Lack of knowledge of real analysis. That a function takes irrational values doesn't mean that it's discontinuous. That statement is, just, wrong.

It would be a good idea to learn real analysis, this: https://www.geneseo.edu/~aguilar/public/notes/Real-Analysis-HTML/ch3-sequences.html might be a good place to start.

Stam Nicolis The function gives only irrational outputs. What about the density of rational numbers in IR??? Who needs some lessons of analysis ? You or me ?

Once more: That a function takes only irrational values doesn't man that it's discontinuous. The reason is that the rational numbers are of zero measure, the irrationals, however, aren't. (That was found by Cantor in the 19th century.) More to the point not all sequences of rational numbers converge to rational numbers, all sequences of real numbers converge to real numbers. This is usually taught in the first year of any mathematics course in any university-just study the material I linked to.

A function can be discontinuous-not, however, because it takes only irrational values.

Stam Nicolis I think that you should prove first that all the irrational outputs of the studied Theta Function has no rational accumulation point. Can you prove that?

This wouldn't imply that the function is discontinuous.

The fact that a function takes only irrational values doesn't imply that it's discontinuous. That statement is, trivially, wrong: The function f(x)=sqrt(2) takes an irrational value for all real values of x. It's continuous, by an immediate application of the definition of continuity.

By the way, theta functions are continuous functions, as a consequence of their definition. That was proven already by Jacobi and is another standard exercise.

This might be useful:

https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://metaphor.ethz.ch/x/2019/fs/401-4110-19L/sc/talk4.pdf&ved=2ahUKEwiQ-66_ttiDAxUyTaQEHfv-BEkQFnoECCgQAQ&usg=AOvVaw076IcJu9R1gYuj174aOAzy

Stam Nicolis I understand what you say. Actually sqrt(2) is a constant function, hence it isn't an appropriate example. My question is that the studied Theta Function choses the set of its outputs from the line of real numbers. We don't even know if all the irrationals can be outputs of this Theta Function. Can you prove that any irrational can be an output of this function? The Lebesgue measure of irrationals is 1 just like real numbers. Is the continuity that you are proposing for this Theta Function a direct consequence of the Lebesgue measure 0 of rationals?

It would be useful to read the notes on the properties of the theta function...

Stam Nicolis Thank you for the link to the important document. The properties in your document never declared that all the outputs of the studied Theta Function are from the set of irrationals. Now that rationals are all excluded from the outputs, what makes you sure that any irrational can be an output for the studied Theta Function ?

By applying Theorem 3.2 in the notes. That it takes all possible values, irrational or not, in any bounded interval in its domain of definition is a consequence of this theorem.

Stam Nicolis You should know that it is demonstrated that the set of outputs of the studied Theta Function is purely irrationals. Do you mean that the article proving the irrationality of the outputs is wrong?

Stam Nicolis I added this paragraph as a part of the conclusion in my article:

This kind of discontinuity is difficult to disprove even if the consequence of the continuity proved in this article is that the studied Theta Function should take all possible values, irrational or not, in any bounded interval in its domain of definition. Actually, the problem persists even if we know that the set of rationals is countable which allows us to use a method in order to choose them isolated with countable domains in the line of real numbers and that this proves the Lebesgue measure of rationals which is zero. This is because we aren't free to choose rationals with the same method from the set of the outputs of the studied Theta Function since the choice method can only be made by this studied Function. This means that we don't know how the studied Theta Function avoids the rational outputs because if this function avoids a set of points which has a rational number as accumulation point, then many irrational numbers will also be avoided as output of this function. Hence, we don't know even if any irrational can be a possible output value for the studied Theta Function since the known properties of Theta Functions in literature never declared that all the outputs of a Theta Function are from the set of irrationals. Now, can we still think that the continuity proved in this article for this Theta Function is not a contradiction and that it is a direct consequence of the Lebesgue measure which is “zero” for rationals and “one” for irrational and real numbers?

