Mathematics - 2018-04-28 (page 1 of 3)

1:25 AM

hey, I said approximately :P

and 1/2 is just a silly constant

who cares about those?

maybe I should have used some big "Oh!"s

can't there be a bijective function for two sets of unequal size? like $f: X \to Y$ by $f(a) = a^{\frac{1}{2}}$ for $a \in X$ and $X = \{4,9,16,25\}$ and $Y = \{-5,-4,-3,-2,2,3,4,5\}$ every element in $X$ gets mapped to two elements in $Y$ so it's surjective and no two domain elements get mapped to the same codomain element so it's injective.

How do you define "size" @Obliv?

though, for most typical definitions, the answer is "No". Two sets are the same "size" if and only if there is a bijective function between them.

i know that's why i said size instead of cardinality

i just meant no. of elements

well, for finite sets, the number of elements and the cardinality are the same thing

well then i'll rephrase the question and ask if the function i defined above counts as being bijective

1:38 AM

That function isn't bijective

a function cannot, by definition, map one element to two different things

in that case, it is not a function at all, but a relation

(or, rather, you could make sense of a "multivalued" function by treating it as a relation)

ohh

Sha Vuklia

@Xander @Ted yea thanks, I guess I expected $\sin z$ to look like a wave spreading in all directions, but I really had no grounds for that, other than the $\mathbb R$ analogy

thanks @XanderHenderson that's very helpful :)

Sha Vuklia

I did not expect this at all (found it online)

but I guess, it is what it is

So, like $$A\mathrm{e}^{i\lambda x} + B\mathrm{e}^{-i\lambda x} = A\sin(\lambda x) + B \cos(\lambda x). $$

Sha Vuklia

1:45 AM

actually this is better

gah... where is the error???

found it

'cause $A$ and $B$ are just constants

@XanderHenderson how did your PDE class go

and, for the record, $A \ne A$

it went fine

we solved a wave equation problem

on some kind of rectangular domain

with a silly initial condition

then I "forgot" to erase the whiteboard and gave them the typo-laden quiz that I was directed to give them

(I am going to get AMAZING evals this quarter)

@XanderHenderson great!

For their homework, they are being asked to solve a heat equation on a circle; their answer is, I think, meant to be the Fourier series of the initial condition (more or less)

which seems reasonable

but (1) I don't know if these guys know much Fourier analysis and (2) even if they do, they are going to write everything in terms of the real Fourier series, which makes me cry :'(

$$f(x) = \sum_{k\in\mathbb{Z}} \hat{f}(k) \mathrm{e}^{ikx}$$

none of this sines and cosines BS :\

1:58 AM

what fourier do you even need to know for that

it's just separation of variables

they'll memorize a formula

GAH!

That makes me even more sadder-er

They should be more gooder

There is an ice cream truck driving through my neighborhood

the were just playing Swan Lake

and now they are playing Christmas muzak

WTF?

4:21 AM

Axiom of choice actually puts a very powerful constraint on what kinds of infinite sets are allowed

Alex D

Hi all

Could someone explain this proof here? math.stackexchange.com/questions/1540739/…

In particular, is the convergent subsequence a consequence of the bolzano weierstrass theorem?

Also, what does the D() mean on this line?

"There exists $e>0$ such that $\forall u,v\in D (|u-v|<e\implies |f(u)-f(v)|<1).$"

Ohhhhh D is the set and the implication is just written in parenthesis, right?

Sorry that makes sense

The Hitchhiker's Guide to the Galaxy (film)

10:09 AM

[Random worldbuilding]

The most powerful superpower is not omnipotence

It is preorder

Why? Because the moment when you say A is more powerful than B, you have already imposed a preordering into the situation

Those who wield the power of a preorder will overpower everything because they can freely rearrange the preordering

(or not overpower everything because it can be reversed

Which then begs the million dollar question: What happens when preorder encounters omnipotence and logical paradox. Who is more powerful

The answer, requires the entire universe to compute

The Hitchhiker's Guide to the Galaxy is a 2005 British-American science fiction comedy film directed by Garth Jennings, based upon previous works in the media franchise of the same name, created by Douglas Adams. It stars Martin Freeman, Sam Rockwell, Mos Def, Zooey Deschanel and the voices of Stephen Fry and Alan Rickman. Adams co-wrote the screenplay with Karey Kirkpatrick but died in 2001, before production began; the film is dedicated to him. The film grossed over $100 million worldwide. == Plot == The film begins with a Broadway-style number, "So Long, and Thanks for All the Fish", sung by...

space, time, reality, mind, soul, power

These are all just that, words

soul > time > reality > mind > space > power

It is all a preordering and is thus subjected to set theory

But what about... $\pi$

$\pi$ is transcendental, thus by definition any circular objects are law breakers

They will not be confined by any categorisation or classification we humans imposed on it and they will break any rules at will

We are part of Cthulhu, and thus the cosmic edritch aabdominon are not subscribed to what we called "rules"

blogs.ams.org/blogonmathblogs/2018/04/22/…

BeginningMath

1:38 PM

can anyone please help me with a simple question?

if i toss a coin 50 times, how can i know what are the odds that i will get 4 times in a row heads?

