Mathematics - 2012-07-20 (page 3 of 5)

7:11 AM

If I correctly read it, you don't even need to worry about the converse, since your proof is a string of biconditionals.

Peter Tamaroff

@CLarue Are you around?

CLarue

@PeterTamaroff Referring to location?

Peter Tamaroff

@CLarue No, I mean if you were online here. You are.

:)

CLarue

aha

Peter Tamaroff

I have to prove that $(1)\Rightarrow (2)$. I do it by proving $(1)\Rightarrow (3)\Rightarrow(2)$

Let $F\subset X$ be closed.

CLarue

7:23 AM

It is my inclination that you do not have to involve $(3)$ here.

robjohn

Can anyone see why this was downvoted?

Peter Tamaroff

@CLarue Let $(1)$ be true. Then $(3)$ is true, so $f$ and $g$ are continuous. Let $F\subset X$ be closed. Then $f(F)=g^{-1}(F)$ is closed iff $F$ is, by $(3)$.

@robjohn No idea.

CLarue

@PeterTamaroff I think that is correct, but a bit long winded. :)

You know that for each open set $O$ there is a unique closed set $F=O^C$

Peter Tamaroff

@CLarue And $(3)\Rightarrow (2)$ follows by $F=C(O)$.

Old John

@robjohn No - can't see why it would be downvoted

Peter Tamaroff

7:26 AM

@CLarue I could have used that before too.

sobri

hi! i've got a line slope problem (the slope is always positive, even when reversing endpoints). is this a sensible place to ask for help?

CLarue

So if you introduce $(1)$, $F(O^C)$ is the complement of $F(O)$, as you said, but $O$ is open if and only if $F(O)$ is open by hypothesis.

Peter Tamaroff

@CLarue So I could just say $(1)\iff (2)$ by ($F $ is closed $\iff$ $F^c$ is open.)

CLarue

I believe so, though it might require a bit more explanation than that. I was focusing on $(1)\Longleftrightarrow(2)$.

Peter Tamaroff

@CLarue Oh, whoops yes.

sobri

7:29 AM

omg. i'm an idiot. i shall retract that question, and walk away in shame. sigh

Peter Tamaroff

@CLarue I have another denumeration in my notes.

robjohn

@PeterTamaroff: @OldJohn: Thanks. I thought there was something obvious I was missing.

Old John

@robjohn it would be nice if downvoters left a comment to say why

Jonas Teuwen

Hi hi!

Old John

@JonasTeuwen Hi Jonas

robjohn

7:37 AM

@OldJohn I have said the same thing many times. However, that would remove the anonymity of the voting.

Old John

@robjohn probably, yes

Do you want to drop me an email about that diophantine stuff?

Matt N.

wheee I got some teddy yesterday : ) : )

robjohn

@OldJohn The Mordell book?

Matt N.

@JonasTeuwen Hi bro : )

Old John

@robjohn yep

Jonas Teuwen

7:41 AM

@MattN. Hi! :-). @OldJohn Hi! :-). @robjohn Hi :-).

Matt N.

Hi robjohn : ) Hi Old John : )

Hey @JonasTeuwen I have a question.

Jonas Teuwen

@MattN. So do I! Shout!

Old John

@MattN. Good morning

robjohn

@MattN. hey there!

Matt N.

I missed the last two weeks of my CA lecture. Then I thought that it followed the book 1:1. Now I realised that he started doing all the weird stuff that is not in the book right after I left. The guy I'd copied off the notes til now seems to be away since he hasn't replied to the email I sent him yesterday.
Now I'm thinking about contacting the lecturer and asking him to allow me to copy. Is that a bad idea?

I'm worried that he might ignore me and then ask only stuff about the last two weeks during the exam.

@JonasTeuwen Well, shoot! What is it?

Jonas Teuwen

7:45 AM

@MattN. It is a perfect idea.

@MattN. The classes are not compulsory and it is your own business. If he does not have notes, he can at least tell you what it was about. Otherwise he is asshole² and I don't think he is.

Matt N.

@JonasTeuwen So you don't think he'll think "What a stupid idiot, I'll fail him"?

Maybe you are right and I should email him.

Jonas Teuwen

@MattN. Nope.

