4 ms·
Indeed, unlike the axiom of choice, which is routinely used by most working mathematicians, the continuum hypothesis is not. (In fact, most working mathematicia
by IngoBlechschmid 8y ago
Indeed, unlike the axiom of choice, which is routinely used by most working mathematicians, the continuum hypothesis is not. (In fact, most working mathematicians can probably not even state this hypothesis precisely, as they're concerned with different areas of mathematics and haven't received the necessary training in logics or set theory; and among set theorists, consequences of CH are explored just as often as consequences of !CH are. (CH does simplify several things, such as arithmetic with infinite cardinal numbers, but the point still stands.))
Yes, the P=NP problem is not yet proven to be independent of ZFC. It's certainly a possibility, and to be honest a quite intriguing and fun one, but we shouldn't make the mistake to think that any hard open problem is independent of ZFC. (I know that you didn't state that, I'm just writing this for the record.) Sometimes hard open problems are simply hard open problems.
- baddox 8y agoI agree on your last point, but I will say the P=NP feels like a particularly good candidate for being independent of ZFC, not only because it seems to be very difficult, but because it feels so adjacent to things like the halting problem and incompleteness theorems. :)