Why are the charge operator $Q$ and the baryon number operator $B$ unbounded? A friend recommended me to read PCT, Spin and Statistics, and All That written by R. F. Streater and A. S. Wightman. In page 5 to 6, here's what the authors of this book have to say: 

[...] In fact the operators $Q$ and $B$, being unbounded, do not lie in $\theta^{\prime}$, but the associated projection operators, which project onto the states of various possible values of $Q$ and $B$, do. [...]

Can someone please explain to me as to why the operators $Q$ and $B$ are unbounded? Also, what could be understood by the term associated operator to an unbounded operator? Here $Q$ is an operator for charge and $B$ is an operator for baryon number.
 A: $\newcommand{\ket}[1]{\lvert #1 \rangle}$The charge and baryon number operators are not bounded because you can create states of arbitrary charge and baryon number:
Let $a^\dagger$ be any creation operator that creates a bosonic charged particle state (let's say with unit charge), and let $\ket{\Omega}$ be the vacuum. Then,
$$\ket{n_e} = (a^\dagger)^{n_e}\ket{\Omega}$$
is, for every $n_e \in \mathbb{N}$, a state with charge $n_e$. If you only have fermionic charged states, you need to choose $n_e$ different momenta $p_i$ and use $\prod_{i = 1}^{n_e}a^\dagger(p_i)$, but the idea is the same.
The same reasoning works for baryons - if you can create a state with one baryon, then you can create a state with $n_b$ baryons.
This shows that the number and charge operators are unbounded, since their eigenvalues $n_e,n_b \in \mathbb{N}$ can be chosen arbitrarily high.
Projection operators, on the other hand, are always bounded, because their highest eigenvalue is 1.
A: For the unboundedness see ACuriousMind's answer.
About the associated projections, for unbounded operators there is the notion of affiliation. An unbounded, closed and densely defined operator $A$ is affiliated with a von Neumann algebra $M$ if all its spectral projections are in $M$, that is
$$A = \int\lambda\text dE(\lambda),$$
with the spectral measure $E$ taking values in the projections of $M$. In the book you have referenced $\theta'$ is the commutant of the C*-algebra of all the observables, and therefore a von Neumann algebra by von Neumann's bicommutant theorem.
