Can you provide an example of unbounded function f:R->R, discontinuous at every point x belongs R, satisfying the following property: f(xn) converges for all real numbers convergent sequence (xn)n?
Hi Dinu Teodorescu,
Maybe I am missing, but I think ihat such function doesn't exist. Suppose it exists. As it is not continuous, there exist two sequences (xn) and (yn) which converge to the same limit, while corresponding sequences f(xn) and f(yn) converge to different limits. The sequense zn=(xn,yn) is converging, but f(zn) is not. It contradicts to the assumption that all sequenses f(zn) are converged. Thus, such function doesn't exist.
I will show, by contradiction, that such a function, say f, does not exist.
Moreover, I will show that the unboundedness assumption is not necessary.
First, I will show that all discontinuities of f are removable and using that and the negation of the theorem regarding the countability of the set of discontinuities of any monotone real function on an interval, I will arrive at a contradiction.
To this end, I assume that such a function f exists.
Let any x0 belongs to R be given and let (xn) and (yn) be any two real sequences converging to x0.
By assumption, the sequences (f(xn)) and (f(yn)) both converge.
If they converge to different limits, then considering the sequence (zn) be defined by z2n-1=xn and z2n=yn, we have that (zn) converges to x0, as the subsequences of odd and even-numbered terms of (zn) both converge to x0, but the sequence (f(zn)) diverges, because its subsequences of odd and even-numbered terms, i.e. the subsequences (f(xn)) and (f(yn)) converge to different limits.
Hence (f(xn)) and (f(yn)) converge to the same limit and since the sequences (xn) and (yn) are any two sequences converging to x0, then for any sequence (xn) converging to x0, the sequence (f(xn)) converges to the same limit.
Then, by the sequential criterion for limits, the limit of f at x0 exists.
Further, by assumption f is discontinuous at x0.
Hence, f has a removable discontinuity at x0.
Since x0 is any real number, we arrive at the conclusion the f has a removable discontinuity at any real number.
Next, for every n belongs to N, the interval [0,1/n] is an uncountable subset of R; then f has removable discontinuities everywhere on [0,1/n], and thus f has uncountably many removable discontinuities on [0,1/n].
Then, since the set of discontinuities of any monotone real function on an interval is countable, f is not monotone on [0,1/n].
Hence there exist xn,yn belong to (0,1/n] such that f(xn)>f(0)>f(yn) (1).
Taking the limit n->oo, the squeeze theorem for sequences implies that both sequences (xn) and (yn) converge to 0.
Then, since the limit of f at 0 exists, the sequences (f(xn)) and (f(yn)) converge to the same limit.
Since the limits of both (f(xn)) and (f(yn)) exist, then by the order limit theorem, (1) implies that lim(f(xn))>= f(0)>= lim(f(yn)), and since the two limits are equal, we arrive at lim(f(xn))= lim(f(yn))=f(0); i.e. lim(f(xn))=f(0) (2).
Next, since f has a removable discontinuity at 0, the limit of f at 0 exists; thus for every real sequence (zn) converging to 0, (f(zn)) converges to the same limit, and by (2), that limit is f(0).
Then, by the sequential criterion for continuity, we conclude that f is continuous at 0, which is a contradiction.
Since we have arrived at a contradiction, the assumption that f exists is false, and thus f does not exist, either bounded or unbounded.
Dear Spiros,
If your proof is valid, then there are no functions f: R->R, everywhere discontinuous, so that f(xn) converges for all convergent sequence (xn)n.
So, consequently, if f: R->R satisfies f(xn) converges for all convergent sequence (xn)n, then f has continuity points.
How many points of continuity does f have and what would they be? Due to your proof, it seems that f must be everywhere continuous !!! Really?
It is known that a real function cannot have removable discontinuities everywhere in R.
A useful discussion along with another proof of that can be found in the following link:
https://math.stackexchange.com/questions/3777/is-there-a-function-with-a-removable-discontinuity-at-every-point
There, a link to the following book https://books.google.gr/books?id=6l_E9OTFaK0C&pg=PA249&redir_esc=y#v=onepage&q&f=false is also given.
So, if the claimed function f has removable discontinuities everywhere in R, then it cannot exist.
By using the sequential criterion for limits, I showed that the claimed function must have removable discontinuities everywhere in R.
Yes, it is known that a real function of one real variable cannot have removable discontinuities everywhere in R( because the set of removable discontinuities is at most countable).
Also, it's obvious that if f has everywhere in R removable discontinuities, then f satisfies the claimed property( transports convergent sequences in convergent sequences).
But, reciprocally, we can not be sure that a function satisfying the claimed property has everywhere in R removable discontinuities!
