@lurker Sort of like $\sin(x)=1$?

e^((i)(x)) where x = positive or negative odd multiple of pi

The fact that there is some connection between $e$ and $\pi$ and even $i$ that yields $-1$ so easily

yeah but again the only reason it even yields -1 is because its calculated in radians

if it wasn't, you would get an irrational result

and of course, radians based on a unit circle of radius 2pi

its all very restrictive

sorry not radius 2pi

circumference 2pi

Well its not so restrictive to base the measurement of the angle in a circle on the circumference

In fact, its quite natural

yeah but its radius = 1

Its not like mathematicians chose the things in such a way that the formula is beautiful

any other radius and it falls apart

No certainly not

For every other radius we measure the angle in the same way

yeah but its not calculated in degrees

but radians

and radians is measured by circumference

1pi is half a revolution

for a unit circle

I cannot follow what point you are trying to make

if its radius is 2, then pi would give cos (pi) 0 and sin (pi) 1

You can prove Eulers formula using Taylor series, if you want to get rid of your problem with circles

okay ill look into that

because all over the internet

it says e ^ ((i)(x)) = cos(x) + isin (x)

so maybe im restricting myself

Yeah, you can prove that using Taylor series, I meant

hey

Hi @KarimMansour

why does this topology converges to more than one point postimg.org/image/fn4l640hb

this is an example in

munkres

I need to understand

What do you mean with topology converges?

We can specify every point with open sets

I just looked up taylor series for e^((i)(x))

its the same thing I just wrote

plus theres a thread on stackexchange proving it equals cos(x)+isin(x)

point holds

So what exactly is your point?

@Krijn I meant to say a sequences converges to more than one point

there example if $x_n = b$

the example given in the book

cos(x) deals with radians or degrees, correct?

@r9m So I assume that proving that inequality is not what you are looking for.

likewise with those other function

functions*

Sure

@Karim Think about the sequence $x_n = b$ for all $n$.

$x_1,...$ of points of X converges to the point x of X provided that corresponding to each neighbourhood U of , there is a positive integer N such that $x_n \in U$ $\forall \ n \geq N$.

But why is it so trivial to you that $e^x$ has to do with circles?

@robjohn I already know how to deal with the inequality analytically .. what I was looking for is if we can associate geometric meaning to the steps of the proof (in the best case) or if that is not possible, if we can have a proof that relies on geometric intuitions rather than algebraic manipulations :)

im relating it becuause e^(i)(x) = cos(x) + isin(x)

otherwise I wouldnt mention it

Yes, but the beauty is in that statement

oh yea

yeahhh but pi isnt the only value for x that yields -1

its infinitely many solutions that yield -1

thats why I dont think its a big deal

@r9m That is what I thought. I have tried many geometric pictures, but I have not found anything. :-(

Its not a big deal, its just a nice mathematical expression

I think its overrated as well

But its still a nice expression

agreed

I just think its misleading thats all

@lurker does the fact that $\pi$ is not the only value so that $\cos(x)=-1$ also bother you?

yes

correct

@KarimMansour However, if $X$ is Hausdorff, can any sequence converge to multiple limit points?

The misleading part may be that $e^{ix} = \cos{x} + i\sin{x}$ holds only for radians

yes that too

two parts are annoying

the impression that only one yields it

and the reliance on radians

@BalarkaSen just want to see if my reasoning is correct we can see that there is open set that contains U = {c,b}, for any n we can see that will contain b all the time

It doesnt rely that much on radians though

no that is the reason

they introduced haussdorff

just a sec @BalarkaSen trying to make sure I understand this

I am little sick so I am a bit slow today

ok, go on

I am ill too.

without the unit circle of radius 1

I'm slow even at my healthiest

Fever, sneezing, sinus. All in all, ugh.

But the definition of $e^x$ is not based of the unit circle

you cannot yield -1

yeah fever and sneezing all the time

.... im not talking about e^x

sickness is stupid

anyway

But the equation is

im talking about e^((i)(x))

which equals cos(x) + isin(x)

@KarimMansour Hah, we're on the same boat, then.

I'm waiting for my midterms to end so I can get my flu vaccination

But your argument is the wrong way around

You can prove that $e^{ix}$ is on the unit circle

That is the nice part of the formula

yeah I know it is. it obviously is

@robjohn I hoped maybe the people who proposed the problem in math monthly had some geometric intuition behind ..

but -1 is a result of the circle having radius of 1

It is not defined using the unit circle

No

It is a result of that function having its image in the unit circle

@r9m The way it was posed, it certainly seemed as if it should be solved geometrically.

Not the fact that we chose a circle with radius 1

We did not choose the circle

@BalarkaSen yeah haha. So according to the definition given in the book lets test if it indeed converges to c. We can see that for any n we can see that $x_n \in U = \{b,c\}$

without the unit circle

right ?

it wouldnt yield -1

?

coz by definition U is an open set that contains c

@KarimMansour You mean $x_n \in U$.

We did not define that function to have its image there?

yeah

@robjohn ya! That's what my itch is all about :-)

yeah haha

Yes, it does converge to $c$.

good and similiarly it will converge to b and a

Correct.

good now I can move to $T_1$ spaces

gotta make tea first with lemon lol

@Karim Good idea.

I usually take some tea with ginger. Works wonders.

