If A is a dense set and $x_1,x_2 \in A$, $x_1 \leq x_2$ and $x_1 \leq x \leq x_2$. The proof follows from  $F_1(x_1)= $F_2(x_1$,  $F_1(x_2)= $F_2(x_2$

and monotonicity

$F_2(x_1) \leq F_1(x) \leq F_2(x_2)$.

I do not know the name at this moment (probably there is no a name).

@george thanks for this (i think I recalled you mentioning something earlier) and you might know more about this. I get a bit annoyed with the definitions of monotone increasing/non-decreasing/weakly increasing and then strictly increasing. The first three seem to be identical to one another and are compatible with F being constant, so when they say monotone increasing do mean strictly so ( Ipresume that strict monotone increasing can be read as an order imbedding, in that its biconditional)

F(x)>F(y) if and only if x>y

F(x=F(y) if and only F(x)=F(y)

F(x)

 I was told that if F(x)=x for merely all dyadic rationals in [0,1] then F(x)=x for real in (0,1) if F is montone increasing, as F(x)=x is continuous and increasing, Is there any advantage of having F(x)=x for all rationals or strict monotone increasing. Moreover, If one specifies that F(0)=0 and F(1)=1 in addition does one have to show continuity at those pts or is just trivially identical to F(x) over [0,1] and thus its continuous or this irrelevant,

Or that one can derive it from the fact that F(x)=x in the open interval approaches F(1)=1 from below in a continuous fashion as it just F(x)=x and one never, as it were gets to x=1, so is sufficient so stipulate F(1)=1 and F(0)=0 and if not if its strictly monotopnic; is there some issue with order presevation

of course if F(x)=x over all dyadics its not a constant function but it still may only be monotonely increasing, If its only monotonely increasing what stopping all of the irrational within each tiny interval from being mapped onto the same number (unless one cannot speak of intervals of irrationals number between any dyadic rational number) or being mapped onto the dyadic rational value at the end of each tiny interval.

it may increasing, with no violation of monotonicity, so long as bigger irrational do not have  as they are have strictly smaller function sumsthan one another of the smaller dyadic on the lhs of the interval; it presume though that the RHS end point of the tiny interval will be the lhs of another and in limit smaller irrationals will attain bigger function values then some dyadic or other irrational which has strictly larger domain value perhaps. Or unless that would constute uncountably many discontinuities which would also be violation of mere monotone increasing

I would presume one would need strict monotone increasing-ness to preserve distinct-ness and to really preserve and reflect the ordering

When they increasing do mean strictly increasing?

ah yes that makes sense otherwise the limits would not line up, and especially if one is dealing with infinitesimal differences (for other reasons).

So Then I presume that the above result would hold with a function being $F$ presumed just 'monotonic increasing'. Does strict monotonic increasing offer any other benefits then in this context. The limits will exist for monotone functions; as to whether they are unique or not is a different matter.

; it explains some of my puzzlement deriving continuity at th end points, as given strict monotonic kincreasing one would not be able to the limit using 1\leq limF(x)\leq 1 if that were instead 1>lim x\to1

@george

And that if F(x)=x for all dyadic rationals for example in [0,1] which are dense in [0,1] then if F is just monotonic increasing, F(x)=x for all reals in (0,1), and continuous on (0,1),  and identical to the identity on (0,1)

Given that the its that F agrees with the identity function, on a dense subset of [0,1] , and supposed if F monotonic increasing, said limits exist at the end points (whether they are unique or not is different),  is continuous and identical to to the identity on (0,1),

As such, can one use the fact that it is now the identity G on (0,1) ,to show continuity at 1, and zero,  if in addition F is  stipulated to satisfy F(1)=1 , and F(0)=0.

In some sense F is already identical to the identity on \cup 1(0,1)\cup 0=[0,1] , as G(1)=1 and G(0)=0, to help show continuity of F @ 0, 1, clearly the identity G(x) =x \to 1, as to 1 , and one never actually gets to the point 1 when passing the limit (one is still in the open interval); or does it have to be shown independently. Which can be done in my context.

Or rather F just is the identity function which is the end goal, and continuity at the endpoints being a consequence.

Insofar as never actually gets to 1, 0 as one take limit, as it were, the identity G(x) clearly is continuous and identical at 1, 0 and does that result hold only once continuity at 1 and 0 for F has been shown independently

Dear William,

In connection with followers  comments,

perhaps, we can also consider a slight generalization of the considered statement:

Suppose that $x$ is in an open interval (a,b) and f and g two monotone functions on (a,b). Let (x_n) and (y_n) be two sequences

in (a,b) such that x_n

@ Dear George, thanks for this

- am I correct in presuming that when you say the one sided limit accordingly you mean the limit from below @ 1 and above at zero for example if the domain is [0,1] and not both unlike the interior.

-, when you say the limits may not be finite, wouldnt this count as an essential discontinuity; Or is a distinction between the case where both exist and are infinite, and count as being defined- and one where it diverges off to infinity and counts as being undefined and thus an essential discontinuity.  Perhaps, you mean the case where both of the two limits exist and is infinite and agrees with the function value (in which case given the other exists it could be said to be a jump discontinuity if the other if finite)

-By point of accumulation; do you mean interior points here or a fixed point/uniquely specified point. I presumed that monotonic functions had these limits on both sides within the interior.?