The text is in contradiction with known properties of theta functions, namely that they're continuous functions, since they're analytic functions, away from their singularities. That's why it's trivial to deduce that they aren't discontinuous. It would be useful to study real analysis, complex analysis and the known properties of theta functions, that are the topic of textbooks and courses, before making claims that are trivially wrong. The properties of the Lebesgue measure aren't relevant for the subject.

Stam Nicolis Stop accusing me of being a bad researcher because that would be a defamation against my reputation. Stop saying to the audience that my obvious easy article is wrong. What's the matter with you??

The statement that the theta function is discontinuous, because it takes irrational values, is, simply, wrong. And it suffices to read any textbook on the subject to learn the background of why this statement is wrong.

The irrationality of the values taken by theta functions is discussed, for instance, here: http://matwbn.icm.edu.pl/ksiazki/aa/aa53/aa5333.pdf

(and, of course, as mentioned in a previous text, the Jacobi theta functions provide a parametrization of the Fermat curve, x4+y4=z4, which contains only irrational points and is a smooth surface.)

I'm not making a statement about a person, but about a technical issue. The former isn't relevant for the latter and only the latter is of interest in a technical discussion.

Theta functions have many remarkable properties; that they're discontinuous, because they take irrational values isn't one of them.

Cf. here https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://people.mpim-bonn.mpg.de/zagier/files/doi/10.1007/BF01245085/fulltext.pdf&ved=2ahUKEwjQzvz98tqDAxUTVaQEHS3yCzY4FBAWegQIAxAB&usg=AOvVaw1ZOLCwgWdRt58wzmTHfs4j for more on some non-trivial properties of theta functions. It's a pity wasting time on issues that are trivially wrong, when there are so many non-trivial issues to understand.

Stam Nicolis Since you couldn't bring any logical answer to the discontinuity of the Theta Function caused by the set of outputs which is only irrational, then your textbooks don't contain any logical answer. Stop giving me your literature but you can summarize the logical answers contained in your literature as a comment here. You should also know that I am selfconfident that I am not mistaken at all. I think that you are the one who should revise his lessons because my articles are very important to the audience.

Correction:

Sorry for using the term of adherence whereas I was trying to use the concept of accumulation point or limit point around the point 10/7. Let me use another example with the definition of continuity in term of sequences. Please have a look on the attached image to understand my point:

Akram Louiz,

Your question and comments are complete nonsense and show your lack of knowledge about theta functions and in mathematics,which tripped you up.

Theta functions are entire and holomorphic functions and thus analytic and differentiable and surely continuous.

I would also like to remind you and call the attention of the audience that your pseudoscientific papers are published in crappy journals whose impact factors are less than zero and are deeply below average.

Issam Mohanna I am not impressed by your arab name. You and your superiors are really pathetitic. A previous article proves that all the outputs of the studied Theta Function are irrationals, hence the studied Theta Function can't be continuous at all. Can you prove the opposit and honour your arab name?

Issam Mohanna if you can't honour your arab name by proving the opposit, then you should at least know that you made an obvious defamation against me in front of the audience and that I will reach the law against you. Unfortunately, your beloved Einstein monks won't save you in front of the court my boy, you will see. Arab names are made only for men and it seems that you can't honour yours.

Correction:

Sorry for using the term of adherence whereas I was trying to use the concept of accumulation point or limit point around the point 10/7. Let me use another example with the definition of continuity in term of sequences. Please have a look on the attached image to understand my point:

Dear Akram Louiz, in the usual topology of R, the set of irrational numbers is dense in R. This means that there exists an irrational number in every nonempty open set in R. Using this statement, it is easily shown that there does not exist an irrational number that is closest to 10/7. Indeed, suppose to the contrary that there exists an irrational number, say r, such that r is closest to 10/7. Since r is irrational, it follows that r is different from 10/7. Hence, the open interval (r,10/7) if r10/7, is nonempty, and thus it contains an irrational, which is thus different from r and closer to 10/7 than r, which contradicts our assumption. Therefore, there does not exist an irrational number that is closest to 10/7.

