Gabor uncertainty and time-frequency resolution

I have a question about Gabor's uncertainty theorem, and how it relates to time and frequency resolutions.

As I understand it, Gabor's uncertainty theorem states that the standard deviations of a signal $$\sigma_t$$ and its Fourier transform $$\sigma_f$$ satisfy

$$\sigma_t \sigma_f \geq \frac{1}{4\pi},$$

the consequence of this is that you can't decrease the standard deviation of your signal to $$0$$, without the standard deviation of the Fourier transform increasing to $$\infty$$. I think I get that.

I have often seen it stated that a consequence of Gabor's uncertainty theorem is that we can't resolve both time and frequency, which I think means that we can't make the temporal and frequency resolution both go to $$0$$. I can see that, because if we were to discretely sample $$N$$ points from the signal, say at rate $$\Delta t$$ (this is the temporal resolution), then the Fourier frequencies will be $$j/(N\Delta t), j=0,...,N-1$$ (with frequency resolution $$1/(N\Delta t)$$. I can see that decreasing the temporal resolution ($$\Delta t$$) will increase the frequency resolution, and vice versa, but I'm not sure how this relates to Gabor's uncertainty theorem.

My question is, how is Gabor's uncertainty theorem, which is a result about the standard deviations of a continuous signal and its Fourier transform, related to the time and frequency resolution trade-off, which is a result seemingly about how we sample from the signal. I just can't see it!!!

Edit: an example of someone making that claim can be found on the wikipedia page of the uncertainty principle. They say "when applied to filters, the result implies that one cannot achieve high temporal resolution and frequency resolution at the same time".

• See this answer for more understanding of what's going on. Aug 30, 2021 at 21:25
• 3b1b Aug 31, 2021 at 0:03