-- Is aone sided limit a name for this; or does it just denote the existence (whether one considers the interior point or a endpoints) of one, (Or both) one sided limits. I presume when it is said, that one sides limits, exist, they mean both limits exist (lhs and rhs or two sides limits actually) exist for the interior pts, and the relevant limits at the end points.

Although if one does not know the function value, or its not specified I presume that It might be difficult to know how to take the limit or rather whether it agrees with function value (I presume however, that given the function is monotonic, that the function cannot have 'removable singularities' on element of its domain; its defined there by definition in a sense, even if you do not know specifically what that value is or whether the function is monotonic or not, if the domain point considered is an element of the domain of the function; even if you do not not know what it is)

- So if that ids the case, I presume then that if 'both limits agree with each other', and at an (interior) point in a monotonic functions domain, then as the function is trivially defined on its domain, that is the function's value at that point for a monotonic function . If they agree, it cannot be a jump discontinuity, and as the function is defined do on its domain it cannot be a removable discontinuity; and essential and removable discontinuities are generally ruled out for monotonic function due to the existence of one sided limits  (at least for the latter),

I presume the former would somewhat contradict monotonic increasing-ness (as that there removable discontinuity or defined function value would presumably be equal in be strictly larger for example,then some other other defined points, close to it in the limits,  with a greater domain value. For arguments sake, if the graph were a linear function, F(x)=x and on [0,1], yet F(0.6) were defined explicitly to 0.8, all else being equal, and lie strictly above 'the continuous monotonic increasing linear function for which both limits agree and equal 0.6'  @  this either would violate the monotonicity via fact that F(0.6) > F(0.7)=0.7 (although its limiting value at 0.6 which is 0.6 may agree with monotonic increasing-ness or if one insisted that F(0.6)=0.6 nonetheless, it would violate function-hood, as would have two elements of the range being being assigned to the same domain value)

 I presume a monotonic jump function, for which F(0

@george

presume, when it is said, that-

-'one sided limits must exist' (for example in the context of a monotonic function, but not just in this) it is meant that both the limits from the right and left (above and below) exist or the interior ptsas otherwise, this would not preclude an essential discontinuity (if for example one of them does),. --

I presume that the existence of one sided limits, two sided limits and limits from the right and left, or both above and below are meant to mean the same thing (and do not refer in explicit way to the rhs end point, or lhs end point of a closed interval, by right hand limits, they just mean limit from above, for any point whether its at the end point, or an interior, and likewise for left hand side limits)

I presume that 'one sided' limits might be just used to emphasize the fact that both 'one side limits', may not agree with each others, and so there may not be a real limit at all , despite the fact that there there exist 'genuine limits' from both of ' each side taken individually,' and there are two of them. although they disagree) , or the fact, that  there is one side to consider on the endpoint of an interval. In some sense two-sided limits could be said to emphazise this possibility for strongly (ie that there two limiting values, indicating the possibility that there are two distinct values on each side).

Is it correct, I read, a that monotonic functions may have non-jump dis-continuities (removable discontinuities) at the end points of closed intervals;

- given that in some sense it makes no sense to say that both the limits from above and exist and are equal to each other at each end point,but not equal to the function value,

-as one only considers the limit from below at the rhs end pt of the closed domain, and the limit from above at the lhs end point of the closed domain, and not both) is this just an other name for saying that the limit from below  say 1 if the domain is [0,1]. is just not equal to the function value. Or do they mean something else

-I presume that this is just as bad as a jump dicontinuity and they are only saying that both limits exist and are equal to each other, because there is only a singular limit, to consider, and it always exists and is trivially identical to itself,- otherwise if the limit from below @1 exists and agrees with the 'value of the existing limit from above' @ 1 what exactly is the limit from below @ 1 it agreeing with, (not with the function value, hence a removable discontinuity)-

I presume it is mean that the limit from below @one, trivially agrees with the other limit (from above @1) because it does not make sense to say take the limit from above @1, thus  there does not exist, a limit from above @ 1  to disagree with the limit from below (and hence it trivially agrees with it)

I presume that this therefore, is just as bad as a jump discontinuity at the end point), if for example one were to extend the function. (past that domain  end pt,one may be able to take the other limit and notice that its not equal and so would still qualify as a jump discontinuity)

despite the fact that generally speaking they cannot (I presume the fact that monotonic functions cannot have jump discontinuities applies only to the interior or the open interval); I presume that depend

In any case in the context of the question. If F(x)=G(x) agrees with G(x)=x for all rationals in [0,1]  andfor example. Ie F(x)=x for all rational x in the domain of x;which is [0,1] to [0,1],  Then Once it is presumed that F is montononic  increasing, and thus F agrees with F(x)=x=G(x) for all reals in (0,1),  g and or strictly montonic increasing in addition,  does the fact that F satisfies F(0)=0 and F(1)=1, and as G(1)=1 and G(0)=0 in addition technically entail that F=G on the entirety of [0,1] (in some sense it does) , or implies does continuity at 1 and zero be used . Or does that have be shown independently,.