Also, what do you mean by the notation (10/7-1/n)? If you mean the set {10/7-1/n:n being a positive integer}; i.e., the set {10/7-1,10/7-1/2,...}, then every adherent point of this set is a rational number; there do not exist irrational numbers among the adherent points of the previous set. Note that the adherent points of a subset S of a topological space X are exactly the points of the closure of S; i.e., the points of S and the limit points of S. In the usual topology of R, the closure of {10/7-1/n:n being a positive integer} is the union {10/7-1/n:n being a positive integer} and {10/7}, since 10/7 is the unique limit point of {10/7-1/n:n being a positive integer}, and all points of this union are rational numbers.

Spiros Konstantogiannis Sorry for using the term of adherence whereas I was trying to use the concept of accumulation point or limit point around the point 10/7. Let me use another example with the definition of continuity in term of sequences. Please have a look on the attached image to understand my point:

is there any remark about this proof using sequential definition of continuity? Here is the image attached again:

Akram Louiz, I have not read your article (or preprint) you cite; so, I cannot verify that the examined function takes only irrational values. However, what holds is that if f is a continuous function from a topological space X to a topological space Y, then the image of every compact set in X under f is a compact set in Y. If you want to use the previous theorem to show that a function is not continuous, you should demonstrate that there exists a compact set in X such that its image under f is not a compact set in Y. In R, by Heine-Borel theorem, a subset of R is compact if and only if it is closed and bounded; it is not sufficient that the subset be bounded only; it must also be closed. Further, it is important to know that a closed and bounded subset of R is NOT necessarily an interval; for instance, any finite subset of R is closed and bounded; thus, compact, and then it can be the image of a compact subset of R; thus, also of an interval, under a continuous real function. Another example, the union of the set {1/n:n being a positive integer} and the one element set {0} is a closed and bounded subset of R; thus, compact, too, and it is an infinite set.

Part of the notation you use is incorrect, and thus confusing. For instance, you write f-1(10/7). The notation f-1(10/7) denotes a number; in particular, it denotes the image of 10/7 under the inverse function of f, which exists if and only if f is a bijection. However, writing f-1(10/7) you seem to mean the inverse image of the one-element set {10/7} under the function f, which is a set and is denoted by f-1({10/7}). It is important to use the correct notation; otherwise, what you state is not clear.

Spiros Konstantogiannis I think that finding a counter-example for the definition of continuity in term of sequences is the easiest way to prove the discontinuity. We supposed in the proof that the studied Theta Function is continuous and since this function is also strictly monotone than the studied Theta Function is bijective and thus we have the right to write f-1(10/7).

Please read the proof of irrationality. It is a short easy and obvious proof.

Issam Mohanna I still wait for your apologies. Like any jerk you believe in Einstein theology and you refuse to accept pure mathematics. Why don't you name yourself Albert (just like Einstein) and then you can defy me without using your fake arab identity.

Akram Louiz, I will read it, but by assuming that 10/7 is the limit of a sequence of function values, as you do in the image you uploaded, you assume what you want to prove by the sequential criterion for continuity, and this clearly invalidates your purported proof. I note again that continuous functions map compact sets to compact sets, and a compact set is not necessarily an interval, as you seem to believe.

By the way, a strictly monotone function is injective (one-to-one), but it may not be surjective (onto), and thus it may not be bijective. Your statement "since this function is also strictly monotone then the studied Theta Function is bijective" is false; you have to prove that it is also surjective.

Spiros Konstantogiannis A Continuous Real Function on Closed Interval is Bijective if and only if this function is Strictly Monotone.

