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16:00
Hey everyone!
@Secret just write it as it is... you can refer to my 2D representation, and use the zig-zag argument
@Ramanujan it's a rebuttal
@Secret or, just write it as (0,1,2,3,...)
Now the next question is this:
@AlessandroCodenotti Yeah, it should be in GP
(my question is not this actual question, but a concept of it)
3 hours ago, by Akiva Weinberger
@Secret Maybe find a bijection with $(\Bbb Z^+)^{\Bbb Z^+}$ and mimic the proof of Cantor–Schroeder–Bernstein?
@DHMO Now since we know $\mathbb{R}$ is uncountable, we cannot write it as what we have just discussed, so how can I have a better look at it to get some idea on how to set up a bijective function
@Secret what is the question?
> Show that there's a bijection $f: P(\mathbb{Z}^+)\to \{0,...,n\}^{\mathbb{Z}^+}$ for all countable $n$
16:06
@DHMO I am trying to prove $|\mathbb{R}|=|\mathbb{N}^{\mathbb{N}}|=|2^{\mathbb{N}}|$. Because I am trying to prove this, I cannot use the fact that $\mathbb{R}=2^{\mathbb{N}}$. Now since $\mathbb{R}$ is uncountable, I cannot even write it down in 2D form as we discussed earlier, so how can I have some idea on how to find the injective map from $\mathbb{R}$ to $2^{\mathbb{N}}$ and vise versa?
@Secret can you biject $\Bbb R$ with $[0..1)$?
@DHMO salut, les suites stable "eventualy constant", sont elles convergente dans la topologie cofinie ?
Can anyone give me any hint on how to find the values for $a_0 > 0$ for which the seris $a_{n+1} = a_0 + \sqrt{a_n}$ is convergent?
@Vrouvrou Tu penses quoi?
je pense que les suites stable converge dans dans tout les espaces
So I'm wondering about one thing. The task is to prove that the set of $m\times n$ matrices of rank $k < \min(m,n)$ is a smooth submanifold of $\mathbb{R}^{mn}$ with codimension $(m-k)(n-k)$
Do you think it'd work to consider determinants of submatrices?
16:13
The series, or the sequence? @DHMO
@Vrouvrou "tout les espaces" est un peu...
@Semiclassical the sequence
mais regarde la réponse que j'ai ici : math.stackexchange.com/questions/2117560/…
comment est un peu ...?
okay. If that does converge to some final value $a$, then it'll necessarily satisfy $a^2=a_0+a\implies a = \frac12(1\pm \sqrt{1+4a_0})$
@Vrouvrou What do you mean by "tout les espaces"?
@Semiclassical agreed
So a necessary (but possibly not sufficient) condition is that $1+4a_0\geq 0$.
16:15
@DHMO i mean in any topological space
@Vrouvrou a claim about "any topological space" is quite... how do you say... bold
@Semiclassical agreed
Not sure where to go from there, though.
For reference, given some $S \subset \mathbb{R}^N$, if at every point $p\in S$, there exists some $r$ and smooth functions $f_1,\ldots,f_{N-k}: \mathbb{R}^N \to \mathbb{R}$ such that $rk(df_1,\ldots,df_{N-k}) = N-k$ and $S\cap B_r(p) = \{x\in B_r(p) \mid f_1(x),\ldots,f_{N-k}(x) = 0\}$
@DHMO i don't understand
If that holds, $S$ is a k-dimensional submanifold of $\mathbb{R}^N$.
16:17
@Vrouvrou I mean that it is very risky to make a claim about all topological spaces
we say that $x_n$ is eventualy constant iff $\exists a\in E, \exists n_0\in \mathbb{N}, \forall n\in \mathbb{N}, n\geq n_0\Rightarrow x_n=a$ @DHMO
@DHMO And, actually, that doesn't really tell us anything: We need $a_0\geq 0$ if the recursion is to be well-defined.
Drat.
