Mathematics - 2012-07-21 (page 2 of 4)

@DavidWheeler Let $Y$ be a subspace of the metric space $X$. Then $O'$ is an open subset of $Y$ if and only if there exists an open ubset $O$ in $X$ such that $O'=O\cap Y$.

David Wheeler

and what is your definition of "open"?

just want to be clear, because there are different possible definitions.

i am guesssing you are using a metric d to define "epsilon balls"

Peter Tamaroff

Sorry.

A set is open if it is a neighborhood of all of its points.

@DavidWheeler Yes.

And in fact I have as a theorem a set is open $\iff$ it is a union of open balls.

David Wheeler

and your definition of "neighborhood"?

Peter Tamaroff

4:20 AM

@DavidWheeler A neighborhood $N$ of $x$ is a set contaning an open ball about $x$

David Wheeler

ok, so "open ball" is your "primitive notion" in this case

and i am guessing that an "open ball" is $B_\epsilon(x)$

Peter Tamaroff

@DavidWheeler I prefer $B(x;\epsilon)$, and if you have to detail the metric $B_d(x;\epsilon)$.

David Wheeler

of all points y where d(x,y) < $\epsilon$

Peter Tamaroff

@DavidWheeler Yes

David Wheeler

now Y is a subset of X, correct?

Peter Tamaroff

4:23 AM

@DavidWheeler Yes.

David Wheeler

given an epsilon ball in Y, consider the same epsilon ball in X.

Peter Tamaroff

@DavidWheeler It is an open subset of $X$.

Yes, that is what Ben just made me see. I was impressed it was that trivial.

David Wheeler

right..but since an epsilon ball in Y only includes points IN Y, the epsilon ball in Y may be a smaller set

Peter Tamaroff

@DavidWheeler Smaller than what?

David Wheeler

the epsilon ball in X

Peter Tamaroff

4:26 AM

Oh, yes, yes.

David Wheeler

in other words, the epsilon ball in X is the set we're going to take the intersection of Y with.

you see, Y may be compact "in X"

Peter Tamaroff

@DavidWheeler Compact?

David Wheeler

for example, imagine the euclidean plane, with the usual distance metric

Peter Tamaroff

OK.

$\Bbb R^2$

David Wheeler

an open ball is a circular disk with no boundary

Peter Tamaroff

4:28 AM

Yes.

David Wheeler

but the subspace Y may be the closed unit square.

Peter Tamaroff

@DavidWheeler OK.

David Wheeler

in which case the set of all points in Y less than some distance from the point (a,b)

might not be a "full" circular disk

Peter Tamaroff

But a section.

I see

David Wheeler

right...the parts "outside of Y" get "chopped off"

Peter Tamaroff

4:30 AM

@DavidWheeler Heheh, yes =)

OK. So let me try and sketch it in paper

David Wheeler

i mean the definition of an epsilon ball in Y should start with $\{y \in Y:...$

and that's the important part

Peter Tamaroff

Yes, yes.

David Wheeler

so we have two conditions: $d(y,y_0) < \epsilon$ AND $y \in Y$

Peter Tamaroff

I can write as follows:

Suppose $O'\subset Y$ is open in $Y$. Then it is the union of open balls in $Y$, that is $O'=\bigcup_{x \in O'} B(x;\epsilon_{x})$

David Wheeler

if you "get rid" of the second part, you have a set in X...and the "and" condition is expressed by "intersection with Y"

correct so far

Peter Tamaroff

4:35 AM

Consider the same open balls in $X$, that is $O=\bigcup_{x\in O'}B(x;\epsilon_x)$. Then $O$ is an open subset of $X$ for it is the union of open balls in $X$.

David Wheeler

now form the same union in X (the balls "in X" will now be "full balls")

Peter Tamaroff

Then we have that $Y\cap O=O'$, and we've proven the biconditional one way.

David Wheeler

yes, and there's the desired open set.

Peter Tamaroff

Yay!

OK. Now I have to prove the converse.

David Wheeler

and all you have to show, really, is that the intersection of an epsilon ball in X, with Y, is an epsilon ball in Y

you may as well consider epsilon balls that actually intersect Y, because the null set is trivially open.

Peter Tamaroff

4:38 AM

Yes. We discussed that with Ben earlier.

OK.

David Wheeler

it's really quite straight-forward....the topology you're considering is called the subspace or relative topology, and it's the only one that makes any "sense"

Peter Tamaroff

So: "Assume $O\subset X$ is open in $X$. Then we may write $O=\bigcup_{x\in O}B(x;\epsilon_x)$.

