If you construct R out of Q using Dedekind Cuts, you can then, for each x in R, prove that there is a Cauchy sequence which converges to it. Then if you construct R out of Cauchy sequences, you can then, for each x in R, prove there is a corresponding Dedekind Cut. If you construct R by any other means, you're doing it differently, and need to prove that your R coincides with the one made out of Dedekind Cuts or Cauchy Sequences. Or at least, prove that it naturally embeds Q, is a field, satisfies the principle that for every x in R there is a natural number n in N with n > x, and that it is order complete (every subset of R bounded above has a least upper bound that is an element of R). So how do you construct your set of real numbers?

I guess i'm not supossed to be here

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Nope, sorry mate, you have no idea what you are talking about.

Is this a social experiment, or are you a real mathematical crank?

Is this a social experiment, or are you a real mathematical crank?

It is a conspiracy of mainstream mathematician lizzards! You should meet with John Gabriel (the new calculus), he will be your friend.

Ten minutes and 24 seconds of delusions about mathematics.

Stay in your lane. You are WAY outside your league here.

Please look up the definitions of the things you try to "disprove". Definition of convergence of a sequence: (a_n)n converges to a :<=> There exist an a in IR such that for all eps>0 we find some N in IN such that for all n that are greater than N it holds that : |a_n - a | < eps. THIS is the definition. Informally it means that if you have a convergent series to a limit point a for every security distance eps around a you can find an index number N such that for all indices n greater than N your sequence members stay within the given security distance eps. And you can do this for any eps you pick. And thats how you proof it: You let eps>0 be abitrary and then choose N depenent on this eps such that the inequality in the definition holds for all n greater than N.

Your logic breaks down because there are no segments in the cantor set, so it is impossible to “converge to a segment.” The cantor set is the intersection of many segments, which leaves behind many points that always have a non-cantor set point between them. Some of these points are end points, and some are not, but none are in segments. As with your other videos, you have not defined what *you* think a limit *should* be, as it definitely does not match the currently accepted definition. Perhaps the current one *is* wrong, but it does not support your claims. I’d be open to hearing your more precise definition that is consistent, but it’s not possible to engage with your claims without such a definition.

8:55 That is not what a limit says. When you find a limit at infinity, that basically means (according to the epsilon-delta definition) that for any margin of error greater than 0, you can find a point at which the sequence/function gets within that margin of error of the limit and stays there. For example, for lim n -> inf (1/3)^n, if our margin of error is E, we need the sequence to stay within E of the limit (if the limit is supposed to be 0, between E and -E). It's pretty simple to find an n that makes this no more than E (ceil(-log3(E)), and since the sequence decreases smoothly after that but is always more than 0, it stays between 0 and E, which is within our margin of error, so it works. That being a limit doesn't mean that there is a value of n where it actually gets there; in fact, the whole reason we try to find a limit is that it never gets there for any finite n, but does get indefinitely close. One can also look at this by plotting 1/2^(1/x) and finding the limit as it approaches 0 (equivalent, but more visible); trying to evaluate it at 0 is invalid, but it obviously approaches a value of 0 as you move towards 0 from the right side.

8:55 Limits at infinity make no such claim. Informally, they only require that the function gets within any non-zero distance of the limit, no matter how far to the right of the real number line you are.

9:38 That's not what the proof claims. The starting statement is that "for all A > 0 the inequality B < A holds", not just that "for some A > 0 the inequality B < A holds". The former statement was proved to be true, so whatever follows should not contradict it. The fact the inequality holds for _any_ A > 0 means that whatever positive real number we replace A by, the inequality must hold. If B > 0 holds, then we could replace A by B/2, leading to B < B/2, which means B/2 < 0, and finally deducing B/2 is both positive (because B > 0 holds) and negative (because B/2 < 0 holds), which is a contradiction. Showing that A cannot be B/2 directly contradicts the starting statement: it says the inequality must hold for any A > 0.

LOL You clearly don't know jack about Math even if you say you like it or whatever. For your own good, stop making these videos, they're ridiculous. You clearly do not understand what you are talking about. Your first example of "does it make sense?" is already a bunch of BS if you have any understanding of the topic. It makes sense because R^3 and R have the exact same number of elements even if one represents a 3 dimensional space and the other a 1 dimensional space. You simply take this equivalence in one sense and speak as if it meant the two things are the same in every sense. Then you proceed to talk about things being counterintuitive... when you don't even understand the meaning of the words and concepts you use. Infinity times 0 is not 0, I don't know where you got that from, because infinity is not a real number... I could debunk the whole video but it's not even worth it. You do NOT know Math and it's not surprising at all that someone who doesn't know jack about Math thinks Math is "delusional" when you just don't get it because you haven't studied it as one should and that doesn't allow you to see your mistakes. And, for the love of god, do you think that such an obvious contradiction would not have been noticed before? This is not deep at all and you can be proven wrong in half a page if the steps are properly explained. If you are actually a professor of neurology, you should know you do not have the level of Math required to speak about this (as I do not have the level to speak about neurology) and I'd suggest you ask anyone doing anything related to analysis in the Math department of your university and they will unequivocally explain to you why you are wrong. But I get it, you won't because you know better than them and they're all delusional because you do not understand the argument.