@Vrouvrou agreed
@DHMO and a is in all his neighborhood , so $x_n $ converge to $a$
It might converge to other things too in non Hausdorff spaces
16:20
@DHMO Looking at a few positive iterates, it certainly seems to converge so long as $a_0\geq 0$.
@Vrouvrou agreed
@Semiclassical but how can I prove it?
Yeah. :/
@DHMO @AlessandroCodenotti ok I have no idea, any hints or advice?
sooo I just read about synthesis of the sounds of musical instruments in a book about music theory and I was so fascinated by it I'm wanting to read Fourier Analysis now
could you guys suggest any good books for that?
@Secret basically suggest a bijective function from $\Bbb R$ to $[0..1)$... <-- @AlessandroCodenotti
16:22
I haven't been following what you're trying to do @secret
I'm actually working on my music composing skills in the trance production scene and I think a bit of Fourier Analysis might be of great help
Maybe the thing to do is introduce $b_n=a_n-a_0$. That might seem trivial, but it means that the boundary condition is now $b_0=0$. So now $a_0$ only enters as a parameter in the recurrence relation $b_n = \sqrt{b_{n-1}+a_0}$.
So now one just needs to prove that $x\mapsto \sqrt{x+a_0}$ is convergent for $a_0>0$.
Actually, if you write that as $b_n=a_n/a_0-1$, then that becomes $b_{n+1} = \sqrt{a_n}/a_0 = \sqrt{a_0(b_n+1)}$ with $b_n=0$.
Which seems even more convenient, since then the mapping is just $x\mapsto c \sqrt{x+1}$ for $c>0$. ($a_0=0$ is trivial).)
@Secret use arctan wisely
16:40
@DHMO Take $f(x)=\frac{1}{\pi}arctan(x)+\frac{1}{2}$ for $x/neq 0$ and $f(0)=0$ for $x=0$. Then $f$ is bijective.
@Secret nice
16:54
How should I find a bijective function between $2^{\mathbb{N}}$ and $\mathbb{R}$ given I cannot enumerate any element of $\mathbb{R}$ to guide my partitions?
@Secret but you can represent them in binary
$\tanh(x)$ also provides a bijection from $\Bbb R$ to $(0,1)$
$\dfrac x{1+|x|}$ also works, and has the benefit of also being a bijection between $\Bbb Q$ and $\Bbb Q\cap(0,1)$.
@DHMO How do we prove that? (yes I am trying to go very basic on $\mathbb{R}$ because the thought process needed for this and the proof $|2^{\mathbb{N}}|=|\mathbb{R}|$ are the things I need to understand how to deal with uncountable sets
@Secret well, irrational numbers are limit points of Q
The standard argument is to give a bijection between $[0,1]$ and $P(\mathbb{N})$
Can anyone help me with surface integral ?
Using the binary representation of any $x \in [0,1]$
Any $x$ will be of the form $x = 0.x_1 x_2 x_3 x_4 \cdots$ where the $x_i$ represent digits.
And since each $x_i \in \{0,1\}$, I'm pretty sure you'll understand the argument now.
in surface integral calculation when do I take normal unit vector when when do i take anynormal vector?
@DHMO There are a few similar question on the main. For example, How to define a bijection between $(0,1)$ and $(0,1]$?
17:27
@Secret ^
It's pretty much always the unit normal vector in surface integral formulas as far as I remember.
I ll show you the example on old exam they did not use unit normal
z = x^2+ (y^2/4 ) , z= 4x
About the only case where I'd imagine otherwise is when you're doing frame stuff, and therefore getting the normal vector from the tangent vector of the trajectory
@SteamyRoot @dhmo meanwhile still reading the planetmath link How do we know that there are countable number of entries followed after the decimal point, and not some larger cardinal?
the surface is ((x-2)/2 )^2 + (y/4) ^2 = 1
17:30
@Secret every real number is defined as, well, that depends on your definition
Although I see that you have moved on $|\mathcal P(\mathbb N)|=|\mathbb R|$ in the meantime. (It is useful to know that instead of $\mathbb R$ we can take any interval of our choice - open, half-open, closed. But for that Cantor-Bernstein is a much easier argument than explicitly constructing a bijection.)
i calculated the rot of the vector field and it was ( -1, -1 ,-1 ) i dot it with ( -4,0,1)
I got that normal from the plane equation
The intersection of those two surfaces, you mean?