@MattN. Happens all the time students don't show up. If they take it personally they have quite some Krank Im Kopf eh.

Matt N.

I'll do it tonight to give my class mate some more hours to reply.

Jonas Teuwen

Just say that "due to circumstances" you are were not able to attend but you would like to know what it was about. If he even would have notes, that would be perfect.

There. If he ignores that he is an Arschloch.

Matt N.

Ok.

But you know, knowing that he is an Arschloch (if he is one) won't help me pass the exam.

: )

Jonas Teuwen

7:49 AM

He will not ask as it can be that say someone very dear to you died for example.

It is phrased as: NOYB.

Matt N.

Ok.

Man, I so want to pass these exams.

I think if I didn't I'd feel so much like an idiot that I won't do any papers at all, even stupid ones with you.

Jonas Teuwen

You can. Do your best.

Matt N.

You did not watch The Rock, I take it.

Jonas Teuwen

I do not write stupid papers 8-).

Matt N.

: )

Jonas Teuwen

7:51 AM

But maybe papers with not-so-big results, but almost everybody does that.

Need to go! Bye!

Matt N.

See you later!

@JonasTeuwen See above for The Rock quote about "doing your best" : )

I'm going too.

Jonas Teuwen

8:33 AM

@MattN. :D.

Rajesh D

@Jonas : What is the joke in it explicitly?

robjohn

@JonasTeuwen Hi there. I was afk and iaw when you said hi :-)

Jonas Teuwen

:-).

@RajeshD Third line?

Rajesh D

then what is said after that

is she a girl friend of Stanely?

Why does he cock his gun?

Frank Science

10:23 AM

I'm so tired.

robjohn

10:42 AM

@FrankScience sleep?

Former_Math_Addict

Math requires alertness, so sleep.

user19161

@Former_Math_Addict Well said!

Former_Math_Addict

Thanks ;-)

robjohn

@Kannappan: Suppose that
$$
1-\frac1{n^{1/3}}\lt\frac{\phi(n)}{n}\lt1-\frac1n
$$
If $n=p$, then
$$
\begin{align}
\frac{\phi(n)}{n}
&=1-\frac1p\\
&=1-\frac1n
\end{align}
$$
If $n=\prod\limits_{j=1}^kp_j^{e_j}$ where $\sum\limits_{j=1}^ke_j\ge3$, then, because $\min(p|n)\le n^{1/3}$,
$$
\begin{align}
\frac{\phi(n)}{n}
&=\prod_{p|n}\left(1-\frac1p\right)\\
&\le1-\frac1{\min(p|n)}\\
&\le1-\frac1{n^{1/3}}
\end{align}
$$
Therefore, $n=\prod\limits_{j=1}^kp_j^{e_j}$ where $\sum\limits_{j=1}^ke_j=2$

Gigili

Pft, why should I see this name everywhere ...

robjohn

10:50 AM

@Gigili what name?

Gigili

That name starting with 'K'.

user19161

@Gigili I think you should try to make up with him.

Gigili

I've got a chill!

@JasperLoy I'm not going to do such a thing, as I told you before.

user19161

@Gigili Too much hatred leads to a chill. QED.

robjohn

@Gigili what happened? I have noticed a problem, but I don't know what is the cause.

Former_Math_Addict

10:55 AM

@robjohn I think there is a "bug" in your post to Kannappan above, the down scroll arrow doesn't work on the right bottom corner :(

robjohn

@robjohn problem? :-)

user19161

@robjohn That's a very big box! And now I am wondering whether you have not gone to sleep or you just woke up.

robjohn

@Former_Math_Addict hmmm

@JasperLoy I have been up for several hours at least; why?

Former_Math_Addict

@robjohn I can manually click and drag the scroll bar down.

user19161

@robjohn Nothing. Just a random thought from my observation.

user19161

10:59 AM

@Former_Math_Addict Have you tried refreshing?

robjohn

@Former_Math_Addict It works for me. Of course, I don't have the arrow that allows me to post a reply

user19161

@Former_Math_Addict I don't see any scroll there. Maybe you are on a small screen.

Former_Math_Addict

@JasperLoy Refreshing doesn't help, and yes I'm on a small screen.

robjohn

@JasperLoy I see a scroll bar. It is there so that the whole message doesn't take up too much vertical chat space.