Consider now $O\cap Y$. This si the same set with the restriction $x\in Y$, so this means each $B$ is "converted" into an open ball in $Y$. Then $O'=O\cap Y$ is the union of open balls in $Y$, so it is open"

David Wheeler

now restrict that union to the x that lie in Y

yes

Peter Tamaroff

The same result but for closed sets follow by De Morgan law's right?

Or am I being too utopic?

Because I have to prove that too. =P

David Wheeler

well, there is a wrinkle

because you can only consider finite intersections

but, what you do is consider the complements, which are open

and its essentially the same argument as before

Peter Tamaroff

4:43 AM

I see.

As a third bulletpoint I have to prove that $N'\in Y$ is a nbhd of $a\in Y$ iff there is a nbhd $N$ of $a$ in $X$ such that $N\cap Y=N'$. It shouldn't be too different, right?

Because I have defined neighorhoods in terms of open balls.

David Wheeler

there's a kind of "duality": suppose Y = [0,1). while [1/2,1) is closed in Y, it's not closed in $\Bbb{R}$

yes, really all you're doing is "restricting to Y"

Peter Tamaroff

But that's not the point, since $[1/2,1)=[0,1)\cap[1/2,3]$

And the rightmost interval is closed in $\Bbb R$, right?

David Wheeler

yes, but as you can see the "openness" or "closedness" of a set isn't inherent...it's totally dependent on the ambient topology

Peter Tamaroff

@DavidWheeler OK. However, I'm still in Metric Spaces. I'm studying from Mendelson's Introduction to Topology, and I'm almost done with Metric Spaces.

David Wheeler

right...but a metric is a special KIND of topology

where we use the idea of metric to define "near"

("in the neighborhood of")

Peter Tamaroff

4:50 AM

This section which I just finished and now I'm solving the exercises is SUBSPACES AND METRICALLY EQUIVALENT SPACES. (It also includes Topologically equivalence)

@DavidWheeler Right.

I have that the open balls form the basis for the nbhd systems in metric spaces.

David Wheeler

the thing about metrics, is...they let us turn questions about open/closed

user19161

@PeterTamaroff In some other books, metric spaces is just one brief chapter and then topological spaces follows where you learn all the results that apply generally.

David Wheeler

into facts about real numbers (inequalities)

that's kind of "special"

user19161

@DavidWheeler What a brilliant observation!

David Wheeler

bah!

user19161

4:52 AM

Seriously I am now beginning to hate this Mendelson book approach which seems to go on and on about metric spaces forever.

David Wheeler

metric spaces are what we're "used to"...the first time you pick up a ruler, you're essentially using the concept

user19161

@DavidWheeler Yes, but there is an excess of it in this case IMHO. Later one needs to reprove all the theorems.

David Wheeler

well, metric spaces are really good for analysis, right? because you can estimate if you're "converging"

Peter Tamaroff

@JasperLoy I think @RagibZaman said Mendelson's book is good. I'm liking it.

user19161

@PeterTamaroff Sure, I'm just saying I prefer other books, since I have read over 9000 books on general topology!

Peter Tamaroff

4:57 AM

And in fact this is basically preparing me for topological sapces, justc as Ben said what I'm being asked to prove is used as a definition in topologica spaces, so it is good to know the motivation and meaning of definitions....

Else what you study is an empty concept.

David Wheeler

@JasperLoy one thing that often confuses people when they first start learning topology is: where did all this stuff come from? what's the motivation?

user19161

@DavidWheeler Exactly. Hence what I meant is one short chapter on metric spaces will do as a motivation, not a lengthy exposition like that.

David Wheeler

sure you can start group theory, for example, straight from the axioms, but if you do...you might wonder: why are we doing this? how can i use it?

well, a lot of stuff you learn in calculus...makes more sense when you know about metric spaces.

Peter Tamaroff

@DavidWheeler That's true.

David Wheeler

i think when i took complex analysis...we spent more than a month essentially describing the topology.

all of which could have been summed up as: "the modulus is a metric".

Peter Tamaroff

5:01 AM

@DavidWheeler HAHAH right

Well, it would also been good to know $\Bbb C$ and $\Bbb R^2$ are topologically the same, right?

David Wheeler

so, sure, there's a LOT more to topology, and some very interesting things you can say in much greater generality

Peter Tamaroff

I don't understand why this questinons are so popular.

