@MarianoSuárez-Alvarez and i appreciate that you are trying to help him...but he still needs to understand what's in that book, if for no other reason than communication with his instructor

@DavidWheeler Honestly my instructor likes to sweep a lot of details under the rug

whitehead invented cw-complexes, so he should now :-)

But that is irrelevant for this conversation, mostly

I am not interested in getting him to talk to his instructor

but to understand cw-complexes and friends

@MarianoSuárez-Alvarez Is there a type of complex that is a little more flexible than simplicial complexes, but is still very combinatorial? It would be nice if it handled convex polytopes (so not just triangular faces and tetrahedral solids, but also squares, pentagons, etc.).

exactly. I have seen my instructor many times but it is not that helpful.

I am pretty sure anyone with a sensible understanding of cw-complexes can then read hatcher's version and understand it

@JackSchmidt, regular cw-complexes are close to that

Hahahahahaha the proof of hatcher prop 1.26 is so handwavy :D :D :D

@JackSchmidt, they are cw-complexes whose attaching maps are homeos onto their images

more flexible than what you described, yet not too much

I remember someone using "polyhedral complexes", obtained by attaching polyhedra... but not with topology in mind

@MarianoSuárez-Alvarez switzer was it?

@MarianoSuárez-Alvarez That is definitely an improvement, but I guess I like homomorphisms of (discrete) graphs better than homeomorphisms of the topological realizations of graphs. In the former you say where a finite list of vertices go, and then the edges follow. In the latter, you have also sorts of flexibility in how the attaching maps connect. Most of the flexibility is not real ("simplicial approximation" at some point has a "straightening lemma"), but it still confuses me.

@BenjaLim, yes

Robert

a lot of category theory I see.

@BenjaLim Is a torus defined to be $S^1\times S^1$?

the category theory is just language

not really defined

you could motly take it off

for instance: $f:[0,1] \to [0,1] : x \mapsto x$ and $g:[0,1] \to [0,1] : x \mapsto x^2$. $f$ and $g$ both attach an edge to an edge, but they do it differently. The difference doesn't matter (and mostly cannot matter), so why not skip to the chase and say $f$ and $g$ takes 0 to 0 and 1 to 1.

but yes you can identify it with $S^1 \times S^1$

@PeterTamaroff that is "a" torus

saying "lots of category theory" when looking at that book is close to saying "lots of ink" :-)

and I don't know how to skip to the chase for open disks, because they don't have vertices :-)

@DavidWheeler Well, but we can regard it as the torus, can't we?

@JackSchmidt put some in?

@BenjaLim How do you define a torus?

you tell me

@PeterTamaroff it is what is usually meant by "the torus", yes

in topology definitions are ("flexible")

what's the difference between a circle and an ellipse? both are closed loops in the plane, but the ellipse is known for its odd habits.

or you could use another "standard definition" of $\Bbb R^2/\sim$ where (x,y) ~ (x',y') iff (x-x',y-y') is in $\Bbb N^2$

@DavidWheeler Well, I'm asked to prove that =P

@PeterTamaroff Are you trying to compute the fundamental group?

@MarianoSuárez-Alvarez You know honestly I am trying very hard at AT.....

@BenjaLim What? Dude, I'm just reading some basic stuff about topology...

You always insists on the more advanced stuff....

@PeterTamaroff the basic recipe is this, label a square on the right and left sides with an arrow a pointing up, and the top and bottom sides with an arrow b pointing right. sew the edges together so that the letters and arrow directions match.

the first "identification" gives you a cylindrical tube, and then you identify the top and bottom of the tube

@PeterTamaroff In algebraic topology we deal with stuff like that all night long

or you could use a parameterization using s,t and trig functions

@BenjaLim Kudos.

@PeterTamaroff more complicated and messed up than that...

one of the most beautiful pieces of music i've ever heard: this

o.o i kilt chat

@DavidWheeler hehe no, went to dine

@MarianoSuárez-Alvarez I understand why the cells when attached to the subspace are disjoint

@JasperLoy i like a lot of different stuff

It's because we do $A \sqcup D_\alpha^n$!!!!!!!!

@JasperLoy Why would it matter?

listening to a killer version of lili marleen by theo bleckmann atm

@DavidWheeler listen to quarter tone pices by charles ivess

@DavidWheeler You said you were "old" before. What is old for you?

old as in...?

i've always thought drummers might appreciate the theory of the integers modulo n

what syncopation essentially is, is picking out the units in the group of beats modulo a rhythm cycle

@DavidWheeler As in age, I suppose you meant, I don't know.

