\left(\prod\limits_{a'}(A - a')\right) \mid \gamma \rangle = \left| 0 \right\rangle$$ (equation 1)
So, we have
$$A \mid a' \rangle = a' \mid a' \rangle$$ (equation 2)
and
$$\mid \gamma \rangle = \sum_{a'} \langle a'\mid \gamma \rangle \mid a'\rangle$$ (equation 3)
Using (3) in the LHS of (1), therefore, yields,
$$\sum_a' \langle a' \mid \gamma \rangle \prod\limits_{a''}(A-a'') \mid a' \rangle  = \sum_{a'} \langle a' \mid \gamma \rangle \prod\limits_{a''}(a' - a'') \mid a' \rangle = \left| 0 \right\rangle $$ (equation 4)
as when $a'' = a'$ inside the continued product, we get a $0$.
(b) From the previous calculation it's clear that:
$$\prod\limits_{a'' \neq a'} \frac {A - a''}{a' - a''} \mid \gamma \rangle = \sum_{a'} \langle a' \mid \gamma \rangle \cdot 1 \cdot\mid a' \rangle = \mathbb{I} \mid \gamma \rangle$$
$$\Longrightarrow \prod\limits_{a'' \neq a'} \frac {A - a''}{a' -a''} = \mathbb{I}$$ (equation 5)
(c) Illustration for $A = S_z$ of a spin $\frac {1}{2}$ system:
Let $\mid + \frac {1}{2} \rangle$, $\mid - \frac {1}{2} \rangle$ be the eigenvectors of $S_z$ operator. The $S_z$ operator can be decomposed as:
$$S_z = \mathbb{I}\cdot S_z\cdot \mathbb{I} = \left(\mid + \frac {1}{2} \rangle \langle + \frac {1}{2} \mid + \mid - \frac {1}{2} \rangle \langle - \frac {1}{2} \mid \right) S_z \left( \mid + \frac {1}{2} \rangle \langle + \frac {1}{2} \mid + \mid - \frac {1}{2}\rangle \langle - \frac {1}{2} \mid \right)$$
$$= \langle + \frac {1}{2} \mid S_z \mid + \frac {1}{2} \rangle \cdot\mid + \frac {1}{2}\rangle \langle + \frac {1}{2} \mid + \langle - \frac {1}{2} \mid S_z \mid  -\frac {1}{2} \rangle \cdot\mid  -\frac {1}{2} \rangle \langle - \frac {1}{2} \mid$$
$$= \frac {\hbar}{2} \left( \mid + \frac {1}{2} \rangle \langle + \frac {1}{2} \mid - \mid - \frac {1}{2} \rangle \langle - \frac {1}{2} \mid \right)$$ (equation 6)
The null operator $\hat{O}= (S_z - \frac {\hbar}{2}).(S_z + \frac {\hbar}{2}) = S_z^2 - \frac {\hbar ^2}{4} \mathbb{I}$ and the identity operators are $\mathbb{I} = \frac {S_z}{\hbar} + \frac {\mathbb{I}}{2}$, $- \frac {S_z}{\hbar} + \frac {\mathbb{I}}{2}$
Where I'm having trouble
The problem is with the very last result, on the last line of the page. I am getting
$$S_z=\frac{\hbar}{2}\mathbb{I}$$ and also $$S_z=-\frac{\hbar}{2}\mathbb{I}$$
which is clearly not correct. My guiding equation has been equation 5 (which seems to be correct). I have put $A=S_z$ of a spin 1/2 system in equation 5. I cannot locate the flaw in the steps.

So, where's the mistake? Please share your views.

Note: I should be sort of double lined and hollow like in the last sentence of "my work", but the command I found didn't work. Also, the arrow should be the same style. The command I found for that didn't work either. I replaced them with a normal I and a normal arrow. Here is the link to the work page image, if anyone wants it.
 A: Here's some notes on the other parts as well as the spin $z$ one:


*

*I would write the $0$ in the vector space as $0$ and not $\left|0\right\rangle$, since the latter suggests it's an eigenvector with eigenvalue $0$, but you wouldn't consider $0$ as a proper eigenvector since it would have everything in the field as it's eigenvalue.

*For part $b)$ note if $\left|\gamma\right\rangle=\sum\limits_a \langle a|\gamma\rangle\left|a\right\rangle$, then acting with $\prod\limits_{a''\neq a'}\frac{A-a''}{a'-a''}$ gives 
$$\sum\limits_a \langle a|\gamma\rangle\prod\limits_{a''\neq a'}\frac{a-a''}{a'-a''}\left|a\right\rangle=\sum\limits_a\delta_{a,a'}\langle a|\gamma\rangle\left|a\right\rangle=\langle a'|\gamma\rangle\left|a'\right\rangle$$
Since if $a\neq a'$ then $a''$ can equal $a$. Or if $a=a'$, then the above is a product of $1$'s. Significance: Does it look like a projection operator?

*We write $\left|\gamma\right\rangle=\langle \uparrow\big|\gamma\rangle\left|\uparrow\right\rangle+\langle \downarrow\big|\gamma\rangle\left|\downarrow\right\rangle$. Then 
$$\left(S_z-\frac{\hbar}{2}\right)\left(S_z+\frac{\hbar}{2}\right)\left|\gamma\right\rangle = 0\cdot\left(S_z+\frac{\hbar}{2}\right)\langle \uparrow\big|\gamma\rangle\left|\uparrow\right\rangle+\left(S_z-\frac{\hbar}{2}\right)\cdot 0\cdot \langle \downarrow\big|\gamma\rangle\left|\downarrow\right\rangle=0$$
Similarly (now there's only one term in the product) I'll pick $a'=\hbar/2$ value:
$$\frac{\left(S_z+\frac{\hbar}{2}\right)}{\hbar/2+\hbar/2}\left|\gamma\right\rangle=\langle \uparrow\big|\gamma\rangle\cdot \frac{\hbar/2+\hbar/2}{\hbar/2+\hbar/2}\cdot\left|\uparrow\right\rangle+\langle \downarrow\big|\gamma\rangle\cdot 0\cdot\left|\downarrow\right\rangle=\langle \uparrow\big|\gamma\rangle\left|\uparrow\right\rangle$$
as expected