David Wheeler

because, for many people, especially lay people, mathematics is "interesting things about numbers"

Peter Tamaroff

@DavidWheeler When I tell people I pursue a career in maths, they usually ask "DO you like numbers that much?

user19161

@DavidWheeler One starts off with numbers, strays away from numbers at a higher level, and then returns to numbers at even higher levels.

Peter Tamaroff

5:05 AM

I should just say : "NO! I hate numbers, but I like math!"

@JasperLoy Explain yourself.

David Wheeler

well, one can return to numbers...but not all mathematicians do.

user19161

@PeterTamaroff For example, if you do analysis, the computations can involve using many explicit quantities. Sure they are all written as letters, but they represent numerical quantities.

Peter Tamaroff

@DavidWheeler For the result about closed sets, I can start with "Suppose $F'\subset Y$ is a closed subset of $Y$. Then $C_Y(F')$ is open..."

David Wheeler

...and...

Peter Tamaroff

"...then a miracle happens"

David Wheeler

5:10 AM

well the complement is just Y\F'

Y-F', if you prefer

Peter Tamaroff

Yes sorry I have old notations. $Y\setminus F'$

I have $C_Y(F')$ from Mendelson, $Y-F'$ from Halmos, and $Y\setminus F'$ from those people here that think the other two are old or lame @anon

[mean face]

David Wheeler

lol

user19161

@PeterTamaroff Yes, I need a number of miracles in my life...

David Wheeler

now you have an open set U in X, such that $U \cap Y = Y\setminus F'$

what can you say about X\U?

Peter Tamaroff

I can take complementssssssssssssss

And it is closed.

David Wheeler

5:16 AM

what do you suppose $(X \setminus U) \cap Y$ might be?

Rajesh D

Hi @DavidWheeler

David Wheeler

Hi @RajeshD

Peter Tamaroff

@DavidWheeler Let me see

David Wheeler

well it's a subset of Y, right?

Peter Tamaroff

The building sentence is $x\in X:x\notin U \wedge x\in Y$

David Wheeler

5:18 AM

ask yourself 2 questions: does it contain every point of F'? does it contain ANY point of Y\F'?

user19161

@RajeshD Hi Rajesh. You didn't say hi to me...

Rajesh D

@JasperLoy Hi Jasper, sorry I missed you

user19161

@RajeshD It's OK. This is what sitting in a comfy chair does!

Rajesh D

Hmm...but I should mention that I am lying on a bed

Peter Tamaroff

@DavidWheeler I'm stuck.

$X\setminus U\cap Y$ seems to be equal to $Y\setminus U$

David Wheeler

5:24 AM

suppose y is in F'. then y is NOT in Y\F'. now $U \cap Y = Y\setminus F'$, right?

Peter Tamaroff

Yes

ami

Hi

David Wheeler

so for y to be in U, it has to be both in Y (which it is) and Y\F' (which it isn't).

ami

I am trying to find the coefficient of [x^ny^m] in \frac{1}{(1-xy)^2(1-x)}\log(1/(1-xy))

David Wheeler

so we conclude y isn't in U

Peter Tamaroff

5:26 AM

$ [x^ny^m]$ in $\frac{1}{(1-xy)^2(1-x)}\log(1/(1-xy))$

David Wheeler

thus y IS in X\U

since y is in Y, we know that y is in $(X \setminus U) \cap Y$.

this shows that set contains all of F'.

on the other hand, if y is NOT in F', then y IS in Y\F'.

and thus y IS in U (since Y\F' is a subset of U).

Rajesh D

@PeterTamaroff If I am stuck at something, then I would try to stop typing symbols in latex and move away from computer and spend time thinking about it cooly.

ami

peter: mathurl.com/c95e8xh

Peter Tamaroff

@DavidWheeler Don't you mean $Y\setminus U$?

David Wheeler

since y is in U, y cannot be in X\U.

and thus cannot be in any subset of it, so $(X \setminus U) \cap Y$ contains NO points of Y\F'.

ami

5:33 AM

David, mathurl.com/c95e8xh , Please provide any pointers.

I am stuck :/

David Wheeler

so...it must be the case that $(X \setminus U) \cap Y = F'$, so X\U is the closed set we seek.

Peter Tamaroff

Let me recap.

David Wheeler

in other words, we take a closed set, complement, use our result on open sets.

complement again, and intersect.

Peter Tamaroff

If $y\in F'$ then $y\notin Y\setminus F'$, but since $Y\setminus F'=Y\cap U$ and $y\in Y$, it must be the case $y\notin U$. Then $y\in X\setminus U$.