Have a look on this link where you find the proof: https://proofwiki.org/wiki/Continuous_Real_Function_on_Closed_Interval_is_Bijective_iff_Strictly_Monotone

Actually my purpose is to prove the contradiction of continuity that happens when we use the summation of Poisson.

I gathered the two short articles above in one better and longer article. This will make you understand my point easier.

Here is the link:

https://www.researchgate.net/publication/377434870_The_contradiction_of_continuity_by_using_the_summation_of_Poisson_which_threatens_Riemann_Zeta_function

Akram Louiz, thank you. A strictly motone function is injective (one-to-one). Further, any function - not necessarily continuous, not necessarily monotone - is surjective on its image, which is the set of values of the function, and is a subset of the codomain of the function. As a result, a strictly monotone function is bijective on its image, not necessarily on its codomain; though. However, for the inverse function g-1 of a function g from a set A (domain) to a set B (codomain) to exist, the image of g must be equal to its codomain B. For the notation f-1(10/7) to be meaningful as the image of 10/7 under the inverse function f-1, the image of f must be equal to its codomain, which is by definition R, since you assume that f is a real function. However, if f takes only irrational values, as you claim you have proved, then the image of f is a strict subset of R; hence, f is not surjective, and the inverse function does not exist.

Spiros Konstantogiannis The irrationality shouldn't be used from the beginning because we want to make a contradiction for the bijectivity of the Function f. We have at the beginning the fact that the Function f is strictly monotone and is continuous in any closed interval which is included in the interval [2; +infinity[. This means that if f(x) = 10/7 then x is obviously included in a closed interval which is included in [2; +infinity[ hence f is bijective in this closed interval of x. Then, we have the right to write f-1(10/7). However, after that, we use the characteristic of irrationality to contradict the bijectivity of f.

Please read the article that gathers the two discussed articles to understand my purpose.

Akram Louiz Dear Akram, I think you know well all properties of exp function, f:R->R, f(x)=ex. f is continuous on R, consequently f is continuous on the interval [1, 10/7] ....let's say! I affirm f(t)=et is irrational for all t in [1, 10/7] ! Can you contradict me? Can you prove it exists u in [1, 10/7] such that f(u) is rational?

So, is exp continuous on R, or not !?

Dinu Teodorescu Yes I can contradict you. We have exp(t)=p/q whereas p/q is a rational in the interval [e; exp(10)7]. This is because Rationals are dense in IR hence it exists a number p/q such as e

Ok! It's true! Clearly exp(ln(p/q))=p/q. Let's go further! Is every exp(y) with y in Q irrational? And, if exp(t) is rational, can we say that t is always irrational?

For this assumption you will have to make a mathematical proof first before believing it.

1. Using examples, can you contradict the questions from my previous comment? Or all examples tend to show my statements are right?

2. Is the set {p+iq / p,q in Q} dense in C, in other words(notations) is Q+iQ dense in R+iR=C ?

3. It's correct if I say that theta function is Theta:C->C, Theta(z)=SUM(for n in Z)exp(iz(pi)n2) for every z in C ?

Akram Louiz Sorry! Now I have seen your statement: "Actually my purpose is to prove the contradiction of continuity that happens when we use the summation of Poisson."

Ok! In fact what do you want to prove? That the summation of Poisson type is not a well posed problem, defining functions being simultaneously continuous and discontinuous!? At the moment I observe you don't use correctly the epsilon delta definition of continuity of a function in a point! When you choose eta, you must take into account that eta can depend only on epsilon and on the point a(which is fixed), not on x!

For theta function, continuity can not be approached in a such way. You must use theta is a holomorphic complex function to obtain continuity.

Now, sincerely, I cannot see your aim!

Yes, you are allowed to use x

when choosing eta but that would be done when you were doing the limit at an unknown point x. This means that we don't use any specific value of x.

Akram Louiz Dear Akram, the things are not always as they seem to be!

Now I understand! You say: the image of R+ by theta function is in R\Q, then theta function is not continuous on R+. I'm wrong?

