It's not a silly question.
There is a fantastic post by
doug293cz that wowed us with some maths, if you haven't seen it its here
https://www.homebrewtalk.com/forum/threads/keg-purging-with-active-fermentation.628658/
Taking the same initial assumptions he did and adding further that
- You spund with 1 Plato of extract remaining.
- Head space after filling the spunding keg is 1 L of air (210000 ppm oxygen)
1 Plato is 1% sugar by weight. Thats 0.2016 kg of sugar, 0.1037 kg of C02 after its fermented. Thats 51.850 L or 518,500 little bubbles.
Plugging that into his final formula gives:
Final O2 Conc = 210000 ppm * ((1 L - 0.0001 L) / 1 L)^518500 = 6.352e-18 ppm
That's nothing.
You'll dissolve some oxygen as you transfer into the spunding keg and as
Hannabrew just said it's really easy to purge with fermentation so that's what I do.
* I've had a few beers tonight so apologies if I made a mistake (do your own calculations - don't trust weirdos on the internet).