Rajesh D

@ami : What are these equations about. What is $H$ could you give more information about the problem.

Peter Tamaroff

5:37 AM

This proves $F'\subset X\setminus U$

ami

@RajeshD H_n => Harmonic number

David Wheeler

yes, but do you understand that?

ami

@RajeshD I am trying to find the coefficient of $[x^ny^m]

anon

@ami the expansion of 1/(1-xy)^2 does not involve harmonic numbers

it is simply $\displaystyle\sum_{k\ge1}k(xy)^{k-1}$

ami

@anon [x^ny^m]$\frac{1}{(1-xy)^2(1-x)}\log(1/(1-xy)) $

anon

5:42 AM

$$\frac{1}{(1-xy)^2}=\sum_{k\ge1}k(xy)^{k-1} \\ \frac{1}{1-x}=\sum_{\ell\ge1}^\infty x^{\ell-1} \\ \log\frac{1}{1-xy}=\sum_{t\ge1}\frac{(xy)^t}{t}.$$

Peter Tamaroff

@DavidWheeler I got it.

Now I'll try to prove the converse and go to sleep

David Wheeler

@ami i don't understand what you're doing...but it seems to me you might have the index range for k wrong...

Peter Tamaroff

It is almost 3 am

David Wheeler

again...it's the same strategy @PeterTamaroff use complements, and the result already proved on open sets.

Peter Tamaroff

@DavidWheeler Yes.

anon

5:44 AM

oh, you incorporated one of the 1/(1-xy)'s into the log 1/(1-xy) function. you could have at least said so @ami

ami

@anon yes.

mathurl.com/cqpyoy8

@anon I should have explained it. But added this step here. mathurl.com/cqpyoy8

Peter Tamaroff

@DavidWheeler So I assume $F$ is a closed subset of $X$, then $X\setminus F$ is open so there is an open subset $V$ of $Y$ such that $X\setminus F \cap Y=V$

I assume the set I'm looking for is $Y\setminus V$?

ami

@anon I used $\sum_{k=1}^n H_k = (n+1)(H_{n+1} - 1)$

at 4th step

David Wheeler

yes, now prove that $F \cap Y = Y \setminus V$

Peter Tamaroff

@DavidWheeler Can't I prove it directly by some "algebra" on $Y-((X-F)\cap Y)=Y-V$

David Wheeler

5:50 AM

again, a null intersection of F and Y is trivial, so you can just focus on non-null intersections.

Peter Tamaroff

@anon Unleash you inner set theorist.

=D

David Wheeler

if that is what you want to do, sure.

anon

@PeterTamaroff I learned enough about ordinals to understand Asaf's answer to my set theory question recently.

David Wheeler

link to question, @anon

anon

http://math.stackexchange.com/questions/46833/how-do-we-know-an-aleph-1-exists-at-all

(it is not a recent question though)

Peter Tamaroff

5:53 AM

@DavidWheeler Next time you should say "PICS OR IT DIDN'T HAPPEN" =P

anon

eh nvm

ami

@anon book's answer looks wrong to me.

I think I did right.

David Wheeler

@anon why would i care when the question was asked?

ami

googling it if I can find somehwere

anon

If there's an error in your calculation I don't see it.

Peter Tamaroff

5:57 AM

@anon How can I prove $Y-((X-F)\cap Y)=F\cap Y$?

Where $Y\subset X$

and $F\subset X$

David Wheeler

@PeterTamaroff i'm dim-witted, so when i have to prove two sets are the same, i always show each is a subset of the other.

ami

@anon lsi.upc.edu/~conrado/research/reports/ALCOMFT-TR-03-50.pdf

please look at page no. 3

right side

Peter Tamaroff

@DavidWheeler By a Venn diagramm that is clear.

ami

$z^nu^m$

anon

@ami Dude. $$(m+1)H_m-m=(m+1)(H_{m+1}-1).$$

David Wheeler

6:02 AM

suppose y is in $F \cap Y$ ( i am tacitly assuming this is non-null).

then y is in F, so y is NOT in X\F.

anon

@PeterTamaroff Clearly both sides are $\subseteq Y$, so apply $Y\setminus$ to both sides (a reversible operation)...

Peter Tamaroff

@anon What do you mean by reversible? I have that $(X-F)\cap Y=V$. I want to prove $Y\setminus V=F\cap Y$.

David Wheeler

so y is not in V, which is a subset of X\F. therefore y is in Y\V.

Peter Tamaroff

@DavidWheeler Yes.