@DavidWheeler hhehehe ok. I recently learned how to play a pretty hard song in the guitar.

basically the rhythm was hard, and fast.

@DavidWheeler Here. THe hard part is 0:30 and you'll notice it ends when it repeats the intro.

The intro is actually pretty easy.

my age is 51 years

well that is very nice, if you can play that yourself

@DavidWheeler if you listen closely the lower strings and the higher strings make the exact same thing but out of "phase"

i mean the exact same "rhythm"

it sounds like 2-hand fretwork

@DavidWheeler it is one

@DavidWheeler can you read music? (I can't but I have a music sheet for that tune)

do you know what i mean? where you use your right hand to hammer-on at the same time you do the same thing with the left hand

@DavidWheeler yes, how else would i be playing that tune ? =P but the tappings are never simultaneous there.

it is kind of a galloping

Suppose $|$ are half notes and $_|$ are fourth notes

yes...alternating

Then in $4/4$ the rythm is

$\bf l$ $|$ $_|$ $_|$ $|$ $_|$ $_|$ $|$ $_|$ $_|$ $|$ $_|$ $_|$ $\bf l$

sort of like taking Z3 x Z4, hmm?

@DavidWheeler $\mathbb Z_3\times \mathbb Z_4$??

Z4 is a cyclic group of order 4

if you "subdivide" the cycles of 4 into 3rds, you get a cycle of order 12

@DavidWheeler Oh, OK.

@DavidWheeler and how do you note the "fast" fourths versus the slow halfs?

well, actual musical rhythm is very complicated, because you have "addition" (how many "bars"), and "subdivision" (if you are in 4/4 time you have "4 beats per bar", but these time-intervals can be whole notes, half-notes, or quarter, eighth, sixteenth in any combination that adds up to 4)

one thing musicians often do is set up a rhythm like 1/2/3/4, and then "drop out beats" (like in reggae) so you get 1/-/3/-

and then, you might hit the beat a little late on the "3": 1/-/..3/-

sort of "borrowing space" from the 4th beat that isn't there

@KannappanSampath Chapter 7 is kind of long. Collins's Representations and Characters of Finite Groups is broken up into smaller pieces. Collins 2.4 is the first part of Isaacs 7 on Frobenius groups, 2.6 is character theory for Frobenius groups. 2.7, 2.9, 2.10 are examples similar to the dihedral centralizer in Isaacs 7. Collins 3 and 4 finish out Isaacs 7, but also include the original motivations. I found Isaacs 7 after Brauer-Suzuki to be pretty unmotivated, but Collins 3 and 4 are great.

@JackSchmidt Are you there?

@DavidWheeler Or maaaaaybe Wheeler?

@anon Hey.

hey

@anon I have aproblem with a proof. Prolly set theoretic.

In general it is the case $f f^{-1}(A)\neq A$.

you sure?

@anon I get my theory mixed up sometimes.

I'll rewrite.

one can increase the domain/codomain while keeping the equality true if desired

for any particular A

hold Q on mainsite

nevermind

go on

@PeterTamaroff $f^{-1}f(A) \neq A$ in general

@DavidWheeler Yes, I'm saying that.

David just reversed the order.

but $ff^{-1}(A) = A$

no, not always

@anon Yes, that's my point.

Only if $f$ is onto.

if A contains any element whose corresponding singleton has empty preimage, then that element will not be in $f f^{-1}(A)$

@anon (i.e. $f$ is not onto)

@PeterTamaroff only if the restriction of $f$ to domain $f^{-1}(A)$ and codomain A

@anon, yes, i see what you're saying...

@anon According to my definition onto means that for $f:X\to Y$, we have $f(X)=Y$.

as i was going to say, consider $f(x) = x^2$

I know what onto means.

@anon I know you know.

See there.

then if A = {2}, then f(A) = {4}, which has pre-image {-2,2}

The obvious function f:{1}->{1} will obtain the equality with A={1}. It will still retain the equality if we append {2} to the domain but not the codomain, as a result A={1} will still obtain the equality but f will not be onto.

Note the author writes $f[f^{-1}(U)]=U$

in that image you posted, it seems like to be rigorous, we should really consider U in the relative topology on f(X).

then the problem evaporates, and doing so doesn't really change the "flavor" of the kind of open sets we have.

hrmm

I don't want to make such a silly question on main =P

@BrianMScott is here to save the day.