Let f:R->R defined by f(x)=sinx/x if x belongs to R-{0}, f(0)=1/2. Clearly f has removable discontinuity at the point x=0. If xn=1/n and yn=0 for all n, then we have
f(xn)->1 and f(yn)->1/2.
A function f which can be also considered is f(x)=2x+1 for x rational, f(x)=sinx/x for x irrational.
Regarding your counterexample f:R->R such that f(x)=sinx/x if x is non-zero and f(0)=1/2 (or generally different from 1), if we consider the real sequence (yn) be defined by y2n-1=xn, where (xn) is any real sequence with non-zero terms that converges to 0, and y2n=0, we have that (yn) converges to 0, because its subsequences (y2n-1) and (y2n) both converge to 0, and yet (f(y2n-1)) converges to 1, since f(y2n-1)= sinxn/xn->1 as xn->0, while f(y2n)=f(0)->f(0) which is different from 1; hence the sequence (f(yn)) diverges, even though (yn) converges.
Therefore your counterexample does not possess the desired property “(f(xn)) converges if (xn) converges”.
The same holds for the other counterexample you mention.
By the same way, we can show that if a real function f has a removable discontinuity at x0, then there exists a convergent real sequence (xn) such that (f(xn)) diverges.
This means that if a real function f with the property “(f(xn)) converges if (xn) converges” exists on R, then f cannot have removable discontinuities on R.
On the other hand, the property “(f(xn)) converges if (xn) converges” implies, as I showed in my first answer using the sequential criterion for limits, that the limit of f at any x0 belongs to R exists; so if f has discontinuities, all discontinuities must be removable, which contradicts the previous conclusion.
So, if a function f has the property “(f(xn)) converges if (xn) converges”, then it must be continuous.
The arxiv link that is given above refers to a function that does not map every convergent sequence to a convergent sequence, since it is left continuous with right-hand limits, and the two one-sided limits are different at countably infinitely many points of (1,2].
Dear Spiros,
1. No, I cannot provide such example! It's the reason for wich I asked about such function!
2. Please be cautios relatively to articles appearing on arXiv or to preprints published somewhere on internet. Such papers are not peer reviewed, so it's possible to contain errors and sometimes serious mistakes! You must trust only in papers published in peer reviewed journals.
3. Please don't consider my answers as attacks on you! I'm only careful that the math statements I talk about satisfy the neccessary math rigor.
4. Examples provided by me were given only to show that your approach with sequences is not ok! Both here and in the case of riemannian sums sequences. I'm busy enough, and I haven't had time to prove that, using counterexamples, which are not easy to find. The sequential approach is ok and it works, but your construction "starting from two sequences to one, considering initial sequences as subsequences of new obtained sequence" is not ok. But we must find conclusive counterexamples!
5. Regarding my above question, I presume that function f must have finite right-hand limit, finite left-hand limit( l and L, with l different from L) at every point in R. Maybe right or left continuous! I don't know! Clearly such example is very hard to find! f cannot have infinite right or left-hand limits, because in this case the condition “(f(xn)) converges if (xn) converges” cannot be satisfied.
All the best!
After more than two months, It seems that the counterexamples that Dinu Teodorescu mentiions have lost their way.
Since this interesting question is still open, I would like to restate my point that if a real function has the given property, then it is continuous, and the opposite.
The proof, which is based on the sequential criterion (or characterization or definition) of continuity, is given in my first post and is further elaborated in my third post.
The answer of Victor J. Аdlucky is correct, but the explanation he gives is not correct.
I would also like to note that it is really dissaponting to see a qualified mathematician to cite an irrelevant preprint and then 3 others to recommend that irrelevant preprint.
Here is one more proof of the implication that if a real function f sends Cauchy sequences in R to Cauchy sequences in R, then it is continuous on R.
It is known that if a real function is uniformly continuous on a subset A of R, then it sends Cauchy sequences in A to Cauchy sequences in R. “Cauchy sequences in A” means sequences that are in A and converge in R (not necessarily in A).
Interestingly, if A is bounded, then the opposite holds too; i.e. if a function is defined on a bounded subset A of R and sends Cauchy sequences in A to Cauchy sequences in R, then that function is uniformly continuous on A. For the proof, see, for instance, [1].
We will use the previous proposition to show that if a real function f sends Cauchy sequences in R to Cauchy sequences in R, then it is continuous on R.
Let any x0 belongs to R be given and let the interval [x0-1,x0+1].
We consider the restriction f1 of f to [x0-1,x0+1].
Clearly [x0-1,x0+1] is a bounded subset of R.
