The whole of theorem $5$ of this paper Some further results on ideal convergence in topological spaces by
Pratulananda Das is unclear to me.
Theorem $\mathbf{5}$. For any sequence $x = (x_n)_{n\in\mathbb N}$ in a hereditarily Lindelöf space $X$, there exists a sequence $y = (y_n)_{n\in\ma...
Answers will probably depend on what $I(C_x)$ means --- perhaps the ideal obtained by adjoining $C_x$ to $I$, or perhaps the ideal obtained by restricting $C_x$ to $I$ --- and therefore depend also on what $C_x$ means. If you supplied those definitions, you might get an answer from someone who can explain the problem but doesn't want to take the time to get the paper and seek out the relevant definitions. — Andreas Blass13 hours ago
@AndreasBlass: OMG, you could have just suggested mentioning those definitions here. But why make the wild guess about $I(C_x)?$ That's ridiculous. And also the people who I know are capable of answering this do not need to go to the paper, they know what those things are and also I've added proper citation. But thanks anyway. — user11849413 hours ago
If you want someone to verify a proof that you wrote, use [proof-verification], otherwise don't; if you want someone to help you with the writing of your proof, use [proof-writing], otherwise don't; if you want someone to explain to you someone else's proof, use [proof-explanation], otherwise don't. — Asaf Karagila13 hours ago
@user118494: I don’t know about Martin Sleziak, but I certainly had to go to the paper to answer one of your earlier question. — Brian M. Scott13 hours ago
@BrianM.Scott: You mean you had to go to the paper to take a look at which one that was ,right? Not that you had to go through it like reading for the first time. That's different from what AndreassBlass means , he/she is not familiar with this topic apparently, judging from his comment. — user11849413 hours ago
@user118494: No, I mean that it was wholly unfamiliar to me. I was seeing it for the first time, and while I have on rare occasions dealt with filter convergence, I had never seen ideal convergence. And I have very, very little doubt that anything along these lines that I can do, Andreas Blass can do at least as well. — Brian M. Scott12 hours ago
@BrianM.Scott: Ok,sorry, my bad. But it still doesn't change the fact that AndreasBals's comment with that wild guess is ridiculous. BTW, I thought you guys already knew everything that I could ask. But anyways, it's just as good. (Off-topic : Do you and Mr. Sleziak know each other in real life?) — user11849412 hours ago
@user118494 A more constructive approach than adding comments why you did not include the definitions would be add them to the question. This helps potential answerers. and it is not much work for you, since you had several questions about $\mathcal I$-cluster points and similar topics recently. You can either copy the definitions between the posts or simply add something link: The definitions of these notions are given in this post (followed by a link). Here I have edit your post to add the definitions (which are definitely not known to everybody), please do check whether my edit was correct. — Martin Sleziak2 hours ago
And, if I may comment on this particular (for the lack of better word) incident, you might consider yourself lucky that users such as Andreas Blass (which is one of the world leading experts in set theory) and Brian M. Scoot (whom I consider one of the best experts on general topology brightest users on this site - judging by his contributions I have seen so far) took interest in your question. So if they are asking for a clarification, it's better simply to explain what's need rather than add comments which might alienate them. — Martin Sleziak2 hours ago
@MartinSleziak: Sir, my comment to AndreasBells was not any excuse to not adding definitions,which is why I thanked him/her, and right after that I had edited my post to add the definitions though, under the heading necessary definitions near the end of the post. I just pointed a finger to his guess,that's all. And your edit is perfectly correct Sir. — user1184942 hours ago
@user118494 My mistake. (The only excuse that I did not notice that you in fact included those definition is that is it 6 A.M. in my timezone. I cannot thing of some more plausible excuse.) Feel free to remove the parts I have added or if you think that they add some useful stuff to the definitions you have already added, then perhaps you can edited them somehow to put them together. (Or, of course, if you are satisfied with the way the post looks at the moment, you can simply leave this version.) — Martin Sleziak2 hours ago
@MartinSleziak: Not at all removing them. As for Mr. Blass and Mr. Scott, if you think my comments have somehow offended(which was absolutely not my intention,may be my wordings somehow and I'm not fluent in this language) them in any way , I'll apologize to them separately.Putting anybody off my posts would be the last thing in my mind. — user1184942 hours ago
@AndreasBlass : Sir, please accept my apology if my comment has , in any way, gave you any kind of bad feeling. That was so not my intention.I in fact edited my question after your comment but unfortunately did not mention that in my comment. — user1184941 hour ago
@BrianM.Scott:Sir, please accept my apology if my comments have , in any way, given you any kind of bad feeling. That was not my intention at all. — user1184941 hour ago
@user118494 I was going to suggest that we could remove the comments starting with my comment which begins with "@user118494 A more constructive approach ..."
However, since you pinged other two users in the meantime, it is probably better to leave the comments there at least until they see them.
If we decide to clean up the comment thread, we can do it later.
In case you come here to the chat room - although I will stayed logged on here, I will probably not at the computer all the time. So I cannot guarantee you that I will respond - at least definitely it might take some time before I respond if you post something here.
@PlasmaVenom just a comment about one of your last questions, is that from my viewpoint the (only and more) important fact is if you will have an advisor. Then he/she will tell you about the best books and articles in analytic number theory and you will do a thesis. In Universidad Autónoma de Madrid there are people working in analytic number theory, also I suppose in hundred of universitites in the world , but the only important key is if you have an advisor. Good luck.
I'm a student of a master program in maths, and I would like to study later in a Ph.D. program. I've read some things in analytic number theory, and have liked.
My question is: In which universities are investigating this theory? (I speak spanish (native), and I practice english everyday).
Any ...
@user243301 No problem. However, posting the same message in Mathematics chat room would not make much difference. As I explained above, the user you want address by the comment would not receive any notification about your message. (Simply because that user was not present in that chat room in recent days.)
But why leave it there if it is irrelevant to the question.
But since you have pinged Brian M. Scott and Andreas Blass, I would probably leave it there so that they can see your response.
That is, unless you decided that you want to remove the comments addressed to them.
Of course, comments can be deleted not only by the user who posted them. Comments can be flagged as obsolete and in that case mods can decide whether they should be deleted or not.
@MartinSleziak : Yes, I agree we have digressed far from the actual problem. My post has been edited and it has now $4$ questions that I'm looking for the answers to. If you could please help whenever you have the time. Thank you.
I'm giving up for today. If I have time, I'll try to look at this problem during the weekend again.
The part I do not see is why this holds:
"The subsequence $x_{k_n}$ has no accumulation point in $L(x)\setminus \mathcal I(C_x)$."
Basically if we can prove $\subseteq L(y)\subseteq \mathcal I(C_x)$ then we are done, since we know that $$\mathcal I(C_y) \subseteq L(y)\subseteq \mathcal I(C_x)=\mathcal I(C_y).$$
I think that can see why $L(y)\subseteq L(x)$.
If we can also show that $L(y)\cap L(x)\setminus \mathcal I(C_x)=\emptyset$, then we get $L(y)\subseteq \mathcal I(C_x)$.
BTW since you are probably new to chat, if you want to get MathJax here in chat rendered, have a look here or here.
I'm a student of a master program in maths, and I would like to study later in a Ph.D. program. I've read some things in analytic number theory, and have liked.
My question is: In which universities are investigating this theory? (I speak spanish (native), and I practice english everyday).
Any ...