Especially for the sore throat.

yeah ginger is very good

yeah

" Let P be a subset of R. If (1) 0 in P (2) sup{x in [0,1] : [0,x] is a subset of P}=1, then [0,1] is a subset of P " Is this true or false?

it sucks that in upper level math classes where its usually 3-4 people if somebody got sick all of will catch it

well maybe i should create a new post...

What does it mean when someone says that two maps are mutually inverse?

Does it mean they are inverses of one another?

Sounds like it.

Probably they're trying to say that $fg = gf = 1$.

Sounds right

actually you know @BalarkaSen thats the definition of isomorphism in arbitrarily category

Yes.

Hi @BalarkaSen

Hello

I am very fond of groups again

Good!

Well I read Dummit and Foote up to solvable groups

and field theory

I am trying to get my head around fund. galois theory

that's nice.

I get the intuition

Lattice of subgroups is the same as the lattice of subfields

In the opposite orientation. Thus, we call $L \mapsto Gal(K/L)$ "contravariant".

Cayley graphs are also very interesting

Cayley graphs are super-interesting.

Is it possible to determine the minimal amount of generators needed for a group?

Hard problem. I am not sure if there is an algorithm.

@Alizter In case you don't know about all this already, read the discuss here.

So if you picked two different generating sets of minimal size would the cayley graphs be isomorphic?

No.

Absolutely not.

Consider the Cayley graphs of $(\Bbb Z, \{\pm 1\})$ and $(\Bbb Z, \{\pm 1, \pm 2\})$.

One is a tree, the other is not.

In any case, the magic word is "quasi-isometry". Just read that discussion.

No but the second cayley graph doesn't have a generating set of minimal size

So it doesn't fit my criteria

Alright. But it's not true anyway.

I am sure you can find an example.

Hmm maybe $(\Bbb Z, \{1, -2\})$

Yes that is a counter example

Yeah, I think that works.

So is it possible to put restrictions on the group

to make it true

Can we find a finite group where this doesn't hold

No, as I said, isomorphism isn't the right thing.

You don't want to think about it.

Just click the link :P

You want to identify two Cayley graphs of groups of two different generating sets. There is an awesome answer to this question if you just forget about graph isomorphisms.

ahh interesting

Is it possible to determine whether a given graph is a cayley graph or not?

@BalarkaSen

@Alizter Good question. Again, I am not sure if there is an algorithm.

But there are loads of such graphs.

If we had such an algorithm

Then we could construct graphs to fit it

And then we could construct countless groups

@Alizter "Is it possible to determine the minimal amount of generators needed for a group?" It probably depends what you mean. But in general, there is no algorithm to determine whether a presentation for a group is a presentation for the trivial group (which requires zero generators), so there is no algorithm which decides if a group has 0 generators or not. This is related to something called the word problem for groups.

Oh, right, @PaulPlummer/

I doubt that you could decide whether a given graph is a cayley graph too, although I think it is a tad more subtle, since you are not asking about what group that graph would correspond to. I guess if you had decided a graph is a cayley graph, you could construct a group, by choosing a vertex, and corresponding generators for the edges connected to the vertex, and add a relation for every loop through the vertex... hmm

I will have to think about that one

@BalarkaSen Hey

Could someone take a look at a question originated from reading a proof? math.stackexchange.com/q/1486535/275935

I miss Mike Miller

Does mike not come by anymore?

He has to work on his own schoolwork

suppose $Y = (0,1] U (3,2]$ consider the subspace topology given on this with Y is the subspace of X, where $X = \mathbb{R}$ what is the interior of (0,1]?

@KarimMansour How do you know what an open set in the subspace topology is

well, the interior is all arbitrarily union of U open in subspace topology that is contained in A.

subspace topology is generated by the topology that has the following as basis

$B = \{ B \cap Y | B \in \mathbb{B}\}$

where $\mathbb{B}$ is the basis for the standard topology

or if you want it in terms of the topology it is as follows

$\tau_Y = \{ U \cap Y | U \in \tau_X \}$

i.e it is the intersection of all open sets of topology X with Y

fuck

I think it is (0,1) again

right ?

Why is it $(0,1)$

because what is the open topology in the subspace topology that contain (0,1]

@robjohn have you seen my attempt at Cody's integral here ? :-)

well in order for the open set to contain (0,1]

to sit inside it I mean

is for $A \cap ((0,1] \cup (3,2] \subset (0,1]$

Give $(0,1]$ the subspace topology, what is the interior of $(0,1]$ (with respect to the subspace topology)? @KarimMansour

I think it is (0,1) right @PaulPlummer?

because interior is the union of all open set U inside the standard topology that such that $U \subset (0,1]$

because the open set is the form of $U \cap Y$

@KarimMansour You need to focus a little more on what the open sets of $Y$ are

and

disagree

ok well what is the open sets of Y ? Suppose that $U_y$ is open in Y then $U_y = U \cap (0,1]$ such that U is open in the standard topology

Not just $(0, 1]$

nvm

so

the interior of (0,1] is actually

(0,1]

Yup

Same idea works for your original problem

yeah

I was trying to find example of the following

Give an example of a subspace $Y \subset X$ and subset $A \subset Y$ such that the interior of A in X intersected with Y isn't the same as the interior of A in Y.

which is trivial

based on the above

Yup. A simpler example would be to take $Y=\{0\}=A$, the interior in $X$ is the empty set and interior in $Y$ is $Y$.

Taking $X=\mathbb{R}$

nice