Why are reciprocal lattice vector periodic, and time-frequency not? k-space vectors are related to each other by $k=k'+G$, where $G$ is the reciprocal lattice vector $G=2\pi/a$. This means that the frequency of oscillation in real space of a plane wave $e^{ikx}$ is bounded (correct?).
I am trying to draw an analogy to fourier analysis of a time-domain signal, where one can expand the signal in terms of $e^{i\omega t}$, however, there there is no reason for the $\omega$'s to be bound, or related by a translation $2\pi/\tau_0$.
In other words, I am trying to understand why $k$ vectors in a periodic lattice are themselves also periodic.
 A: I try to give you an intuitive reason for this:


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*As has already been said in the comments, the time-frequency DFT of a signal is also bounded in therms of the maximum frequency that can accurately be measured/reconstructed. This limitation stems from the sampling frequency of your hardware. It is therefore not a fundamental limitation, but merely imposed upon your measurements by yourself (e.g. you could spend more money and get better hardware with a higher sampling rate). So here the "spacing of the sampling points" is relevant.


Now comes the solid state physics part:


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*The periodicity of r-space comes from nature, since she decided to create crystals the way they are. You have different periodicities given by the Wigner-Seitz-cell of the lattice, which is nothing different than a Voronoi-cell. If you step over its edge, space looks the same as in the cell before, seen from its opposing side. You "alias" from one side to the other in real space. 

*The periodicity of the k-space vector comes now from the fact, that the periodic lattice from space is converted to a periodic lattice in reciprocal space by the Fourier transformation, i.e. k-space. The cell there is called a Brillouin zone, which again is nothing different than a Voronoi-cell in k-space. If you step over its edge with your k-space frequency, the reaction of the crystal looks the same as if you'd impose a k-space frequency upon it from the other side of the Brillouin zone. You once again get the aliasing effect.
There is a simple intuitive way to feel the $k$-periodicity if you think in terms of wavelenghts. Because the real-space lattice is discrete and not continuous in space, two wavelenghts might be different but carry the exact same physical information. You can see it on this picture : while the red and black wavelenghts are clearly different, the black atoms cannot distinguish between the two. There is thus a periodicity in terms of $\lambda$, which produces a periodicity in $k$-space.
I hope from this reasoning you've got a more intuitive understanding of what is going on. Certainly, all the statements can be cast into more or less beautiful mathematical equations, but I don't find them very teaching for this particular question.
A: The reciprocal space has a periodic structure only if the real-space potential is also periodic. This is because of Bloch's theorem: if you have a periodic hamiltonian, e.g.
$$\hat H=\hat T+V(\hat x)=\hat T+V(\hat x+a),$$
then you are guaranteed an eigenbasis of functions of the form
$$\psi(x)=e^{ikx}u(x),$$
where $u(x)=u(x+a)$ is periodic. In this case, $k$ is the quasimomentum of the state, and it can (only) be extracted from $\psi$ via its eigenvalue under a translation by $a$, which is $e^{ika}$; as such, it is only defined up to a multiple of $2\pi/a$, i.e. quasimomenta separated by $2\pi n/a$ are equivalent.

Something exactly analogous happens if you have a hamiltonian that's periodic in time, e.g. something of the form
$$\hat H=\hat T+V(t)=\hat T+V(t+T).$$
Here you know that if $|\psi(t)⟩$ is a solution of the Schrödinger equation, then $|\psi(t+T)⟩$ must also be one, so time translation is a symmetry of the system and we can hope for solutions of the form
$$|\psi(t)⟩=e^{i\varepsilon t}|\varphi(t)⟩ \tag 1$$
where $|\varphi(t)⟩$. As in the space-periodic case, the phase $e^{i\varepsilon t}$ is required because a time translation by $T$ must give you an equivalent state, but that only means equal up to a phase and not necessarily exactly equal. 
The states in $(1)$ are known as Floquet states, and they are studied by Floquet theory, which is well-established but for which introductory resources are relatively scant. Each Floquet state has a quasienergy $\varepsilon$, and these do indeed have the same periodicity properties as crystal momenta; in particular, changing $\varepsilon$ to $\varepsilon +n \omega$ will yield a state of the same form, since $e^{in\omega t}|\varphi(t)⟩$ is also periodic.
Moreover, you are also guaranteed a basis of TDSE solutions of Floquet form, though here you need to go a bit beyond the workings of the spatial case (where it suffices to show that $[\hat H,e^{ia\hat p}]=0$), by taking the Floquet hamiltonian
$$\hat H_F=\hat H-i\frac{\partial}{\partial t}$$
on an expanded Hilbert space $\mathscr H$ given by the tensor product of the original Hilbert space $\mathcal H$ and the space of periodic functions on $[0,T]$; Floquet TDSE solutions then map into eigenstates of $\hat H_F$ and you can use its eigenbasis.