Ilya

I'm happy we do not have such kind of questions on MSE

robjohn

11:04 AM

@Ilya Play with different people?

user19161

@Former_Math_Addict I'm on a super huge screen, not a Mac though!

Old John

@robjohn I think that book you wanted can be downloaded

(probably not legally)

robjohn

@OldJohn It can be downloaded legally, but it will cost me.

user19161

@Ilya But maybe we have such players on MSE who treat this like a game? :-)

Old John

@robjohn It is freely available on an ftp server in the Ukraine ;)

robjohn

11:07 AM

@OldJohn If I find a place where UCLA has a library agreement, I can download it there.

@JasperLoy It isn't a game? :-)

user19161

@robjohn Actually, life is one big game!

Gigili

@robjohn It's hard to put my feelings into words. It's quite old and partially relevant to the recent mod election and such, thinking about it just makes me want to vomit.

robjohn

@Gigili As I remember, this started quite a bit before the elections.

Gigili

@robjohn Well, yes but my point is, he and @Thechaz did their best to sabotage my election chances.

No point in discussing it right now and here, so pft will suffice.

David Wallace

11:22 AM

OH, I remember. They were both really horrible, if I recall correctly.

Frank Science

11:42 AM

@robjohn I have been working on proof and analysis of a particular algorithm.

robjohn

@FrankScience what does the algorithm do?

@FrankScience Ah, that is why you are tired -_-

Frank Science

@robjohn I failed to analyze it.

Longest palindromic substring

Juering's algorithm.

robjohn

@FrankScience does that mean you've given up?

Frank Science

@robjohn I have no idea to analyze.

@robjohn It's not an end because I have to check my proof.

robjohn

@FrankScience By analyze do you mean find the order of the time of execution?

Frank Science

11:53 AM

@robjohn No, the expression of the exact running time for each operation.

@robjohn here

robjohn

@FrankScience Heh: [8] Iverson bracket

is 8 true or false?

Frank Science

@robjohn Why did you say that?

robjohn

@FrankScience That appears in the bibliography. It looks like an Iverson bracket in the reference for an Iverson bracket :-)

Frank Science

@robjohn I only use Iverson bracket once in my article.

robjohn

@FrankScience and the reference I quote appears on page 8

Frank Science

12:00 PM

@robjohn It's not Iverson bracket. And notation $a[k]$ is used for programming in my article.

robjohn

Just reading from the pdf: i.stack.imgur.com/fviFJ.png

Frank Science

[8] is not Iverson bracket.

robjohn

@FrankScience as I said, I was just reading from the pdf of the paper you pointed to.

Frank Science

@robjohn I meant that [8] in that page is not Iverson bracket. It's only a label.

robjohn

Wrong link :-(

@FrankScience I know that. That is why it was funny.

CLarue

12:07 PM

An empty reference of sorts, I like it.

Frank Science

12:38 PM

It seems that the analysis of algorithms is very difficult for me.

Mats Granvik

1:22 PM

Sum[N[1/n^(1/2), 20], {n, 1, 48}] + 1 + 2/3
=14.134762112427871445

Mats Granvik

2:01 PM

1 + 2/3 = 1 + 1/2 + 1/3 - 1/6 = 5/3

Mats Granvik

2:17 PM

Sum[N[1/n^(1/2), 15], {n, 1, 109}] - (1 - 1/2 - 1/3 + 1/6) - 5
=14.1347798759772

What are you doing?

What I always do.

Could you explain?

The sixth value of the von Mangoldt function is equal to zero.
The Dirichlet series for the von Mangoldt function has numerators that are a repeating sequence. 1, -1,-2,-1,1,2, ... repeat. By letting the denominator be 1,2,3,4,5,6, 2,3,4,5,6,7, 3,4,5,6,7,8... and so on, one gets a number -2.549127729379167407581967029267929878... which appears to roughly relate the 1:st , 2:nd and the 3:rd zeta zero to the 18:th, 33:rd and 42:nd zeta zero. But only roughly. This suggests that the zeta zeros could have something to do with this kind of sum. The Dirichlet generating function for the Mangold…

Old John

@MatsGranvik and that is what you always do?