Not the unit normal, I note.
17:31
in the solution of the exam they used that normal with out dividint by sqrt 17
hmm.
What are they integrating on said surface?
I got everything right up to that factor of sqrt 17
same region
@MartinSleziak That theorem still need injective functions, but since I cannot really write out what $\mathbb{R}$ look like before proving (and that I cannot be certain the decimal expansion has countable number of entries), I don't have any pictorial thing to help me to set up a partition that my injective function can use
@Secret How could it not be countable?
No, I mean, what's the vector field? It sounds like they're doing $\iint \vec{F}\cdot d\vec{S}$ but I wanted to be sure
17:33
I mean, you can just construct the $x_i$.
okay F = ( y , 5y^4+z , x+y^5 )
RotF = ( -1 , -1 ,-1 ) all these calculation i did right
@Secret Well, we have $|\mathbb R| = |(0,1)| \le |[0,1)| \le |[0,1]| \le |\mathbb R|$; so we simply apply Cantor-Bernstein to this.
@Secret do you know any definition of real numbers?
Actually, I should've said $\iint (\text{rot }\vec{F})\cdot \hat{n}dS$.
yes i got that part
17:34
Which presumably means they initially asked for $\oint \vec{F}\cdot d\vec{l}$?
Just wanting to make sure.
yes
they put it in standard form
pdx+qdy +R dz
what they did were they got the normal from parametrization
I'm a bit nonplussed, I'll confess. $\hat{n}$ should be a unit vector, but $(-4,0,1)$ isn't.
@DHMO I have ZERO background in real analysis. This is why I have postpone any investigation about uncountably infinite structures until I bump right into them in Munkres
17:36
(x,y,4x)
that was their reason
@Secret do you want to go through its definition?
@Secret If you want - for some reason - avoid "drawing pictures", you can just formally verify that $x\mapsto \frac{x}{|x|+1}$ is a bijection between $\mathbb R$ and $(-1,1)$. But I have to say that I do not see what the advantage of being very formal against plotting a simple graph of a function is.
I'm not sure if going through Munkres with zero background in real analysis is a good idea
Did they get a finite result at the end?
In cofinite topology


Please there two type of convergent sequence: the eventualy constant sequence which converge to a unique limite, and the injective sequences which converge to any limite . right ?
17:37
yes 24 pi
Hm, so that sqrt(17) really does matter.
i got 24 pi / sqrt 17
And I'd tend to agree with you.
i mean its comfusing why in this example they did not use unit normal
One (infuriating) possibility was that it was a typo/oversight.
17:38
is there a way to get the right solution and be sure?
@Vrouvrou does 0,1,0,1,... converge?
it could be but this is a corrected exam
In principle, yes.
so should not be the case
Aye. So either we're missing something or they really missed something.
17:40
@DHMO no not alwayse but this sequence is not injective
this is Teds field
I wish he were here
But if we wanted to do the line integral directly, hmm.
Hmm it would be way harder
thats why we do stokes
@Vrouvrou that's not completely correct, consider the cofinite topology on $\Bbb N$ for simplicity, does $1,1,2,3,4,5,...$ converge to something?
they give examples to force us to use stokes
17:41
What we'd need to do is find a parametrization of the curve, which is a particular cross section of a paraboloid.
Well, certainly. But I don't see another way to check the answer.
@BalarkaSen can we discuss something ?
hmm okay thanks semi
I ll do some other examples and see whats going on
this topic really driving me crazy =p
@AlessandroCodenotti i don't know
17:43
take a point $x\in\Bbb N$ and an open nbhd of that point, does it contain points from the sequence?