David Wheeler

so $F \cap Y \subset Y \setminus V$

anon

6:05 AM

By reversible I mean $A=B\iff Y\setminus A=Y\setminus B$ when $A,B\subseteq Y$.

(which follows from $Y\setminus(Y\setminus A)=A$)

ami

@anon lol

@anon Thanks

anon

@PeterTamaroff By far the easiest route imo is setting $X=Y\cup Z$ with $Y\cap Z=\emptyset$.

Peter Tamaroff

@ami I have to show then that if $X\subset Y$ then $X\setminus F\cap Y=Y\setminus (F\cap Y)$

anon

$X\setminus F=(Y\cup Z)\setminus F=(Y\setminus F)\cup (Z\setminus F)$, and $Z\setminus F\cap Y\subseteq Z\cap Y=\emptyset$

Therefore $(X\setminus F)\cap Y= Y\setminus F$, and you can do the rest.

Peter Tamaroff

But I have to prove $(X\setminus F)\cap Y=Y\setminus (F\cap Y)$

anon

6:13 AM

Same thing.

Peter Tamaroff

You proved something else.

anon

You don't see the equivalence of $Y\setminus F$ and $Y\setminus (F\cap Y)$?

David Wheeler

on the other hand, suppose y is in Y\V. y is not in V. since y IS in Y, the only way y can be in $(X \setminus F) \cap Y$ is if y is in X\F. but $V = X \setminus (Y \cap F)$.

anon

Very easy if you think in terms of venn diagrams. But if you want to prove it formally, you can use the very same idea I just pointed out. Set $F=(F\cap Y)\cup A$ with $A$ and $F\cap Y$ disjoint, then $Y\setminus F=Y\setminus ((F\cap Y)\cup A)=Y\setminus(F\cap Y)\cup Y\setminus A=Y\setminus (F\cap Y)$.

moral of the story: decompose things into binary partitions to distribute out unwanted things through setminuses.

actually that should be $\cap Y\setminus A$ rather than $\cup Y\setminus A$. edit interval is gone.

David Wheeler

so y lies outside of $Y \cap F$, and y certainly isn't in the part of F that lies outside Y.

Peter Tamaroff

6:19 AM

@anon I don't =(

David Wheeler

so y is not in F, but IS in X, so IS in X\F.

done.

anon

Make a venn diagram with Y on the left, F on the right, and F\cap Y in the middle. The left side not including the middle is both $Y\setminus F$ and $Y\setminus (F\cap Y)$.

Peter Tamaroff

@DavidWheeler Thanks.

David Wheeler

the partition i think anon is referring to is:

Peter Tamaroff

@anon Got it.

But should a Venn Diagram suffice?

anon

6:22 AM

actually I should have specified $A$ and $Y$ disjoint, but whatever

it does for me!

David Wheeler

$A = (A \setminus B) + (A \cap B)$

anon

yes

David Wheeler

where the + in this case indicates we have a disjoint union

(or "symmetric difference operation")

but, Venn diagrams are useful, because you can draw the situation, which guides your logical "worded" explanation.

and yes, if you really, really wanted to, you could make a first-order wff, and then apply de morgan's laws.

Peter Tamaroff

@DavidWheeler What is first order wff?

anon

well-formed formula, propositional stuff

David Wheeler

6:27 AM

well-formed formula

Peter Tamaroff

I got stuck in the other direction. I proved $F\cap Y\subset Y\setminus V$.

But I got stuck in proving the other inclusion.

I should know more set theory. Fuck me.

David Wheeler

think of it this way: F is the "bigger set", right?

Peter Tamaroff

Bigger than what¿

David Wheeler

Bigger than Y\V.

F lies in X, where open balls may be "bigger" than open balls in Y.

Peter Tamaroff

Yes.

David Wheeler

6:31 AM

F splits into 2 parts, clean as a whistle: the part in Y, and the part not in Y.

Peter Tamaroff

Yes.

David Wheeler

the complement of F (our open set we use our earlier result from), also splits into 2 parts as well, the part in Y, and the part not in Y.

Peter Tamaroff

Ja.

David Wheeler

the part of X\F in Y is still going to be the complement (in Y) of the part of F in Y, see?

draw a picture....draw Y as say a square on a piece of paper.

Peter Tamaroff

@DavidWheeler $Y\setminus F\subset X\setminus F$

David Wheeler

6:37 AM

draw a closed set (say a closed disk) that intersects Y

Peter Tamaroff

Yes, I see it.