I presumed that once can can then take G's limit from below at 1, and show that it equals G(1)=1=F(1)=1, (which it will ) as , G  if linear and  continuous, likewise with from above, at 0; G(0)=F(0)=0,  as F's limit would presumably just be G's limit (not given the fact that F is continuous on (0,1) as this does not entail this,  but rather given the fact F is identical to G on (0,1) as so its one and same function as one takes the limit (as never actually gets to one).

Unless it only entails that the values of F are the same, and may not have the same functional form (unless there are other functions which distinct from G but almost identical on [0,1]);  and so it has to that F is continuous @ 0 and 1 independent (in my context I can do that) if necessary.

However, I presume the argument is generally that F just is G on (0,1) and thus continuous on (0,1) as G is continuous, and in my context I am trying to show that F=G, this is what trying to do, so would continuity @ 1 and 0 even be sufficient for that (unless considers some infintisimal differents at the boundary despiteif F(1)=1 and G(1)=1 , or would continuity of F on [0,1] just follow from the fact that F=G everywhere on [0,1] given F (1)=1 and G(1)=1 , G(0)=0 and F(0)=0 and G is continuous (or would at least the fact that F=G everywhere on [0,1] which is really the whole purposes given the addition of the endpoints)

An easy proof of the result you really want: F is monotonic. Let F(x)=y, y/=x. First assume y>x.

Then there is a rational number r such that x

@ F. Leyvraz

Thanks for this

I see, that as F(x)=y>F(r), violating monotonicity; although the derivations sounds like use with an involution . Does F(x)=x over [0,1]

Given F(1)=1 and F(0)=0 or just on (0,1) whereF(x) is strictly monotonic increasing F:[0,1] to [0,1] where F(x)=x for all rational x ( as a result of symmetry F(1-x)+F(x)=1, F-1(x)+F-1(1-x)=1  F(0.5) =0.5which is entailed by it and midpoint convexity at 1 and zero). F(1/2x)\leq1/2F and F(1/2(1+x))\leq1/2+F(x)/2

Given symmetry, the latter two collapse into equalities (before one applies any kind of monotonicity or continuity like argument) It also entails F(2x)=2F(x) , among other thingsand thus maybe some kind of rational homogeneity over irrational x

I did not think my result was that symmetric (that will give you any arbitrary transcendental number) I presume you did not use that. if the irrationals are uncountable. As I said given mid-pstar convex at 1 and 0 and symmetry  one can get F(r)=r for all rational I think, but I cant see how, given even strict monotone increasing, (unless its a theorem of mathematics, that between any two irrationals, there is a rational, are there nott there uncountably many more)

I mean if F(x)=y , y>x, and both x,y,are irrationals, wont there by can cases so finely grained that there are not rationals between them. I know the rationals are dense, but they are only countable in number, so one would presume that at some abstract, fine grained levem level that there could be whole series of irrationals with nothing but irrationals in between them; in virtue of their cardinality (in infinitesimal interval). I may be quibbling over technicalities which may only hold (if at all) in some weird non-standard analysis framework. Strict monotonicity will ensure that their order is preserved and they are well approximated but presumably they, could within themselves, be slightly out at the numerical level (one could squeeze their values into a smaller and smaller interval whilst preserving their rank and the fact that their numerical values are tightly bounded between the 'closest rational,x where F(x)=x, upper bound and lower bound outside this interval).

But if that means they will be out at most the infinite decimal place which means that they wont be out at all (because thats how they are defined/approximated) I suppose it does not matter)

if that makes sense there exists an irrational number) . I can see how it works for the irrationals with an involution, (i see it as the strongest functional equation , as one read off the transcendental not just approximate them).

. I hope it does not have that result without symmetry (ie one just put in F(1)=1 and F(0.5)=0.5 which are entailed by but are not the only thing entailed by symmetry (it entails jensens equation so one may not need monotonicity but I am willing to use strict monotonicity- as i want to weaken the convexity condition as much as possible)

have many a proof of my result F(x)=x given symmetry under which conditions did you use- midpoint convexity at 1 and 0, or just symmetry and F(0)=0 and monotonicity/strict or notiont F:[0,1] to [0,1]

dear Miodrag Mateljević, does one have show continuity at end point separately or does this follows given F(1)=1 and F(0)=0 and F is continuous on identical to F(x)=G(x)=x on (0,1) to (0,1) given F strictly monotone increasing, Given that G(1)=1 and G(0)=0. Can one just say then that F is identical to G, or otherwise use G's limiting behaviour for continuity at 1 and zero or does just attaching the endpoint finish the job ie F(x)=x on all x in [0,1]

Between x and y arbitrary there is a rational. Indeed, let x0. By the Archimedean axiom, there is an N such that N(y-x)>1. Therefore

Ny - Nx > 1

Therefore there is an integer M such that

Nx < M < Ny

and hence

x < M/N < y

which was to be proved.

Cardinality, in such issues, is a bit of a red herring, as shown above,