But that is not important regarding the continuity! You examples use bijections and in the case of theta function the things are not so simple, because we must see the all, the theta function as a complex function.

The continuity in a point signifies that when we are with x very close to the point a, then the values f(x) are very close to f(a).

All my comment can be a nonsense, if I understood bad your intention!

Dinu Teodorescu We have the right to study Theta Functions on IR and this is the case of my articles. I still didn't start the step of studying this Theta Function as a complex function. Indeed, the continuity in a point signifies that when we are with x very close to the point a, then the values f(x) are very close to f(a). However, the only solution to discuss continuity is with the Epsilon-Definition or with counter-examples. Do you suggest that the problem of continuity in IR doesn't affect complex numbers even if IR is included in the set of complex numbers?

Akram Louiz No, I don't suggest anything!

Let's consider the general case! If E,F are metric spaces and f:E->F is a continuous function, then every restriction of f to a subset D of E is continuous, i.e. every g:D->F, g(x)=f(x) is continuous. So, if it exists a discontinuous restriction of f, then the function f is not continuous.

Sometimes it's more useful to work with equivalent definition with sequences of continuity, i.e. f is continuous at the point a of E if f(x(n)) converges to f(a) for all sequence x(n) from E converging to a. The above property with restrictions can be better understand if we use sequences to characterize the continuity. Indeed, if a is a point in D and x(n) having all terms in D converges to a in D, then x(n) converges to a also in E.

To expose another phenomenon, consider the function f from R to R , f(x)=1 for x rational, f(x)=0 for x irrational. f is discontinuous on R, but his restrictions to Q, respectively to R\Q are continuous as constant functions.

Clearly the Theta function is correctly defined, i.e. the Laurent series defining Theta function are finite for all z in C. So, if you prove that theta function is continuous, but his restriction to irrationals(to R\Q) is not continuous, it is guaranteed you have mistakes somewhere in your proofs!

Dinu Teodorescu my proof is not about a restriction but it regards the interval ]2;+infinity[. The image of the studied Theta Function by using a summation of Poisson is a set of irrationals. Hence this fact leads me to the contradiction since without the summation of Poisson we prove that the same studied Theta Function is continuous. And thus I blame the summation of Poisson. Please read the article where I gather and summarize my two previous articles to understand my point. Here is the link :

Deleted research item The research item mentioned here has been deleted

Akram Louiz, you write in Section 3, page 4 (you should number the pages), of your preprint Deleted research item The research item mentioned here has been deleted

Spiros Konstantogiannis and Marina Klykova

Thank you for the valuable feedback. I noticed that the proof of irrationality is not sufficient in my article. Hence, in the new version of the article I just discussed the possibility of irrationality and made an attempt of proof that leads in an infinite loop. You will find in this version also a new demonstrated approximation that can be very useful in this matter. I will wait for your collaboration. Here is the new link: Preprint A discussion of a rationality and continuity related to Thet...

Akram Louiz, you are welcome. Your starting point; i.e., if the sum of the series whose terms are 1/2 raised to the power of n squared is rational or irrational is an interesting question.

Do you have any remarks about the proposed approximation and its usefulness ?

The inequality (37) results from the inequality (36) by taking the limit as n tends to infinity. Since |1/B|1/(n+1)2, for every positive integer n; however, in the limit n->00, the last inequality becomes equality (0=0). This result is known as Order limit theorem; see, for instance, https://math.byu.edu/~bakker/M341/Lectures/Lec08.pdf, page 3. Consequently, the inequalities (43), (44), and (45) are also non-strict, and so is the right inequality of the inequalities (46).

Spiros Konstantogiannis I am aware of the order theorem but this case is different that's why I chose the strict order. In equation (36) the series of the left is strictly increasing with terms which are strictly smaller then the terms of the series of the right and all the terms of both series are positive, and thus I judge that it is enough to choose the strict order despite the Order limit theorem. Hence, the inequality stays the same as it is shown in the attached image. Please remember that the number M is note infinite. I will wait for your opinion again about the strict order and about the usefulness of the inequation.