@anon Could you ping me when I leave the room?I wanna try something.

are you planning on leaving now, or later sometime?

@anon ping me when I leave

@PeterTamaroff ping

@anon Didn't seem to work "offline":

the ping takes some time to get to the mainsite inbox

@Peter: Did you get your question taken care of?

@BrianM.Scott Kind of. I had one, and I could work it out. I had another, which seems pretty trivial.

Which is?

@BrianM.Scott 1 sec.

"Hence, since $e$ is a homeomorphism, it is clear that $X$ has the weak topology induced by the functions $\pi_\alpha e=f_\alpha$."

You can click on the image to make it large.

Since $e$ is a homeomorphism, each $f_\alpha$ is continuous. Then each, $f_\alpha^{-1}(O)$ is open in $X$ with $O$ open in $X_\alpha$

Okay.

Now $X$ has the weak topology induced by $f_\alpha$ iff the sets $f_\alpha^{-1}(O)$ are a subbase, $\alpha \in A$.

It is clear they union up to $X$.

But can't $X$ have some other topology containing this weak topology?

Oh wait.

The homeomorphism works both ways.

@PeterTamaroff It does indeed.

@BrianM.Scott I just "showed" the weak topology is contained.

Now I want to show the other "inclusion".

@PeterTamaroff You showed that the weak topology is contained in $\tau(X)$, yes.

@BrianM.Scott Now, let $O\subset \tau(X)$.

I want to show it is contained in the weak topology to conclude.

@PeterTamaroff My first thought is to let $U$ be an open set in $X$ and $x\in U$, and try to show that for some finite $\{\alpha_1,\dots,\alpha_n\}\subseteq A$ there are open $U_\alpha$ in $X_\alpha$ such that $x\in\bigcap_{k=1}^nf_{\alpha_k}^{-1}[U_{\alpha_k}]\subseteq U$.

@BrianM.Scott Doesn't a $e$ being a homeomorphism produce a bijection between the opensets of $X$ and those of $\prod X_\alpha$?

Not quite: it’s between those of $X$ and those of $e[X]$, which needn’t be all of the product.

@BrianM.Scott Maybe this helps.

It is left as an exercise there that $e(X)$ has the weak topology induced by $\pi_\alpha \mid e(X)$. I.e. as a subspace of $\prod X_\alpha$

@PeterTamaroff Which is pretty straightforward, I think.

@BrianM.Scott OK. But the

Seems to be based on that.

$e$ is a homeomorphism of $X$ with $e(X)$

And $e$ has the topology induced by $\pi_\alpha {\bf i}:e(X)\to X_\alpha$

@PeterTamaroff Sure: $e[X]$ has the weak top. induced by the $f_\alpha\upharpoonright e[X]$, and $e$ is a homeomorphism, so you just use $e$ to ‘pull back’ the topology.

@BrianM.Scott $\pi_\alpha \mid e(X) =f_\alpha \mid e(X)$?

OK.

I have that $\pi_\alpha e=f_\alpha$

@BrianM.Scott And why really don't care about what happens outside $e(X)$, right? (I mean, in $\prod X_\alpha \setminus e(X)$

@PeterTamaroff Oops; I meant the top. induced by the $\pi_\alpha\upharpoonright e[X]$.

@PeterTamaroff No, we don’t.

Hang on a minute now while I write something.

@BrianM.Scott OK.

Given $U$ open in $X_\alpha$, $e[X]\cap\pi_\alpha^{-1}[U]$ is a subbasic open set in $e[X]$. Since $e$ is a homeom. $e^{-1}\left[e[X]\cap\pi_\alpha^{-1}[U]\right]$ is a subbasic open set in $X$. But it’s equal to $f_\alpha^{-1}[U]$.

Thus, the top. of $X$ is the weak top. induced by the $f_\alpha[U]$.

@BrianM.Scott =D So neat.

Sometimes I tangle myself up in theory before looking at the basics.

@BrianM.Scott Can I ask another question?

@PeterTamaroff A lot of basic theorems about products and maps are pretty straightforward if you can just keep track of where everything is; the details look messier than they really are.

@PeterTamaroff Sure.

@BrianM.Scott Yes, that's very true.

He says $ff^{-1}(O)=O$.

Okay.