@AlessandroCodenotti I have a question about the validity of cofinite topology at all
btw semi
if we work on the parametrized curve
@AlessandroCodenotti let's say we have the cofinite topology on R
it does turn out to be the case , we dont need normal vector
i mean unit normal
in that case, {1,1.4,1.41,1.414, 1.4141, ...} converge to anything right
17:44
Well, sure.
because from the formula NDS take care of the length
You don't have to worry about the surface normal if you never bother with a surface integral.
but the thing i got the right normal from the equation of the plane
You'd still need to do a unit tangent vector, though.
17:45
@AlessandroCodenotti then how is sqrt(2) defined?
So some level of normalization can't be avoided.
@MikeMiller I'm pretty sure I solved the $\Bbb Z^n\hookrightarrow B_{n+1}$ thing (if you haven't already)
@DHMO as the limit of a sequence of rationals in the standard topology on $\Bbb Q$ (probably an equivalence class of such sequences? I'm not very familiar with the Cauchy completion construction)
or a Dedekind cut if you prefer
@AlessandroCodenotti but the limit can be anything
hi @MikeMiller
17:48
in the standard topology was the key part
@AkivaWeinberger Someone else suggested to me to use $\sigma_1^2$, $(\sigma_1 \sigma_2)^3$, etc
Which works.
@AkivaWeinberger @MikeMiller can we discuss the following problem ? No solution though if you guys have it.
Hi @SemiC
@AlessandroCodenotti hmm... thanks
if you're willing to change the topology $\Bbb R$ is just an uncountable set
you could take the cofinite topology on any other set of the same cardinality and get an homeomorphic topological space
17:49
Suppose A is closed subspace of X. Show that X has HEP with respect to A iff $(I \times A) \cup (\{0\} \times X)$ is a retract of $I \times X$.
suppose we go the side <--
We know that there exists homotopy $F$ satisfying conditions of the retraction.
@AlessandroCodenotti like 2^N
@MikeMiller Oh. In any case, my idea was to view $B_n$ as the group of rearranging the points $\{0,1,\dots,n-1\}\subseteq\Bbb C$
That was their idea too.
@AlessandroCodenotti let $x \in \mathbb{N}$ an open neighborhood of $x$ is $G$ such that $\mathbb{N}\setminus G$ is finite , but i don't know what is the limite
The first swaps two points twice, the next rotates three of them 360 degrees, the next rotates four 360, et
17:51
and for every nonzero point $i$, let $e_i$ be a 360 degree counterclockwise rotation around the origin. @MikeMiller
I see. Similar idea.
@Vrouvrou does that nbhd intersect the sequence?
No I mean, for bijective constructions between countably infinite sets and between finite sets, we can easily write down the resulting sequence to work out how to construct the bijective function

I know that I can verify that some real function is bijective. What I am trying to learn is how to find them no matter how crazy the uncountable set is. Unlike countable sets, the reals cannot be wrote into a sequence, and if you don't know the function beforehand, the only help from the diagram is you know what the required image of that function should be.
@DHMO I think we can, do you want to do it here or in some other room?
nvm actually it is quite simple
@Secret you want to create a new room again?
I think I saw that you proved that we can't have $\Bbb Z^n$ or higher in $B_n$? @MikeMiller
Involving cohomology?
17:53
@DHMO nope, I think we can chat in number theory if it gets too busy here?
@Secret ok, let's go there
@AkivaWeinberger Right.
@Adeek Sure
There is also set theory chat room @Secret @DHMO. (The topic would fit if you plan to talk about this.)