Akram Louiz, you are right; in this case the inequality (37) is strict*, and so are the subsequent inequalities. Since both rationals and irrationals are dense in the reals, it follows that between the leftmost sum and the rightmost sum of the inequalities (46) both rational and irrational numbers exist. I cannot see how you can use (46) to derive if the sum of your series is rational or irrational.

* Proof

Since the inequality (36) is the sum of n+1-M>=1 strict inequalities, and since if a

Spiros Konstantogiannis Yes, when M is finite, and B is rational the Theta Function is ordered between two rationals. However, we can't although find any real number B with f((B) rational. Can you find it?

Akram Louiz, I do not know how one can calculate the series f(B). As I see in WolframAlpha, all those series are calculated in terms of a theta function. For instance, https://www.wolframalpha.com/input?i=sum+1%2F2%5En%5E2, https://www.wolframalpha.com/input?i=sum+1%2F3%5En%5E2, https://www.wolframalpha.com/input?i=sum+1%2F4%5En%5E2.

Peter Breuer Thank you for your explanations. However, my article proves the continuity of the studied function. Hence, if the function output values are all irrational then this contradicts the continuity of the function. This is what I explain in the article. What do you think?

I think that the best way to show that a continuous real function takes both rational and irrational values is by invoking the intermediate value theorem (IVT), as Peter Breuer writes, but for the IVT to hold, the function must be continuous on an interval of R - it is not enough that the function is continuous on an arbitrary subset of R, not even on a arbitrary compact subset of R - and the function must be non-constant on that interval.

Note also that a real function is a real-valued function; i.e., a function that takes real values, not necessarily a function whose domain is a subset of R.

https://mathworld.wolfram.com/RealFunction.html

Peter Breuer in order to avoid the discontinuity proved in the article, can we say that only the irrational values of the outputs of the studied function are transcendental and that the rational values of the outputs of the studied function are only difficult to calculate ?

Peter Breuer unfortunately I didn't prove the transcendence but I propose a method of approximation in the formulas 46 and 47. What do you think ?

I just mean that a good approximation can abolish some possibilities of transcendent values. I still wonder about the good example to use as a value of the variable B

my approximation didn't determine yet if a Theta Function takes a transcendental value or not. However, it is obvious that with my approximation you can for example prove that the Theta Function hasn't the value Pi/5 or e/4 which are transcendental values.

Marina Klykova

Spiros Konstantogiannis and Marina Klykova

https://www.researchgate.net/publication/377658268_The_proof_that_the_rational_numbers_as_input_values_in_Theta-3_Functions_give_only_irrational_numbers_as_output_in_these_functions

Peter Breuer Theta Functions are transcendental functions however their output values are not all transcendental. Hence it is not proved that Theta Functions give only irrational outputs. If Theta-3 Functions give only irrational outputs then this will contradict their continuity. Who should revise his lessons my boy?

Peter Breuer The wikipedia article says only that the Theta Function output is transcendental for the lemniscate values and not algebraic values. Furthermore, Theta Functions are not Rogers-Ramanujan continued fraction. What's up with you Peter ?

Akram Louiz, you write in the first page of your preprint Preprint The proof that the rational numbers as input values in Theta...

"If there are infinitely many distinct rationals p and q such that, for some strictly positive real number δ we have: |x− p/q|

Spiros Konstantogiannis You proceed only with implications in your proof that is why your proof doesn't lead you directly to the rationality of the number x. The Dirichlet theorem by using a sequence Pn/Qn explains the needed steps to demonstrate the result.

Akram Louiz, where the statement "If there are infinitely many distinct rationals p and q such that, for some strictly positive real number δ we have: |x− p/q|