Mats Granvik

2:30 PM

Mostly, when I am not at work.

Old John

@MatsGranvik OK

Mats Granvik

2:51 PM

Sum[N[1/n^(1/2), 15], {n, 1, 36}] + (1 - 1/2 - 1/5 + 1/10) + 10
=21.0227859325892

anon

so, partial sums of the p-series at s=1/2 plus sum kind of alternating sum of reciprocals of divisors of a number plus some integer, is approximately a zeta zero?

Mats Granvik

I don't know. I should have left the last one unposted. The resolution in the numbers that are partial sums of reciprocals of square roots, is so high that I guess anything could be matched.

anon

plus the fact that signed-sums-of-reciprocals and integers together are fairly dense even for small choices of integers

but I think it's the shrinking resolution that allows arbitrary approximation too

Frank Science

3:14 PM

Generating function $\sum_{n\ge0}z^{\lfloor n\phi\rfloor}$ seems interesting, where $\phi=(1+\sqrt5)/2$.

user19161

@frank Where you come from, do you take the argument of a complex number clockwise or anticlockwise in high school?

Matt N.

3:33 PM

@anon Btw, before I forget: You mistook my previous reaction for something else. "Furiously sad" is more like it. : )

anon

I never said your previous reaction was otherwise. I just figured it happening twice might entail ferocity of a more pure quality :)

Matt N.

Fair enough : )

Rajesh D

What happened at Denver incident?

robjohn

4:43 PM

@HenningMakholm: I have been looking to see if Rofler seems to be downvoting the competition. I see that you got a downvote on this answer, which seems to be a good answer (lots of upvotes and it was accepted). I assume that you see no reason that you should have been downvoted; is that right?

Peter Tamaroff

$$f(x,y) = {e^{ - 3({{(0.5 + x)}^2} + {y^2}/2)}} + {e^{ - {x^2} - {y^2}/2}}\;\cos (4x)$$ people.

Old John

@PeterTamaroff Hi

@PeterTamaroff That's a nasty expression you have there

Peter Tamaroff

@OldJohn Plot it and you'll understand

Rajesh D

post the plot pic @PeterTamaroff

Old John

@PeterTamaroff Guessing it looks like "HI" ?

Peter Tamaroff

4:54 PM

bit.ly/QaNJp9

Old John

@PeterTamaroff Neat

Henning Makholm

@robjohn I don't see any particular reason for the downvote (was it you who just upvoted?). Whether it was Rofler I cannot say -- notice that the third answer to that question does not have any downvotes.

Rajesh D

@PeterTamaroff splendid

robjohn

@HenningMakholm I'm guessing whoever is only downvoting those answers that might be in the running for acceptance.

Ed Gorcenski

For some reason, WA won't show me the plot =(

Henning Makholm

4:58 PM

@robjohn For what it's worth, the downvote was cast on the original version of my answer, which was a bit brasher than what is currently visible.

robjohn

@HenningMakholm Ah, since you've edited it, the downvoter can retract. (and yes)

@JasperLoy we did neither; we took it counterclockwise ;-)

Henning Makholm

@robjohn There doesn't seem to be any downvotes on competing answers here, here or here

In fact, unless I'm missing something, the total number of dovnvotes cast on any answer that competes with one of Fofler's is two. That's not enough to infer foul play, I think.

robjohn

@HenningMakholm That's true, and I had looked at those. There are only two suspect cases, so I am only asking out of mild curiosity.

user19161

@robjohn Same here. And it went from $-\pi$ to $\pi$.

Peter Tamaroff

Can I ask for a proof check?

robjohn

5:06 PM

@HenningMakholm he has not answered many however, but you are probably right.

@PeterTamaroff where?

Peter Tamaroff

@robjohn Here =D

It is 3 lines.

Henning Makholm

I have probably downvoted ten times as many competing answers.

user19161

@PeterTamaroff Shoot!

