Genady

I don't think your equation is correct.

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Bonaire has Carnival, too. The nursing home has its own parade.

Please, stop. These were not my questions. I want to continue using my Skype number the way it used to be under Skype, but not under Skype now. That's all. I think I know the answer now. Consider this thread closed.

My request was not about the procedure, but about recommended or not recommended providers.

Thank you.

Do UK members have knowledge / opinion about UK provider Phonely?

Microsoft will kill Skype in a couple of months. I have a Skype number, which is a normal US phone number for people calling me, and I don't want to lose it. So, I need to port it to somebody else. One candidate is RingCentral. I've checked with them, and they confirmed that my number can be ported to their system. Google Voice, OTOH, accepts only cell numbers. Any recommendations before I make a move? Thank you.

I am not missing it. I am telling you that your metaphor / analogy is wrong.

No, it does not "act upon masses". Firstly, its effect does not depend on masses of whatever it "acts" upon, it is the same for massless particles as for massive particles and regardless of their masses. Moreover, it does not "act". Particles just stay on spacetime geodesics.

Don't you know that gravity is not a force?

Special relativity applies with acceleration as well. Albeit the application is more involved.

The text I'm reading proves uncountability of the Cantor's set by showing that the Cantor's function is a surjective map from the Cantor's set onto [0, 1]. I think that it can be shown by direct use of the Cantor's diagonal argument for the Cantor's set, i.e., without use of the Cantor's function. Am I right or am I missing something?

How do you know this? I asked you to show how they apply specifically to mathematics. I still await this demonstration.

I guess then that I'm just lucky to be helped by the other 1%.

Yes, but it did not help to start using an article in Hebrew and two articles in English. P.S. Papiamentu doesn't bother with articles either.

So, the solution is not to extend 'length' to ALL subsets of real numbers, but rather to ignore "weird" subsets like the one used in the proof above and to extend 'length' to as broad as possible set of subsets which might appear "naturally" in the analysis (Lebesgue measurable sets.)

Something like it is actually often done in books, e.g., OTOH, if it were by some king's decree, it would. So, the lack of a clear answer is a partial answer.

You are not alone: The death of capital letters: why gen Z loves lowercase | Young people | The Guardian (I know that you are not in gen Z.)

I don't believe the two reasons you've quoted, but later on the same page they say, which goes along my original hypothesis. I compare it with Russian, where it is not capitalized (я). It is as "difficult to read" and has as much "importance of the writer" as in English, but unlike in English, there are plenty of other one-letter words in Russian, e.g., 'in' (в), 'at' (у), 'to' (к), 'about' (о), 'and' (и), 'with' (с).

Why the pronoun 'I' is capitalized in English? Is it so because of it being a one-letter word?

Klein Bonaire in the morning light

Notice that the axiom of choice has been used in the proof above.

I've started studying measure theory by this book: Measure, Integration & Real Analysis (Graduate Texts in Mathematics) (It is free on Kindle, btw.) It pretty much begins with proving that it is impossible to simply extend the concept of length from a real interval to an arbitrary set of real numbers. The proof is a bit formal, and I want to make it more intuitive while still rigorous. Would like to hear if the following description is good and if it can be improved. So, we want to have a real-valued, non-negative function 'length' defined for any set of real numbers with the following desired properties: a) it gives length of b-a for an interval [a, b], b) if set A is a union of disjoint sets A1, A2, etc., then length(A) = length(A1) + length(A2) + ..., c) length of a set does not change if the set shifts as a whole along the real line right or left. Let's assume that such function exists. Consider interval [0, 1]. It can be partitioned so that each number x belongs to a subset consisting of all numbers whose distance from x is a rational number. IOW, if y-x is rational then y and x are in the same subset. Let set V to contain exactly one element from each subset of the partition above. If we shift V by a rational distance r, we get a set Vr which contains different elements from the subsets. Since all elements in each subset are separated from each other by rational distances between 0 and 1, the union of all sets Vr which are shifted by all rational distances r between -1 and 1, covers the entire interval [0, 1]. So, the sum of lengths of all sets Vr is greater than or equal to the length of the interval [0, 1], i.e. greater than or equal to 1. Since every Vr is just a shifted V, they all have the same length, which is therefore greater than 0. OTOH, all Vr are shifted from the original V by maximum 1 to the right or to the left. Thus, their union is covered by the interval [-1, 2] and therefore the length of the union is less than or equal to 3. By taking enough of the shifted sets Vr we can get the sum of their lengths to be greater than 3. Then we get disjoint sets with the sum of their lengths being greater than the length of their union. This contradicts the property b) above. Thus, a function with the properties a), b), and c) does not exist. Is it clear enough? How can this explanation be improved?