David Wheeler

the proof is more complicated than the direct apprehension of it being true.

Peter Tamaroff

I know

But I can say since $Y\subset X$ that $(X\setminus F)\cap Y=(Y\setminus F)\cap Y$

But $(Y\setminus F )\cap Y=Y\setminus F$

David Wheeler

i think you need some parentheses, somewhere

Peter Tamaroff

Now?

David Wheeler

6:41 AM

ok...but you seem to have omitted V entirely

Peter Tamaroff

No, wait.

Now I have that $Y\setminus F=V$

So Taking complements rel. to $Y$ gives $Y\cap F=Y\setminus V$

David Wheeler

remember, we took the complement just to get to open sets.

we're dealing with 4 sets here: F (the closed set in X), X\F (an open set in X), V (an open set in Y derived from X\F), and Y\V, the closed set we finally want to get from F.

Peter Tamaroff

OK. All this became a set theory problem now.

I'm off to bed. It's almost $4$ am.

I got it.

I'll finish the third bullet tomorrow.

David Wheeler

wake up in the morning, and drink some coffee, and write it all out. it will be straight-forward, I promise.

Peter Tamaroff

@DavidWheeler Yes, I think it will.

Still, my sloppy set theory knowledge is something I have to work on if I intend to study topology.

David Wheeler

6:47 AM

indeed. because when you get to the general definition of continuity, you have this:

f:X-->Y is continuous if whenever U is open in Y, $f^{-1}(U)$ is open in X.

Peter Tamaroff

Do you have any reccomendation? I am reading Naive Set Theory by halmos by it is scarse in exercises.

David Wheeler

i'm the wrong person to ask

Peter Tamaroff

OK.

Have a good one, Wheeler.

robjohn

@BillDubuque: Ack! I meant to hit "discard" and accidentally hit "post". I do like my formatting better than Arturo's :-)

anon

7:04 AM

user19161

7:14 AM

@anon Is that you?

anon

that's >mfw

(it's a relatively widespread gif)

user19161

@anon I had to look up UD for MFW, you are really a UD guy!

anon

and do you know why I put a > in?

user19161

@PeterTamaroff Like I said, that will suffice unless you intend to do formal logic and set theory, in which case read Mendelson's Mathematical Logic.

user19161

@anon No, in fact MFW is so ill-defined that I don't bother to fathom its mysteries.

anon

7:18 AM

(>greentext on 4chan!)

Clark Kent

7:28 AM

Hi. Is it ok to ask math questions in here?

anon

...

user19161

@ClarkKent Yes, but it is better to post on main if it is nontrivial.

Clark Kent

I have posted it on main.

user19161

OK, then wait for the answer.

user19161

@anon Now people will be wondering why CK and JL coexist!

Clark Kent

7:30 AM

I have an answer but I don't understand the comments to the question.

I don't understand why $G_{\frac12}$ is not a subgroup of $G_{\frac14}$.

Where $G_{\frac12}$ is the subgroup of $\mathbb Q$ generated by $\frac12$.

anon

it's not?

Clark Kent

Well this comment here says that it's not a submodule.

But submodules correspond to (additive) subgroups.

anon

that comment says 1/4 is not a submodule of 1/2. however your post says 1/2 is a submodule of 1/4. both are correct.

Clark Kent

Yes, right.

Oops. I messed it up in my post : (

I didn't realise I got it the wrong way around.

Thanks man.

anon

yup, as Jack and Joe say

user19161

7:38 AM

@ClarkKent You are welcome and you can't be sure anon is a man.

anon

we've been over this already. they can assume that because I drip with manly manliness. flexes muscles

David Wheeler

anon is a man? he's not one of the asgard? i am disappoint.

Clark Kent

Right, sorry. Thank you anon.

I think I can work it out on my own now. See you!

user19161

@ClarkKent Yeah, by the way I used to be Clark Kent...

user19161

@anon Yum yum.

anon

7:46 AM

heh heh.

Clark Kent

8:18 AM

@JasperLoy Probably after you saw me and liked the name because when I signed up I checked and there was no one else with my name.

Jonas Teuwen

Sup!

Rajesh D

OMG! Gortaur = Ilya!

Jonas Teuwen

8:41 AM

@RajeshD No shit.

Rajesh D

@JonasTeuwen ? What do you mean, Its not okay to discuss such things?

Jonas Teuwen

@RajeshD More than a year already?

Rajesh D

@JonasTeuwen I found out just now

Transcript for

$Mathematics$ Mathematics