@AsafKaragila I guess my answer is too short :-) or the fact that the other answer was 3 minutes ahead. I got the banner saying that another answer had been entered just before I entered mine, but I guess that banner is not immediate.
@Srivatsan I might have spent more time on the question, but it as it was a numerical question I just put down my immediate thoughts. When I saw that it had been answered, I decided not to go any further.
@robjohn Sure. It does convey the right intuition; but in going from the asymptotic statement to "I would say that...", I am a little uncomfortable. Perhaps you should at least add a small note there.
@robjohn Well, when you close a question as off-topic, you will be given set of options. Either it is just off-topic, or it is off-topic here, but can be moved to one of {stats.SE, meta.math, physics.SE}
@robjohn Doug left a comment; unclear if it was them who downvoted.
This brute-force solution... I appreciate the boldness, but it's pleasantly surprising that no one downvoted it claiming it does not satisfy some unsaid and dubious assumptions.
Alright, I have a bug that's bugging me on MSE. Sometimes I'll be writing when all of a sudden the Chrome tab freezes up. It was alright on an answer (because of draft saves), but not editing a question. How does one overcome or prevent this hurdle?
Unrelated question: if a picture on deviantart has a watermark on it, does that mean the artist doesn't want it being downloaded as original because they want to sell it (and therefore I can't get a clean version)?
If it's just a statement of copyright then it wouldn't be overlayed over the direct center of the image; the positioning means it's sort of a blockade in my mind. Are you familiar with DA?
@Daniil I did not mean that $\{ 0^n 1^n \}_{n \in \mathbb{N}}$ is not regular; it is not regular but not relevant.
Still you could use $0^n 1^n$ for the pumping lemma. Ok, what do I mean? Pick $n$ o be a large finite number (more than the number of states in the finite state machine).
@Daniil I think the proof goes something like this: assume it's a set. Then the union over all its elements is also an ordinal. Do plus one, which is also an ordinal but not in the set. Hence you get a contradiction.
@Daniil Note that you use the assumption that it's a set where you take the union over all its elements. I think you can't do $\bigcup C$ for proper classes $C$.
@RajeshD Because $\{ x \in \mathbb{Q} : x \leqslant \sqrt{2} \}$ does not have a l.u.b. in $\mathbb{Q}$. (It does have a lub in $\mathbb{R}$, namely $\sqrt{2}$.)
In fact, in one sense, the answer to your question is "no". The fact that $\mathbb{Q}$ does not satisfy the l.u.b. property is morally equivalent to the existence of a set $S \subseteq \mathbb{Q}$ such that the l.u.b. of $S$ is irrational.
Mh. My intuition is failing me. I've always thought that the task of recognizing if something is empty or if two things are equal were pretty much the same.
But my notes say that the task of deciding if a context-free language is empty is decidable, while deciding whether two PDAs accept the same language is not.
Ah. Now I see why. CFLs aren't closed by difference!
@BenjaminLim I think my answer got downvoted by Doug Spoonwood, but to make things as rigorous as he wants, you need to do more than the other answers do. They simply claim that $\left(1+\frac1x\right)^x$ is less than $e$ or $99$ of something.
@robjohn Well, isn't that "common knowledge"? Of course, I can belt out a well-rehearsed proof if someone wants me to, but why do we need to prove it at all?