@AlessandroCodenotti $\{u_n, n\in\mathbb{N}\}=\{1,1,2,3,4,5,6...\}$ for example $G=\{1,3,4,5,7,...\}$ is neighborhood of 1 and $\mathbb{N}\setminus G$ intersect the sequence at 2 , but the itersection stay finit then 1 is a limite
@AlessandroCodenotti so i think that $u_n$ converge to any limite
17:59
it does, because the neighbourhood of any point intersects it in infinitely many points
yet it's not injective
@SteamyRoot Well I think one question I can ask is why when we talk about limits in analysis, sequences and so on we never need to worry about the cardinality of the infinity we are using?. Factorign that (and assuming I have not overlook any justification in the proofs, I see no reason why we cannot have uncountable number of digits after the decimal (or perhaps the justification is we are representing rela numbers with rationals and rationals are countable)?
No, it's pretty much a completely different infinity.
You pretty much just have $\mathbb{R}$ which you extend to $\mathbb{R} \cup \{\infty,-\infty\}$
Which is useful if you want certain limits to be defined, or certain sets to have an infimum, supremum, etc etc.
But on the other hand, it doesn't mix well with the operations $+$ and $\cdot$ on $\mathbb{R}$.
And back to having an uncountable number of digits after the decimal: how would you even define such a thing?
You can't write it as $0.x_1x_2x_3x_4\dots$, since that notation alone implies that the $x_i$ are countable.
Not to mention you'd need to define the sum over an uncountable set...
@AlessandroCodenotti so i can't find a sufficient condition for the convergence of a sequence ?
@TedShifrin hi ted :D
hi @Kasmir
18:09
@TedShifrin i solved a question about stokes theorem , can you please verify with me the result ?
Okay
F = ( y , 5y^4 z +z , x+y^5 )
@Vrouvrou you can, but it's not injectivity
hi @Ted
z = x^2+(y/2)^2 , z= 4x
hi @Alessandro
18:11
intersection of those
What type of object is the infinity we use in analysis, pretty much the only thing I knew is that it is defined to be larger than all finite numbers ($\aleph_0$ also have the same property), Does it have ties with set theory at all?

We can have uncountable index set. Thus the representation may be written as $\sum_{i \in I}b^ix_i$ where $b$ is the base. Granted, such expansion cannot be even written down as some sequence, but I don't see anything inconsistent here nor it is explicitly ruled out (and we can then have real number bases....? )
@AlessandroCodenotti what is please
oriented how, @Kasmir?
I calculated the rot F to ( -1,-1,-1 )
Positive sense
I don't know what that means
positive sense relative to what perspective?
18:12
counter clockwise
if you look from up
Okay last step is this
@Vrouvrou think about what's the difference between a sequence with a unique limit and one converging to everything. You can worry about not converging sequences later
(-1,-1,-1) . ( -4,0,1 )
surface integral of that
I got the normal from the equation of the plane
@Secret what is $2^{\aleph_0}$?
18:13
z= 4x
UNIT normal
this is an old exam with solution , they used that normal not unit
@Secret Even if you could define all of that (I'm extremely sceptical of that), it doesn't matter.
I know what i got was 24pi / sqrt17
their answer is 24 pi
@DHMO That's $\mathfrak{c}$. What $\mathfrak{c}$ is depends on CH
18:14
what they did was parametrization of surface like this
They're wrong. But there's more to do.
(x,y,4x )
It should be clear that the binary represenation of any real number can be done with countably many digits.
if you do it that way you do get 24 pi
but what is wrong with my way of thinking ?
You haven't told me your way of thinking.
18:15
okay
I found the normal from the equation of the plane
since its easiar
You still need unit normal.
Then what?
then I calculated the abs of that to sqrt (17 )
But my answer only differs with that multiple
Keep going.
1 / sqrt 17
thats it , they cant be both right
Huh? All you've done is the dot product. How do we find the surface integral?
18:16
Rot F . ndS
Steamyroot: I see
hi @AndrewT
I did that on the surface ((x-2)/2 )^2 + (y/4)^2 < 1
Hi @TedShifrin!
its an ellips with area
4*2 * pi = 8pi
the dot product of rot and N = (-1,-1,-1 ) ( -4,0,1 ) = 3
18:18
But that's not the surface. That's the projection of the surface into the $xy$-plane. Is area in the plane $z=4x$ the same as area in the $xy$-plane?
hmm what do you mean ?
the ellips is flat
does it matter where it is ?
