Why is the Fourier transform of a Dirac comb a Dirac comb?

This doesn't make sense to me, because the Heisenberg inequality states that $\Delta t\Delta \omega$ ~ 1.

Therefore when you have something perfectly localized in time, you get something completely distributed in frequency. Hence the basic relationship $\mathfrak{F}\{\delta(t)\} = 1$ where $\mathfrak{F}$ is the Fourier transform operator.

But for the Dirac comb, applying the Fourier transform, you receive another Dirac comb. Intuitively, you should also get another line.

Why does this intuition fail?

Answer

I believe that the fallacy is to believe that a Dirac comb is localized in time. It isn't because it is a periodic function and as such it can only have frequency components at multiples of its fundamental frequency, i.e. at discrete frequency points. It can't have a continuous spectrum, otherwise it wouldn't be periodic in time. Just like any other periodic function, a Dirac comb can be represented by a Fourier series, i.e. as an infinite sum of complex exponentials. Each complex exponential corresponds to a Dirac impulse in the frequency domain at a different frequency. Summing these Dirac impulses gives a Dirac comb in the frequency domain.