Peter Tamaroff

I am presented with the following thorem:

**THEOREM 7.10** Let $(X,d)$ be and $(Y,d?)$ be two metric spaces. Let $f:X\to Y$ and $g:Y\to X$ be inverse functions. Then the following four statements are equivalent:

$(1)$: $f$ and $g$ are continuous;

$(2)$: $O\subset X$ is open $\iff$ $f(O)\subset Y$ is open

$(3)$: $F\subset X$ is closed $\iff$ $f(F)\subset Y$ is closed

$(4)$: For each $a\in X$ and $N\subset X$, $N$ is a nbhd of $a$ $\iff$ $f(N)$ is a nbhd of $f(a)$.

In its proof the author proves $$(1)\Rightarrow (2)\Rightarrow(4)\Rightarrow(1)$$

and I'm left to prove $(2)\iff (3)$.

Suppose $O\subset X$ is open $\iff$ $f(O)\subset Y$ is open. Then $O$ is open $\iff C(O)$ is closed. By $(1)$ then $f(C(O))=g^{-1}(C(O))$ is closed. But $g^{-1}(C(O))=C(g^{-1}(O))$ is closed $\iff$ $g^{-1}(O)=f(O)$ is open.

robjohn

@HenningMakholm I guess I am sensitive because I believe in commenting first and downvoting if that doesn't work (and I've never downvoted). Also you've answered more than 10 times the questions he has :-)

Peter Tamaroff

5:09 PM

I was told by Benjamin Lim that $g^{-1}(C(O))=C(g^{-1}(O))$ but I still have to prove that.

robjohn

@PeterTamaroff $C$ being closure?

Henning Makholm

I'm also one of the 3 or 4 users with the highest fraction of downvotes cast.

Peter Tamaroff

@robjohn Complement.

I use $\operatorname{cl}A$ or $\overline A$ for closure.

robjohn

@PeterTamaroff Oh, then yes, and it is pretty simple

@PeterTamaroff and that would finish the proof; so work on that :-)

Peter Tamaroff

@robjohn The proof I give directly proves equivalence, am I right?

robjohn

5:12 PM

@MarianoSuárez-Alvarez: Good day!

anon

If $f,g$ are inverse functions, in particular they are bijective, so e.g. $f(X\setminus F)=X\setminus f(F)$.

Peter Tamaroff

@MarianoSuárez-Alvarez Buenas.

anon

(i.e. $f,g$ commute with complements); this makes $(2)\iff(3)$ immediate by rephrasing $(3)$ as $X\setminus F\subset X$ open $\iff X\setminus f(F)=f(X\setminus F)\subset Y$ open

user19161

@PeterTamaroff The idea is there but the phrasing is awkward. For example, "then" in the second line suggests that that is a consequence of the first which is not the case. So when I read your proofs, I often have to mentally rearrange things to get it. In fact, I am not sure if your idea is absolutely right if you write it this way!

robjohn

@anon that was the simple proof I was mentioning :-)

Peter Tamaroff

5:15 PM

@JasperLoy I think I should change it with "But $O$ is open..."

@anon Yes. I had written in my notes verbatim: "$(3)\iff (2)$ by $F\subset X$ is closed $\iff$ $C(F)$ is open and $f(C(F))=C(f(F))$."

It was so short I didn't understand myself before.

robjohn

@PeterTamaroff This is a bit confusing. It is hard to tell what you are deriving and what you are using as given.

anon

the use of $C(\cdot)$ for complements: also lame

robjohn

@PeterTamaroff Let me see if I can reword, unless you want to use anon's idea

@anon I've used $E^C$ for complement

Peter Tamaroff

@robjohn I'm using four things. $(1)$ is equivalent to $(2)$. $F$ is open/closed iff $C(F)$ is closed/open. The preimage of a closed/open set under a continuous function is closed/open. And $f^{-1}(C(F))=C(f^{-1}(F))$.

robjohn

@PeterTamaroff Ah, you are using $(3)$ but proving it equivalent to $(2)$

Peter Tamaroff

5:21 PM

@robjohn I have "he preimage of a closed/open set under a continuous function is closed/open. " as a theorem already.

user19161

@PeterTamaroff Also, in the third line you brought in $O$ but it is not clear what $O$ is. Is it an open set? Your first line only says a subset $O$ is open if and only if some other condition holds. So another glitch there.