Further, if (xn) is any Cauchy sequence in [x0-1,x0+1], then (xn) is a Cauchy sequence in R and thus, by assumption, (f(xn)) is also a Cauchy sequence in R, and since f(xn)= f1(xn) on [x0-1,x0+1], then (f1(xn))) is also a Cauchy sequence in R.
Then, by the previous proposition, f1 is uniformly continuous on [x0-1,x0+1].
Consequently, f1 is continuous on [x0-1,x0+1].
Consequently, f1 is continuous at x0 which is an interior point of [x0-1,x0+1]]; hence f is continuous at x0.
Finally, since x0 is any point of R, then f is continuous on R.
[1] J. A. Fridy, Introductory Analysis: The Theory of Calculus, Second Edition, Gulf Professional Publishing, 2000, pages 59-60.
It is interesting to note that in Fridy’s proof by contradiction, two sequences are combined to form a new one, in the same way as this is done in my previous posts.
Spiros Konstantogiannis
f:(1,100)->R, f(x)=sin(1+1/x). The interval (1,100) is bounded.
f sends Cauchy sequences in Cauchy sequences.
{Indeed, if x(m)-x(p) converges to zero(n,p->infinity), then
/f(x(m))-f(x(p))/R is uniformly continuous, then for all x(n), y(n) in A with x(n)-y(n)->0 it results that f(x(n))-f(y(n))->0] [easy to prove]
Due to THEOREM, f is not uniformly continuous, because x(n)-y(n)->0 and f(x(n))-f(y(n))->-1
Dinu Teodorescu ,
your test sequence 1/(-1+2npi) is NOT in (1,100), which is the domain of your function f; hence your example is NOT a counterexample. Actually, you cannot find a sequence converging to 0 that is in (1,100), since 0 is not a limit point of (1,100). This is elementary.
The restriction in the sequential criterion that the two test sequences (xn) and (yn) must be in the domain of the examined function is obvious; otherwise the sequences (f(xn)) and (f(yn)) are not defined.
Besides, the function sin(1+1/x) on (1,100) is continuous, as the composition of 1+1/x and sinx, which are both continuous on (1,100). Further, the right-hand limit at 1 and the left-hand limit at 100 of sin(1+1/x) both exist in R. Hence sin(1+1/x) can be extended on the closed and bounded interval [1,100], so that the extended function is continuous on [1,100]. Then, since the extended function is continuous on a closed and bounded interval, it is uniformly continuous on that interval. Consequently, the extended function is uniformly continuous on (1,100), since it is a subset of [1,100], and then sin(1+1/x) is also uniformly continuous on (1,100).
CONCLUSION: the function sin(1+1/x) IS uniformly continuous on (1,100).
You should try harder!
PS: Gulf Professional Publishing belongs to Elsevier.
https://www.elsevier.com/books-and-journals/gulf-professional-publishing
Spiros Konstantogiannis
Yes, you have right, my counterexample is wrong! Sorry!
I was surprised by the result from Fridy's book: "the opposite holds too; i.e. if a function is defined on a bounded subset A of R and sends Cauchy sequences in A to Cauchy sequences in R, then that function is uniformly continuous on A"
The most important result regarding uniform continuity is : the continuity on a compact set C implies uniform continuity on C. The proof, by contradiction also.
So the result from Fridy's book is surprising because just the boundedness is enough, for obtaining uniform continuity. Please send me the proof of Fridy, if you can!
However, I still suspect that the result is wrong. I didn't see it anywhere! How the closedness of set A is tricked!
The great books and monographs on Mathematical Analysis for real functions do not contain this result!
However, it is a challenge for you and me, to be cautios and to try obtain a counterexample. A counterexample for a result appeared in a book edited by Elsevier can be published anywhere in a high quality journal!
Dinu, it was surprising for me too, since I haven’t found it anywhere else yet, and I have searched a lot.
Fridy’s proves the lemma by contradiction, using the negation of the definition of uniform continuity, then the Bolzano-Weierstrass theorem (to take advantage of the domain being bounded) and then he combines two sequences to form a new one to eventually reach a contradiction.
I have attached the lemma and the proof in a pdf I created from the two pages containing the statement and the proof of the lemma.
You are welcome, Dinu. I will read your remarks with interest.
Dear Spiros,
I consider the proof from Fridy's book being clearly valid. Consequently, your proof is perfectly valid. So you proved that: If f:R->R transports all convergent sequence in a convergent sequence, then f is continuous on R.
The result is remarkable! Thus the asked example cannot be given and all is clear now regarding the above question.
Many thanks for your contribution!
Dear Dinu,
thank you very much. I’m really glad to see that you are strict but fair concerning mathematics.
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