Your equation is in the $xy$-plane. The actual curve is in the plane $z=4x$.
The ellipse isn't flat if $z$ is different for different $x,y$.
Consider this question: Take a cylinder $x^2+y^2=1$ and slice it with $z=4x$. You get an ellipse in that plane. What is its area?
hmm
okay let me think for a second
18:20
Does the infinity used in analysis have a set theoretic meaning even though it is not a cardinal?
What infinity used in analysis?
that symbol appeared in the limit symbol
as well sequences and related things
it doesn't mean infinity
It is symbolic. There is no number involved.
It's a way of saying "x gets arbitrarily large"
$\lim\limits_{x\to\infty} f(x) = L$ means $\forall \epsilon > 0: \exists N: \forall x > N: |f(x)-L| < \epsilon$
18:23
I see
sqrt 17 ?
I don't like using $\delta$ in that context, @DHMO. To me it suggests a small number.
don't try to guess the answer. think geometry.
I did this r = ( x,y,4x)
18:23
@Kasmir: times $\pi$. How did you get that?
guys I'm in deep trouble
@TedShifrin done
partial of x times partial of y
and took absolute value of the cross product
That's even worse, @DHMO. $\Delta$ signifies "change in".
OK, @Kasmir, or you can see it geometrically. So the area in the tilted plane is $\sqrt{17}$ times the area of the projection in the $xy$-plane.
Much better, @DHMO.
I was applying to a combinatorics program in Vienna and tonight is the deadline and although my application will be complete by tonight the letters of recommendation also have to sent to the program before tonight, and that will take at least one day more
18:25
So, @Kasmir, you have $3/\sqrt{17}$, but then the area is $\sqrt{17}$ times the area of the ellipse in the $xy$-plane.
@AlessandroCodenotti what do you think about this answer : math.stackexchange.com/questions/2117560/…
I did not think that area of tilted ellips would be counted differently
It's important to think correctly!
I hadn't caught on to that either, though I should've.
18:26
yes :D
The area is in the plane of intersection, not the $xy$-plane.
looks correct @Vrouvrou (the 1 and 2, I haven't read the rest of the reasoning)
@Kasmir: I'm going to put up a picture here. But I have to find it first.
Pretty obvious in retrospect.
@TedShifrin that would help alot thank you Ted :D
18:28
@Kasmir @Semiclassic: Consider the area of $S$ versus the area of the projected rectangle.
Use the angle between the two normals to relate the area.
Alternatively, the two are related by a shear transformation.
Hmm i see it more clear now , this is how the proof of the absolute value of the cross product gives us the element area DS
@AlessandroCodenotti thank you, please if you think that it is an interesting question make +1 please
18:30
But use the cosine of that angle, @Kasmir, and you can get your formula and understand it geometrically.
Thanks alot Ted ! now I can say I finally know how to do stokes theorem questions on exam :D
Wish I knew a good way to show the projection to the z=0 plane as well.
When I wanted such pictures, I programmed the parametric curves into Mathematica, @Semiclassic.
Yeah. This is using ContourPlot3D
But, I mean, it's easy enough to get the contour plot in the xy plane. I meant to show it in the same picture as above.
Analogously to the diagram you gave.
18:33
Oh, draw the two separately, and then do Show[Fig1,Fig2].
I'd be fine with that if both 3D objects. But a contour plot of the xy projection is a 2D object.
So the issue is that I don't know how to make that 2D graphic into a 3D graphic.
Yeah, you can't mix those two.
I'm sure there's a way---I think the Texture command might work---but it's not something I know off the top of my head.
