@DanMKatz Well I am not sure if Asaf has thoughts about this to share. But, I do know he hates questions like, where is Mathematics applied and things like that.
Haha well in that case @AsafKaragila, what do you think? I'm going to be heading to grad school soon, and I'm considering research in the area of mathematical logic (I believe you've seen me ask a few questions on this site on the topic :-) ), but I'm a bit nervous about career opportunities. What's your opinion?
@AsafKaragila well, that's possible, but Fremlin says the same thing, and I think Kunen, too, so probably you're right... It's really only this $0$ thing I'm confused about :)
@DanMKatz My opinion is always to go for what you enjoy doing. If you like logic, you should study it. Enjoying what you do helps being good at it, therefore helps in finding a job later on.
@AsafKaragila I couldn't agree more, but I guess my question is whether there ARE any jobs out there in the field...Or am I going to be like the Art History major ;)
@tb Yes. Wouldn't want to come home to find that you left and I couldn't say bye, would I : )
@KannappanSampath What about the second proof you did yesterday? I redid that this morning too and I think it would be good for you if we went over it again.
Not currently, but I can't count on the fact that 5 years from now I'll still want to be a professor. Suppose I were to decide to go back to being a programmer...I suppose I could always leave the PhD off of my resume in that case, right?
And what about the world within academia? I feel that there are not too many professors in the United States who specialize in mathematical logic, which is probably indicative of a lack of demand.
@DanMKatz Well, sure. The point is that if you want an academic life, then you should go for it (job is not the issue) and if you don't want an academic life, then you can go either way.
@DanMKatz More the reason to study it, and do it good. If you'll be awesome perhaps logic will return to the central stage again.
Now, mind you all folks. I have to finish this alg. top. or my degree is going to dissipate faster than spit on the midday of August 1st in the Sahara.
@KannappanSampath They didn't really understand the material either. He didn't really teach us from books either and no CW-complexes too. Then he asked us questions which probably originate in obscure Russian books from the early days of alg. top.
@KannappanSampath Now note that for each point $e_n$ you can find an $\varepsilon_n$ such that there is no other point of the sequence in $B(x_n, \varepsilon_n)$. Why is that?
No... don't you know that I already have a Generalist?? I'm active in other tags which are not set theory and axiom of choice! (although the questions would often have these two tags on them... :-))
@KannappanSampath Because if you have some that are the same you can throw them away. There will only be finitely many same ones because if you have infinitely many same ones you'd have a convergent subsequence (namely, the constant sequence).
@AsafKaragila well, but there's two I really would like to have, and there's a whole lot of work involved: golden FA (ninety answers short, but 780 votes, approximately) and silver topology (twenty answers short)
@AsafKaragila I haven't noticed. At the moment my procrastination on math.SE is strictly confined to chat and an occasional peek if there's an interesting question. But without much luck by that.
My laptop sounds like a broken fridge since yesterday : / Even though I had the fans replaced last year. But of course, the warranty is no longer valid, I'm sure.
@MattN Yes, that's good enough. Take the closed balls of radius $\varepsilon_n$ and pick $\varepsilon_{n+1}$ small enough that it doesn't intersect any of the previous balls and doesn't contain any other element of the sequence
Are you planning to pop by tomorrow? Or are you too busy?
@KannappanSampath Where are you lost?
Mmm. There is a faint smell of freshly ground coffee coming from the kitchen. I really like this smell. Can't wait to make it into a cup of coffee tomorrow morning.
@MattN I don't know, yet. I'll probably be around at some point in the afternoon, but not too long. If all works out well by end of next week, things should be less hurried :)
@MattN I'll wait with that until the relief has kicked in for sure :) There's not so much I can actively do about it. But things look good, otherwise I wouldn't have made this announcement...
@KannappanSampath The function you want to define is only well defined it the balls are disjoint. To make them disjoint you need to say that because the points are all pairwise different you can pick the radii of the balls small enough so that no other points lie in the ball around $x_n$.