Peter Tamaroff

It is independent of T7.10

@robjohn So since $(2)$ and $(1)$ are equivalent I can use it

@robjohn I fact I also wrote here in my notes I can prove $(2)\Rightarrow (3)$ by $(2)\Rightarrow (1)\Rightarrow (3)$

Henning Makholm

There should be a silver "Backstabber" badge for downvoting 100 competing answers.

3

anon

100?

Peter Tamaroff

@HenningMakholm Hhahahah maybe.

user19161

5:25 PM

@HenningMakholm LOL.

Peter Tamaroff

@anon You don't simply....

Henning Makholm

@anon It mustn't be too easy to get.

user19161

@anon Sportsmanship badge is for upvoting 100 competing answers.

robjohn

@PeterTamaroff yes, but you are showing a biconditiional where each direction needs something slightly different

user19161

@HenningMakholm In this way, one loses 100 rep as well.

anon

5:26 PM

@JasperLoy sure, upvote/downvote ratios are generally very high

robjohn

@PeterTamaroff For example you all of a sudden assume that $O$ is open without stating it

Peter Tamaroff

@robjohn Maybe I can just use that $(3)\iff (1) \iff (2)$

@robjohn I don't understand what you mean.,

robjohn

@PeterTamaroff You sayI will have to type something up.

Peter Tamaroff

@robjohn Why can't I assume $O$ is open?

OK. I have proven $f^{-1}(Y^c)=f^{-1}(Y)^c$ already.

robjohn

5:46 PM

@PeterTamaroff Sorry, I had to take a phone call.

Peter Tamaroff

@robjohn I think I got it.

I'm going to write it down now.

I got it.

robjohn

Okay first you are assuming $(2)$ and you want to show $(3)$. Later we go the other way

@PeterTamaroff okay, I'll wait

Peter Tamaroff

@robjohn I'm proving it as follows. I have that $(1)\iff (2)$, so I'll prove $(3)\iff (1)$ and I'm done.

robjohn

@PeterTamaroff sounds good

Peter Tamaroff

6:06 PM

@robjohn OK, I've finished.

robjohn

@PeterTamaroff 8-) Let's see

Peter Tamaroff

@robjohn

robjohn

@PeterTamaroff Then for each $F\subset X$ that is closed, we have that $g^{-1}(F)=f(F)$ is closed, so that $g$ is continuous.

@PeterTamaroff you should have "that is closed" in there

Peter Tamaroff

@robjohn Oh, yes. Thank you.

Doing Exercises Remains Problematic.

robjohn

@PeterTamaroff I would duplicate the proof for $g$ to prove $f$ is continuous. The one for $f$ sounds iffy.

Peter Tamaroff

6:19 PM

@robjohn What does "iffy" mean?

anon

suspect, sketch, dubious

robjohn

@PeterTamaroff questionable

Peter Tamaroff

@robjohn Why? Because I pick $N=f(F)$?

robjohn

@PeterTamaroff Never mind, it is okay

Peter Tamaroff

@robjohn Darn! I had just crossed it over! XD

robjohn

6:25 PM

It would be nicer for the reader if the directions of the biconditional were in separate paragraphs. Sometimes people put $\Rightarrow$ and $\Leftarrow$ in front of each direction in the proof.

Peter Tamaroff

@robjohn Yes, I do that too. I forgot to do so here. I usually put $(\Rightarrow)\text{ blah blah }$

@robjohn So picking $f(F)\subset Y$ a closed set and saying $f^{-1}(f(F))=F$ is closed so that $f$ is continuous is OK, right?

I mean, my hypothesis is $F\subset X$ is closed $\iff f(F)\subset Y$ is closed.

robjohn

@PeterTamaroff You have to show that for any closed set $K$, $f^{-1}(K)$ is also closed.

Peter Tamaroff

@robjohn Could I argue any $K$ is the intersection/union of sets of the form $f(F)$?

robjohn

@PeterTamaroff No that is not necessary and will only complicate things.

Peter Tamaroff

@robjohn I'm good at complicating things.

Henry T. Horton

6:38 PM

What is the question here, Piotr

Peter Tamaroff

@HenryT.Horton See 544614

Henry T. Horton

@PeterTamaroff I'm good at complexifying things: $\mathrm{Things} \otimes \Bbb C$

Peter Tamaroff

@HenryT.Horton Did you see?