Yeah, Texture would work if I could fix some technical issues. But that'd be work :P
Ted I got another question about the proof of , if the line integral is path independent then the vector field F has a potential
the proof starts with this function
U (x,y) = line integral P (s,t) ds +Q (s,t) dt
I know am not giving you enuf data but we take two points (a,b) and (x,y)
and we then go along x -axis then y -axis
I am familiar with the proof. :)
18:40
:D
Bah, why do I persist in writing silly/wrong things in Mathematica.
Can you tell me how did they define that function U (x,y) ?
btw our notation U , is the potential function
You just typed the definition up there.
Hmm yes the book says we define U (x,y) as such
When you want to compute $\partial U/\partial y$, you do $\int_a^x P(s,b)ds + \int_b^y Q(x,t)dt$.
When you want to compute $\partial U/\partial x$, you do a different path: $\int_b^y Q(a,t)dt + \int_a^x P(s,y)ds$.
Path-independence says both give the same answer.
Draw a picture of the paths in your notebook.
18:43
Hmm i did
But the thing is U (x,y) gives only information about the end point
When I accepted that defition , the rest of the proof was not hard to follow , integral mean value and all
Only got stuck on how they came up with that function and why does they equate it with line integral F.dr
Oh, I wouldn't do that. I'd use the Fundamental Theorem of CAlculus to differentiate the integral.
Draw the paths and parametrize them.
hmm the way they proved it was to show partital of U wrt x = P
partial U wtf y = Q
(U (x+h , y) - U (x,y) ) /h
and then same but along the y direction U( x , y+k ) ..
in one case dt = 0 and other ds = 0
Better just to use the FTC that tells you how to differentiate an integral from constant to $x$.
I am sure you got a better proof but dont think they will accept other ways =p
So you need to take time to figure out where I got my two different expressions for the line integral.
18:48
is anyone here from Austria or Germany?
You need to figure it out.
hmm okay but one thing i did not get is the U (x,y) = line integral over the curve
does that statment make sense?
U (x,y) = = line integral F.dr = line integral P (s,t) ds + Q ( s,t) dt
@Kasmir: I'm getting annoyed. I took the time to type out the detailed formula. And you're ignoring it. Talk to me after you've understood what I typed.
Okay I did not ignore it
But I did not want other way to prove it just all
they will not accept other ways of proof exept the ones from the book on exam
Anyway thank you for all your help :)
I'll have a question for you, @Ted, if I can figure out what the heck I'm doing. :/
19:07
"Proof" that $|\Bbb R| = |\Bbb N|$:
1. It is a fact that $|\Bbb Q| = |\Bbb N|$.
2. In the Dedekind cut definition, a real number is defined as a partition of $|\Bbb Q|$ where the larger set is not empty.
3. For example, if you have 5 objects in a row, you would have exactly 5 ways to partition them such that the right set is not empty.
4. By analogy, since $\Bbb Q$ is ordered, consider it as a row, and the number of ways to partition it would be $|\Bbb Q$.
5. Since each partition defines a real number, the result follows.
@Secret ^
Switzerland
19:49
Let $f:\Bbb R\to\Bbb R$ be a real function. If it is differentiable, we write:
$\displaystyle \forall x \in \Bbb R: \exists L \in \Bbb R: \lim_{h \to 0} \frac{f(x+h) - f(x)} h = L$
If its derivative is continuous, we write:
$\displaystyle \forall a \in \Bbb R: \lim_{x \to a} \lim_{h \to 0} \frac{f(x+h) - f(x)} h = \lim_{h \to 0} \frac{f(a+h) - f(a)} h$
A counterexample is $f(x) = \begin{cases} x^2 \sin \left( \dfrac 1 x \right) & x \ne 0 \\ 0 & x = 0 \end{cases}$.
I am currently trying to prove if $f:X \longrightarrow Y$ is a bijective map and $X$ is hausdorff and $Y$ is compact then $f$ is a homeomorphism.
I know that this means $X$ is compact and $Y$ is hausdorff but I don't know where to go from there
Would I show that $f(U)$ is open in $Y$ if U is open in X?

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