Von Neumann algebra. Question regarding a proof in Kehe Zhu. Theorem 17.1 i Kehe Zhu's "An introduction to operator algebras" is the following statement: 
Suppose $A$ is a von Neumann algebra acting on $H$. If {${T_{\alpha}}$} is an increasing net of self-adjoint operators in $A$ which is bounded above, then {${T_{\alpha}}$} is strong operator convergent to a self-adjoint operator $T$ in $A$. Furthermore, $T$ is the least upper bound of {${T_{\alpha}}$}.
I have two questings regarding the proof: 
1) The proof starts out by claiming, that we can assume "without loss of generality" that there exists a constant c>0, such that $-cI \leq T_{\alpha} \leq cI$ for all $\alpha$. But why is this so?
2) The proof proceeds by first showing that {${T_{\alpha}}$} has a subnet which converges in the weak operator topology. Then the book claims, that (as net {${T_{\alpha}}$} is increasing), this implies that the whole net is convergent in the weak-operator topology. But why is this so?
 A: *

*Because to test convergence you only care about "the tail" of the net. So you can start on a fixed $T_{\alpha_0},$ and now $$-\|T_{\alpha_0}\|\leq T_{\alpha_0}\leq T_\alpha\leq c$$ for all $\alpha$. 

*I'm  not sure what this has to do with the proof, but assume a subnet $\{T_{\alpha_\beta}\}_\beta$ converges wot to $T.$ Fix $x\in H$. For $\varepsilon>0$, there exists $\beta_0$ such that $$\langle Tx,x\rangle-\langle T_{\alpha_{\beta_0}}x,x\rangle<\varepsilon.$$ Then, for any $\alpha\geq {\alpha_{\beta_0}}$, we have $\langle T_{\alpha_{\beta_0}}x,x\rangle\leq\langle T_{\alpha}x,x\rangle$, and so 
$$\langle Tx,x\rangle-\langle T_{\alpha }x,x\rangle\leq \langle Tx,x\rangle-\langle T_{\alpha_{\beta_0}}x,x\rangle<\varepsilon.$$ So (by polarization), $T_\alpha\to T$ wot. Now, using that $T-T_\alpha\geq0$, 
$$
\|(T-T_\alpha)x\|^2=\langle (T-T_\alpha)^2x,x\rangle \leq\|T-T\alpha\|\,\langle (T-T_\alpha)x,x\rangle\to0.
$$
The last inequality, if not obvious, is justified as follows: for a positive operator $S$, we have, by Cauchy-Schwarz, 
$$
\langle S^2x,x\rangle=\langle S\,S^{1/2}x,S^{1/2}x\rangle\leq\|S\|\,\|S^{1/2}x\|\,\|S^{1/2}s\|
=\|S\|\,\|S^{1/2}x\|^2=\|S\|\,\langle Sx,x\rangle.
$$