Henry T. Horton

No I don't know how to use a chat room

Gigili

@HenryT.Horton Just leave the tab open, it'll automatically use itself.

Peter Tamaroff

6:42 PM

@HenryT.Horton Just go up till you find long message of my part with THEOREM 7.10

2 hours ago, by Peter Tamaroff

I am presented with the following thorem:

**THEOREM 7.10** Let $(X,d)$ be and $(Y,d?)$ be two metric spaces. Let $f:X\to Y$ and $g:Y\to X$ be inverse functions. Then the following four statements are equivalent:

$(1)$: $f$ and $g$ are continuous;

$(2)$: $O\subset X$ is open $\iff$ $f(O)\subset Y$ is open

$(3)$: $F\subset X$ is closed $\iff$ $f(F)\subset Y$ is closed

$(4)$: For each $a\in X$ and $N\subset X$, $N$ is a nbhd of $a$ $\iff$ $f(N)$ is a nbhd of $f(a)$.

2 hours ago, by Peter Tamaroff

In its proof the author proves $$(1)\Rightarrow (2)\Rightarrow(4)\Rightarrow(1)$$

and I'm left to prove $(2)\iff (3)$.

@HenryT.Horton

Henry T. Horton

WTF you make me look for it then after I find it you paste it? You little monkey

3

Hi folks

Hey Dr. John's

Hi Johny.

This is the slowest-moving Friday in recent memory =(

Old John

6:47 PM

@EdGorcenski Then we should liven it up :)

Ed Gorcenski

Agreed!

Henry T. Horton

<puts shirt on backwards> How's that for wild and crazy!?

Gigili

Not handsome enough to tempt me.

Ed Gorcenski

Wait, shirts don't normally get worn backwards?

Gigili

waits

Henry T. Horton

6:52 PM

I've been waiting my entire life.

Bill Dubuque

@robjohn I didn't check the correctness of your answer, but the only other reason I could imagine for the downvote is that someone thought it was much more elegant to use the approach in the other answer. I have seen what I presume are downvotes based on that before, perhaps with the goal of raising up what they consider the better answer. One dimensional votes are problematic.

Peter Tamaroff

@HenryT.Horton Duuuuuuuuuuude.

robjohn

@BillDubuque Thanks for looking at that.

Bill Dubuque

@robjohn That's just a guess based on a quick glance. There are probably other possibilities too. As I'm sure you know, inferring reasons for downvotes is more difficult than proving GRH.

robjohn

@BillDubuque Yes. I have discussed such with Henning :-)

@BillDubuque Indeed

@PeterTamaroff I think you should stick with $(2)\Leftrightarrow(3)$ :-)

user19161

7:20 PM

@HenryT.Horton Wait no more, for I am here.

user19161

@form Nice new avatar!

Gigili

@JasperLoy I fail to see the relevance.

user19161

@Gigili I am just talking rubbish as usual, don't you know?

Gigili

@JasperLoy Uh, I thought we all were talking nonsense.

user19161

@Gigili Good for you. But some nonsense may have hidden meanings. It is up to us to interpret what we want.

Jonas Teuwen

7:27 PM

Haha a Backstabber badge.

I would totally go for that one.

Alex Becker

@JonasTeuwen What's this?

user19161

@JonasTeuwen Why stab someone in the back when you can do so from the front?

Jonas Teuwen

@AlexBecker Check the starred message of Henning.

@JasperLoy Because I'm a sneaky bastard.

Alex Becker

@JonasTeuwen I like it. But 100 might be a high bar.

user19161

@JonasTeuwen Good for you. I am a sneaky bustard, note the different spelling.

Jonas Teuwen

7:29 PM

@AlexBecker Yeah, you would have to work hard to get it.

Alex Becker

@JasperLoy Besides, if you play spy then a stab in the back is a 1 hit KO.

Gigili

@JasperLoy Nonsense has only one meaning which is no meaning and that is why it is called nonsense which means they are meaningless which is ...

Jonas Teuwen

@JasperLoy A... bird?

user19161

@AlexBecker And it's competing answers only!

user19161

@JonasTeuwen Ding ding ding!

Matt N.

7:30 PM

I like turtles!

Hey bro!

