Suppose Riddle 1. For all real number A > 0, A > B. B is ≥ 0. Then, what is B? B = 0 is good. But we want to prove that there is no B that satisfies A > B > 0. There is a proposition that “the only non-negative real number which is less than every positive real number is 0”. This proposition is tricky because there is a conflicting proposition that “every positive real number has a smaller positive real number”.

There is infinitesimal which is not a real number and less than every positive real number. I will stay away from infinitesimal in this discussion. Riddle 2. For all real number A > 0, A > B. B is ≥ 0. If A = B/2, what is B? This is a nonsense riddle because A cannot be B/2 when A > B. There is no B that satisfies B/2 > B regardless of whether B/2 is 0 or > 0. Thus, A = B/2 is an invalid assumption.

The proof is to exclude B > 0. Reasoning of the proof is like this: For all real number A > 0, A > B. B is ≥ 0. If we assume “A = B/2 > 0”, this assumption of “A = B/2 > 0” leads to a contradiction that B/2 > B. Thus, “B/2 > 0” is contradicted. Thus, B = 0. However, “A = B/2” contradicts A > B regardless of whether B/2 > 0 or = 0. “A = B/2” is an invalid assumption as in Riddle 2.

Let’s see what happens if B = 0. If we assume “A = B/2 = 0”, this assumption of “A = B/2 = 0” leads to a contradiction that A = 0 and 0 > 0. Thus, we have to conclude “B/2 = 0” is contradicted, thus B > 0. Thus, this proof shows that both B/2 > 0 and B = 0 are contradicted, thus B is both 0 and > 0, which is nonsense. Like Riddle 2 is nonsense, this proof is nonsense because of an invalid assumption “A = B/2”.

Instead of concluding that the proof is nonsense, they are trying to salvage the proof by removing B = 0 in the proof with an expression that “B = 0 would not be a vaild choice”. But what they are doing is covering up the nonsense which arises from an invalid assumption A = B/2. Rather than covering up this nonsense with “B = 0 would not be a valid choice”, we should conclude that the proof is nonsense because of an invalid assumption A = B/2 which is obvious in Riddle 2.

We can have B = 0 by Riddle 1. However, Riddle 1 does not exclude B > 0. The proof tried to exclude B > 0 to prove B = 0 as the only solution, however, the reasoning in the proof is nonsense. In this discussion, I do not discuss whether the proposition is true. I just discuss that the proof is wrong. Only in later posting math.stackexchange.com/questions/4857638/… , I discuss that the proposition is wrong.

@user1286486 There is a proposition that “the only non-negative real number which is less than every positive real number is 0”. This proposition is tricky because there is a conflicting proposition that “every positive real number has a smaller positive real number”. That is precisely why this proposition is not tricky - if you assume that there exists a positive real number less than all other positive real numbers, then you immediately arrive at a contradiction.

@user1286486 In any case, I'm sorry to hear that you have not understood the proof. I thought perhaps based on your comments to my answer to your other question, you might have accepted the possibility that you have a series of significant misunderstandings about this topic, but it appears that you are convinced that you are right, and are not actually trying to learn about the subject. I only answer questions asked in good faith. Best of luck.

Both proposition are true. Thus, I see that they are conflicting. Am I wrong?

@user1286486 Which two propositions are you referring to?

“the only non-negative real number which is less than every positive real number is 0” and “every positive real number has a smaller positive real number”. Thank you very much.

@user1286486 Those propositions do not conflict at all - the former follows directly from the latter! If $B$ is a non-negative real number, then either $B>0$ or $B=0$. $B$ cannot be $>0$ precisely because your second proposition is true, leaving only the possibility that $B=0$.

I am sorry to argue. But since the second proposition implies that there is always a smaller positive real number for any positive real number, it contradicts the first proposition which states the only non-negative real number which is less than every positive real number is 0 because there has to be a smaller positive real number for any positive real number. Am I mistaken? Is it the interpretation of any vs every that makes a difference? Thank you.

@user1286486 I don't understand why you think that is a contradiction. There is no positive real number which is less than all other positive real numbers, precisely for the reason you say. But zero (which is not positive) is less than all positive real numbers.

I am sorry again. But isn’t it true that all positive real numbers have a smaller positive real number by the second proposition? I am sorry that I am hard-headed.

@user1286486 Yes, that is true. And that is why, if $B<A$ for all $A>0$ and $B\geq 0$, the only possibility is that $B=0$.

Now I understand the problem. The second proposition is in the set of positive real numbers without 0, and the first proposition is in the set with 0. Because they are dealing with different sets of real numbers, I cannot state that there is a contradiction between the two propositions. However, may I be able to say the following? There is a proposition that “every positive real number has a smaller positive real number”. Thus, it is tricky to state that there is no B that satisfies A > B > 0. Thank you very much. You are most helpful.

@user1286486 I see the problem. Zero is neither positive nor negative, so when we say positive we mean strictly greater than zero, and when we say non-negative we mean either positive or zero. It is correct to say that there is no positive $B$ such that $A>B$ for every positive real number. But that is not what I, or the proof given above, said. When dealing with limits and their rigorous proofs, there is a crucial difference between $>$ and $\geq$ for precisely this reason.