Shouldn't that happen iff $f$ is onto?

@BrianM.Scott I mean, I was expecting that to be part of the proof.

You mean iff every element of $O$ has nonempty preimage, which is not equivalent to $f$ being onto, like I said earlier.

But I missing something.

He’s assuming that $f$ is onto. Alternatively, he should say that the topology on $f[X]$ is the quotient topology. (There are a few bugs in Willard.)

tsk tsk!

@anon What does that mean?

It means I'm wagging my finger at Willard.

@BrianM.Scott Where should I realize he's assuming that?

@PeterTamaroff It’s what’s needed to make the argument work.

He really should have stated it; that he didn’t is one of those bugs that I mentioned.

@BrianM.Scott Yes. But an open continuous map need not be onto, right?

@BrianM.Scott Oh, OK.

@PeterTamaroff No it needn’t, though its range does have to be an open set in the codomain.

@BrianM.Scott Because one induces a quotient topology on $X$ by $f:Y\to X$ given $f$ is onto, right?

@anon That is like a tongue clicking sound right? =P

@BrianM.Scott I might be asking too much but I think there is another, maybe...

@PeterTamaroff Yes. If $f$ isn’t onto, you induce a topology only on $f[Y]$.

http://www.youtube.com/watch?v=-imfjwhDM0A

Is that an English-only thing? :P

@anon No no. It's what I said.

@PeterTamaroff Okay; what about it?

@BrianM.Scott The weak topology is induced by a subbasis, correct?

@PeterTamaroff Every base is also a subbase.

@BrianM.Scott Yes. Well, just checking. But then why not say subbase? What are we losing?

@PeterTamaroff In Theorem 8.14 you’d be losing the information that if the functions separate points, the sets $f_\alpha^{-1}[V]$ actually form a base for the topology. That’s much nicer to work with than a random subbase.

@BrianM.Scott The only extra condition I recall is that the intersection of two basic sets contain a basic set, right?

That means open sets are unions of basic open sets.

Which would the "dope" thing about basic versus subbasic?

@PeterTamaroff $\mathscr{B}$ is a base for the topology $\mathscr{T}$ iff for each $U\in\mathscr{T}$ there is a family $\mathscr{B}_U\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{B}_U$.

If all you have is a subbase, all you can say is that each open $U$ is the union of some family of finite intersections of subbasic sets. That’s much messier to work with.

@BrianM.Scott Yes. Well, there are two equivalent formulations IIRC. I think that one though, is pretty enlightening.

@BrianM.Scott OK. I got your paper on inhomogenous spaces. Hope I can read it one day.

@PeterTamaroff It may be a while yet, simply because you’ve not seen that kind of argument, but it’s not actually terribly hard or tricky.

@BrianM.Scott Brian, the Klein bottle has how many dimensions?

It’s a surface: it’s locally $\Bbb R^2$.

@BrianM.Scott Willard says "it can't be faithfully represented in 3 dimensional space without self intersection"

@PeterTamaroff Perfectly true: it can’t be embedded in $\Bbb R^3$ without self-intersection.

A Möbius strip is also locally $\Bbb R^2$, and while it can be faithfully embedded in $\Bbb R^3$, it can’t be in $\Bbb R^2$. The Klein bottle is just a little nastier even than the Möbius strip.

@BrianM.Scott Aha. He says the Möbius strip has interesting properties that cannot be proved without combinatorial or algebraic theory. What is he referring to?

@PeterTamaroff I don’t know exactly what he had in mind there; that’s very much not my kind of topology.

@BrianM.Scott Oh, OK. What topology do you "do"?

@PeterTamaroff What’s usually called general or point-set topology, and set-theoretic topology. Not geometric or algebraic topology.

@BrianM.Scott The kind of topology I like, Brian :)

Or maybe I should say "am better at than algebraic topology"

@FortuonPaendrag Too much algebra. Not enough set theory. But what you like is undeniably more mainstream nowadays than what I like.

The kinds of arguments I like are compactness, connectedness, finiteness. I delight in the Lindelof property

@BrianM.Scott I think he is talking about General Topology.

@PeterTamaroff Yes.

I'm off to bed for the day. Thanks for the talk.

Do you deal with Lindelof spaces, @Brian?

@FortuonPaendrag Sometimes; they’re a fairly basic class of spaces.

Ah, my topology class last semester discussed it.

@FortuonPaendrag Probably either right after discussing compactness, or when showing that in metric spaces separability, second countability, and the Lindelöf property are equivalent.

