If space is not continuous, then is calculus wrong?

[D H]

\mathbb Z is definitely not the intersection of all the hereditary sets containing 1; \mathbb N is a hereditary set containing 1, but it doesn't contain the rest of the integers. And the interval \left[1,\infty\right) is a hereditary set containing 1, so \mathbb N is not the only hereditary set containing 1.

Got it. I just didn't follow your directions the first time around. I didn't stare at it long enough.


Anyhow, we are getting quite far afield from the OP. Then again, the original post represented a misconception that was dealt with in the first few posts of this thread. Whether space is discrete or continuous has nothing to do with the validity of calculus. Or the reals for that matter.

 

[Studiot]

Lugita, thank you for these clear and handy explanations.

 

[Studiot]

Here is another way to look at the question
(This was inspired by another recent thread)

If we roll a die n times the probability of a specific sequence is

{P_n} = {\left( {\frac{1}{6}} \right)^n}

Now let n tend to infinity

{\left[ {{P_n}} \right]_{n \to \infty }} = \left( {\frac{1}{6}} \right)_{n \to \infty }^n = 0

That is the probability of any particular sequence becomes vanishingly small.

Yet we assert that if we add all of these up we get a finite total.

\sum\limits_{n = 1}^\infty {\left( {{P_n}} \right)} = 1

Which is essentially the same process as the probabilisitc calculation/view in quantum mechanics.