Want to talk about Fourier series with me?

Or not in the mood?

Gigili

Yuck!

Matt N.

Ooh, Alex is here, too!

@AlexBecker What's the addition on $\mathbb T$? I'm still confused.

Alex Becker

@MattN. Hey matt!

Matt N.

Hey Alex : )

Alex Becker

@MattN. If I read your question correctly, it's $x+y\mod 1$, right?

Matt N.

7:33 PM

@AlexBecker Well, that's my guess. But when I wrote that I was thinking of $\mathbb T$ as $[0,1) = R/Z$. The "mod 1" doesn't make sense on complex numbers.

Wait let me read your answer again.

Btw, that ping didn't ping me for some reason.

Alex Becker

@MattN. Remember $\mathbb R/\mathbb Z\neq [0,1)$.

@MattN. That's odd.

Matt N.

@AlexBecker Yes, exactly, that's why I'm confused. If the torus looks like $S^1$, what does "mod 1" mean?

Jonas Teuwen

@MattN. Sure.

@MattN. But first... I need to tinkle.

Matt N.

@JonasTeuwen Awesome. Meanwhile I'll discuss something with Alex.

Alex Becker

@MattN. mod 1 maps 0 and 1 to the same place, so it basically takes $[0,1]$ and glues the endpoints together. The result is a circle.

Jonas Teuwen

7:37 PM

No, it takes the circle and glues the endpoints together, so it is a map... etc 8-).

Matt N.

@AlexBecker So the addition happens in $[0,1]$ and is then mapped to $S^1$?

Wait, Dylan's answer says it I think.

Peter Tamaroff

Posted on main.

Ragib Zaman

Hey dudes.

And dudettes.

Peter Tamaroff

@RagibZaman Hello.

I just posted something on main, about a topological equivalence of $\Bbb R^n$with itself.

Maybe you're interested.

Ragib Zaman

I'll check it out now.

@PeterTamaroff Just as I finished reading, Arturo seems to have hit it on the head

experimentX

7:47 PM

hi

Peter Tamaroff

@RagibZaman Your input is welcome too!

Gigili

@OldJohn: Did you find my greeting off-putting?

experimentX

anyone help me interpret this

what does that mean ?? convergent??

or divergent??

Matt N.

@AlexBecker Ok, I think I understand it a bit better now.

experimentX

anyone free or a min??

Matt N.

7:50 PM

@JonasTeuwen Let me know when you're back : )

Jonas Teuwen

@MattN. Wait, stressing my ass of for a moment :-).

experimentX

what happened to my link

the series was sum of $$ \cos n \over n^{2/3}$$

anon

@experimentX What exactly are you trying to evaluate? Whatever you've inputted, it's so sloppy that W|A just made up something to talk about instead.

Alex Becker

@MattN. That's good. The important thing to remember is that requiring addition to be continuous on $\mathbb T=[0,1)$ (as sets, not as topological spaces) induces a topology on $[0,1)$ that makes it homeomorphic to the circle.

experimentX

@anon hello ... nice to see you

anon

7:52 PM

@experimentX Input Sum[Cos[n]/n^(2/3),{n,1,Infinity}] instead. (Also, hello.)

experimentX

yup ... does it converge??

Matt N.

@AlexBecker I think it's the other way around: we require addition to be continuous with respect to the product topology on $S^1 \times S^1$, no?

anon

@experimentX It doesn't say.

experimentX

Oh ... what do you think??

Old John

@Gigili No, not at all - I was just busy editing one of my answers (although it seems to be getting zero interest) ;)

anon

7:55 PM

Dunno offhand. I'll try an experiment.

Alex Becker

@MattN. Right, but we already know what the addition is, and a priori don't know the topology (when we're looking at $\mathbb T$ as the image of $[0,1)$ under mod 1). So we must choose a topology such that addition is continuous. In general, we choose the corsest one. This not the only choice we could make, but is the most useful, so we define $\mathbb T$ to be this topological group.

Gigili

@OldJohn Good, good.

Matt N.

@AlexBecker Ooh, ok! Now it all makes slightly more sense : )

whistles

Jonas Teuwen

@MattN. Yes.

Transcript for

$Mathematics$ Mathematics