Yeah, your guess shows that it is fairly standard. We began with problems like showing $\mathbb R^n$ is lindelof and stuff like that. My professor was Agnes Szilard.

Do you know her?

’Fraid not.

Ah, she was my professor in Budapest. I suppose the topology world is wide :)

I just happen to come back now. Agnes Szilard taught me complex analysis when I was at Budapest!

and I took number theory from Balog, which had a lot to do with how I got to the number theory I do now

@FortuonPaendrag Not all that wide, actually, but judging by the title of her PhD dissertation, she and I wouldn’t have all that much in common. (Isn’t she really Ágnes Szilárd? :-)

@mixedmath YAY! BSM for the win!

BSM=Budapest semesters in mathematics

@mixedmath Don’t know Balog, but there’s a very good set-theoretic topologist named Zoltan Balogh.

Antal Balog is best known for contributing to the Balog-Szemeredi-Gowers theorem of additive number theory

@BrianM.Scott Yes, my apologies, she IS Ágnes Szilárd. Hungarians have so many letters though!

which was really slick for a while, until Gowers revisited it and blew the result out of the water

and then, not to be outdone, I think Terry Tao improved upon the result again

@FortuonPaendrag did you pick up much Hungarian while you were there?

for that matter, did you take C&P?

Nem. Nem Beszelek Magyarul is about 30% of my total hungarian knowledge. I learnt all my letters reading roadsigns. I did! Did you too? @mixedmath

when I was there, I tried to pick up as much as I could, but I can't really say anything now. In the optional pre-semester language program, we learned a song called Borond Odon (every vowel double-dotted), and I happen to remember that song

not so much, I guess

@FortuonPaendrag Surely you must have picked up kérem and köszönöm. Oh, and eszpresszo! (I was there for a week and a half back around 1980, I think, for a topology conference.)

Haha. Of course. The metro would always say "Kerem vigyazzanak" with assorted dottings over the letters.

Anyway, Goodnight folks! And to be pseudo-hungarian, Viszontlatasra! :)

@FortuonPaendrag G’night!

goodnight!

In the above proof, Ruding is assuming that since ${p_{n}}$ is converging to $q$, $q \in E^{*}$

But this may not be true, since we do not know whether $E^{*}$ is complete.

What does it matter?

$E^*$ is defined as the set of all limits subsequences of the $\{p_n\}$...

@anon we have to to prove two things here. One the limit point exists, and two it is in $E^{*}$

The first is proved using the $\delta$ argument.

Wait, I will read again and then ask if I have a problem. Think I just got it.

You agree that $\{p_{n_i}\}$ converges to $q$, right?

@anon yup

Meaning $q$ is the limit of a subsequence of the $p_n$'s, which by definition means it's $\in E^*$

@anon yeah, right. Got it when I was typing it out.

By the way, doesn't this fact that we can take the limit points of the sequences and take its limit, depend on Zorn's Lemma?

It looks just like induction; the infinite stuff seems like it should come for free by hypothesis. (Then again, I don't have experience thinking about these questions at all.)

Okay

wait, hmm

it's tricky

yup, I am here.

still seems like all of the infinitary choices needed come from hypothesis

@anon yup, axiom of choice or zorm's lemma whichever form is more convenient.

I mean you don't have to use them, because the existence of $x\in E^*$ and the $n_i$ are obtained from the hypotheses alone, without needing to make more than a single choice at each step.

Hmm. Right.

$^{131}\text{I}$ does not sound like a healthy snack.

@JonasTeuwen whaa?

Whaa!

Okay, got it. I write it with the superscript on the right.

So, you had some kind of scan taken?

@JonasTeuwen what is your field of research?

@JayeshBadwaik Harmonic analysis and mathematical physics (kinda).

@JasperLoy Nope.

Hi folks

Hi.

Now then. Time for half an hour break before I'm off back to the school counselor to pick up the letter for the extension.

Hello.

@Matt Hi Matt

Crap, forgot to badger the teddy about my Fourier question.

He said something and I wanted to investigate.

Then he popped up and I completely forgot about it.

I think I'll have to write a stickie note or something. Otherwise I'll forget again next time he pops up.