@MattN All this is just there in order to avoid using something like Tietze's extension theorem. Otherwise you could just define $f(x_n) = n$, say that it is continuous on the closed subspace $\{x_n\}_{n \in \mathbb{N}}$ and then extend it to a continuous function on all of $X$.
@MattN well, it's what I did in this answer where I showed that a metrizable space is compact if and only if all metrics are bounded, but I was sure that there had to be an explicit way, so I concocted this construction :)
Oh. @tb I wanted to ask whether or not there are natural metric topologies on normed vector spaces which are not the norm-topology (or even such that the norm is discontinuous with respect to them)
@tb I haven't read that post because first I was in retreat from maths because I hadn't managed to help Kannappan without your help and once retreat ended I was busy redoing the two proofs.
But I'm going to read and upvote it in the near future, so your rep is certainly not stationary : )
@AsafKaragila Well, it's a common misconception: the weak$^\ast$ topology on unit ball of the dual of a separable space turns out to be metrizable. But on the entire space it isn't...
@AsafKaragila But if you just want to have a metric topology that isn't the same, take a discontinuous functional $\varphi$ and put $\|x\|' = \|x\| + |\varphi(x)|$.
(they aren't the same because $\varphi$ is continuous with respect to the new norm)
I also started to read "Categories" by Aristotle for my liberal arts self-education, frankly, I don't like it. It's hard to understand what he is saying, but when you finally do understand you realize how non-deep the underlying meaning is.
@Daniil Yes, I had something else in mind. Either way, you forgot that I am Jewish and as such I am certified to answer questions with more questions. However questions are not positive or negative. So I have a way of evading your trap easily.
@AsafKaragila there's a joke about that: An American SQL server returns correnct answers when queried. A Russian SQL server returns angry answers when queried. An Israeli SQL server returns more queries when queried.
the categories is supposed to come first, i know, but i also found it dead boring
the nichomachean ethics contains the most quotable and pithy statements per cubic centimetre, i find
but it's when he uses euclidean geometrical arguments to examine what are fundamentally not geometrical ideas that he really shines, in my book
but, all told, roger penrose makes a very good argument for platonic idealism in the road to reality... so much so that i no longer consider myself an aristotelian
@MattN Thanks for pinging. I am to do the following. Let $(K_n)$ be a decreasing sequence of non-empty compact sets in a metric space $X$. If $diam(K_n) \not\to 0$, then prove that $\cap K_n$ has at least two elements.
I realise from the Complete intersection property, we have it has atleast one element.
Since $K_{n+1} \subset K_n$ we have that $\bigcap_{n=1}^N K_n ) K_N$ and hence $\lim_{n \to \infty} \operatorname{diam}{(K_n)} = \lim_{N \to \infty} \operatorname{diam}{(\bigcap_{n=1}^N K_n)} = \operatorname{diam}{(\lim_{N \to \infty} \bigcap_{n=1}^N K_n)} $ where we can swap the limit with $diam(\cdot)$ because $diam(\cdot)$ is continuous. So we have $\lim_{n \to \infty} \operatorname{diam}{(K_n)} = \operatorname{diam}{(\bigcap K_n)} = diam(\{x\}) = 0$.
@KannappanSampath Well, $\operatorname{diam}{X} := \sup_{x,y \in X} d(x,y)$ and we know that $d(\cdot, \cdot)$ is continuous in each argument and $\sup$ is also continuous (I think!) hence the concatenation of the two is continuous.
I think this will fix your proof @Matt. From $X \to \Bbb R$, define $x \mapsto diam(\{x\})$ is a constant $0$ function and hence continuous. So, swapping limits is justified?
@MattN Thanks for pinging. I am to do the following. Let $(K_n)$ be a decreasing sequence of non-empty compact sets in a metric space $X$. If $diam(K_n) \not\to 0$, then prove that $\cap K_n$ has at least two elements.