@Matt Its very inconvenient of these people not to be around when they are needed :)

@OldJohn Oh, no no: that's not at all what I mean. He's always there when I need him. At least that's the way it feels. : )

@Matt Its OK - I was joking

Ah good. : )

@OldJohn Btw, I was being deliberately obtuse when I wrote "would you prefer "impostor"" : ) Just for the record.

Of course native speakers know all the words : )

@Matt Yep - I assumed so - no problem :)

I really don't like sun. And today is one sunny day. Feels and looks like 28 degrees. Just burning it down.

Not having been here very long, I hadn't identified who "teddy bear" actually was

@OldJohn Here, in case you want to know what he looks like in real life.

@Matt Lovely!

@Matt What did he say?

@Matt Feeling a bit blue, that one...

@JonasTeuwen Don't know yet.

@J.M. No, it's grumpy bear : )

@Matt What is the question!

I-131 does not sound like a good snack. Does it?

No.

Son of a crack. I am getting trolled by doctors.

Troll back.

One looked at my like this when I asked: so what now? while he was saying "I don't know".

@Matt I'd like to, but...

@JonasTeuwen Here in case you were referring to my question.

@JonasTeuwen But?

Oops, got to run. See you later!

@Matt I know the answer.

See you.

@Matt Let me type something up.

@JonasTeuwen Took me a while to see what you meant. I'm more accustomed to it as $\require{mhchem}\ce{^{131}_{53}I}$

Don't worry, as long as you don't have to have it everyday...

For the convergence of Fourier series you also have to mention which mode of convergence (e.g. in which topology/norm) as you know. Pointwise is probably what you want, and there you have examples of $L^1$ functions which diverge everywhere (due to Kolmogorov) for example. Convergence in norm, especially $L^2$ will be the one you are interested in is an easy exercise which you have done before.

@J.M. Right. I don't know why they want to do it actually. TSH is low, but not ridiculously low.

Probably in the fashion of "I don't know so let us try something else!".

And these radically leftish "nonconventional medicine" claim Candida causes low body temp! But I have not really found a real reference.

@JonasTeuwen Nah, that's bollox.

And I don't really fancy getting hit by Ampho too often 8-).

@J.M. I figured so already, yes.

Maybe complete systemic infections do? But then I would not be typing.

@JonasTeuwen Well, it does a number on your liver and heart, so you'd want to avoid having to use that...

@J.M. Yep. But! I seem to be quite resistant to other stuff. For some very odd reason which cannot be explained yet!

@JonasTeuwen They use that radio-iodine in two ways: diagnostics and actually nuking the thyroid. For diagnostics, the radiation dosage is about the same as what you'll get from an X-ray, so not much to fear there.

@J.M. Oh, right. Yes, it is for diagnostic use first.

Well. If nothing can be found it means I am even more crazy than initially thought! 8-).

So test is negative I am like oh good feel much better. But after a couple of hours... that fades away.

Not necessarily crazy. Just that the body's an even weirder place than was previously thought. :)

Natural hypothermia!

Yes. The immunologist seems to think it is quite funny.

So do I now.

...for all we know the immunologist is preparing a new paper based on his observations...

I would like to be included in the statistical analysis. I would like to have my name on such a paper 8-).

"Atypical Fungal Growth in Young Belgians."

Yep. Very atypical. He stated there are only case reports as of now. But mine does not really fit the pattern. What the...

"Belgians and Shrooms". Yep.

Anyway, off to the gym. Perhaps exercise helps! 8-). Have a nice day @JM (and the others).

See you later!

@JonasTeuwen Later!

Hey, not-so-young John!

@J.M. Hi there

@JonasTeuwen hey

Can I ask you something?

Is Hardy's A Course of Pure Mathematics could be a reference text(Supplemented to Courant's or Spivak's) for calculus prior to analysis ?

@XingdongZuo Hardy's Pure maths is very old-fashioned now - but still interesting. I used it 45 years ago

I prefer modern treatments for the first course in calculus, especially those which have a lot of computational motivation.

@OldJohn Since the purpose is to remain 2-3 excellent books in Calculus for future self-reading after passed this course, will Hardy's fits it ? Now as I could think the very good ones including Courant's and Spivak's highly reputed

@XingdongZuo I would think that Hardy is too old-fashioned for that purpose really

@OldJohn Even though it revised to 10th edition in 1952 ? In addition, if I remember correctly, Courant's Intro to Calculus and Analysis had the newest revision in 1965.

@XingdongZuo So that makes Hardy's book 60 years old - a lot of ideas have changed about teaching calculus in that time