(1/2)^4 times the number of permutations in 50 slots where HHHH appears

BeginningMath

2:00 PM

thank you very much for your answer @Secret

so i'm trying to understand - for instance,the chance of getting 4 heads in the first 4 tossings out of 50 is equal to the chance of getting 4 hads in the last 4 coin tossings?

heads*

is equal to the chance of getting 4 heads in the last 4 coin tossings out of 50?

you only want the probabbility to get 4 heads in a row, not where you get 4 heads in a row, thus the probabbility should be the same regardless of where that string of HHHH are, thus you add those all up as mutually exclusive events

BeginningMath

2:18 PM

thank you very much, i understand

2:47 PM

[Ordinals]

The 3 Tuple (a,b,c) captures the growth by indicating the easiness of transversing from a starting element x to the destination y. a is the operation that is used to get from x to y, b denotes what is being used each step, and c denotes how many repetitions

For example $1 \to 3$ with the notation $(+,1,2)$ means $(((1)+1)+1) = 3$

As we shall see later, b, c also have dependence on a which is why the nature of a is important

$0\to 1\to 2\to \cdots < \omega : (+,1,1)$

$\omega\to \omega 2\to \omega 3\to \cdots < \omega^2 : (+,\omega,1)$

$\omega^2\to \omega^3\to \omega^4 \to \cdots < \omega^{\omega} : (+,n,\omega)$

Note that in the 3rd case, $c = \omega >$ finite, thus the ordinals in the sequence are additively indecomposable

$\omega^2\to \omega^3\to \omega^4\to \cdots < \omega^{\omega} : (\times,\omega,1)$

3:28 PM

if $K = \Bbb Q(\sqrt{-17})$ and $K_1 = K(\sqrt{17})$, is there some obvious reason that for some prime $\mathfrak{p}$ of $K$, we have $2 \notin \mathfrak{p}$ and $17 \notin \mathfrak{p}$?

Certainly if $2 \in \mathfrak{p}$ then $17 \notin \mathfrak{p}$ else $\mathfrak{p}$ is not prime, but if $2 \notin \mathfrak{p}$ I can't see how to show that $17 \notin \mathfrak{p}$ (which would be the suggested outcome by the exercise I'm doing)

loch

I'm not sure if I understood your question.. if you're asking for primes in $O_K$ that don't contain 2 and 17 - they clearly exist (by taking any prime lying over a prime not 2 or 17)!

Q: Hilbert class field of $\Bbb Q(\sqrt{-17})$

3:44 PM

@loch I've asked it as a question on the main site, I may have missed the point of the exercise completely though!

0

Let $K = \Bbb Q(\sqrt{-17})$. I wish to show that the Hilbert class field $H$ of $K$ is the extension $H = K(\alpha)$ where $\alpha = \sqrt{(1 + \sqrt{17})/2}$ (following an exercise in Cox's Primes of the Form $x^2 + ny^2$). I have already shown that $\operatorname{Cl}(\mathcal{O}_K) \cong \Bbb...

Let (P, <=) be a partially ordered set.

Say an element x in P is in the "centre" of P if for every y in P we have x <= y or y <= x.

Is the supremum of a collection of elements in the centre of P also necessarily in the centre of P, assuming that the supremum exists?

Mary Star

Hello!!

Let $\mathbb{R}$ provided with the metric $d(x,y)=|x-y|$. I have shown that the collections of sets $$S_1=\left \{\left (\frac{x}{2}, \frac{3x}{2}\right ): 0<x<1\right \}, \ \ \ \ \ S_2=\left \{\left (x-\frac{1}{2}, x+\frac{1}{2}\right ): 0<x<1\right \}$$ are open covers of $A=\left \{\frac{1}{n} : n\in \mathbb{N}\right \}\subset \mathbb{R}$.

To check if they have a finite subcover, we have to check if $S_1$ and $S_2$ respectively, have a finite subset which is also a cover of $A$, right?

loch

@ÍgjøgnumMeg I think you can use K_1 = K(\sqrt{-17}) = K(\sqrt{-1}), i.e. let u = -1

@loch well $K = \Bbb Q(\sqrt{-17})$ already but then $K_1 = K(\sqrt{17})$ means $\sqrt{17}/\sqrt{-17} \in K_1$ so $K_1 = K(\sqrt{-1})$? Is this what you mean?

loch

Yes

3:52 PM

Nice, that's probably (definitely) the way to do it, since it's a similar trick to a worked example given in a previous section lol

The Spacelords - Water Planet (2017) (New Full Album)

►

dee oh pee ee

@BalarkaSen why should I care about the Poincare-Lefschetz duality

what good is $H(X,\partial X)$

It's the right Poincare duality for manifold with boundaries. I don't know what else you want to know.

$H_k(M, \partial M)$ is non-stupid in $k = \dim M$ unlike $H_k(M)$, for example (if $M$ has boundary)

So I can prove that $H^k(M)\cong H^{n-k}(M,\partial M)$, both de Rham

but I'm not sure what the RHS is telling me

$H^k_{dR}(M, \partial M)$ is made up of $k$-forms which vanish on $\partial M \subset M$. Think of it like compact support; if you had a noncompact manifold you'd want it to vanish at infinity.

4:05 PM

I know the definition, thanks

I have no interest in helping you with imprecise questions, then. Bye.

Meh, I'm drowned in workload and cracking. Sorry for that, @0celo7, I realize I'm probably being unhelpful. My point was more, having compactly supported cohomology in noncompact manifolds is what gets you the right Poincare duality, eg.

Not a very intelligent point probably.

I don't know what would be a more intuitive description of that group.

@BalarkaSen It's ok.

I know why that group shows up in PD, like I said I have a proof in the de Rham case using Hodge-Morrey.

But what's an application of this duality, say.

Okay I see. Hm.

Hodge-Morrey generalizes the Hodge decomposition to allow for boundary conditions.

Ah

I have never seen that.

4:18 PM

That's because the proof is quite a bit harder.

Is there any way that i get points here, as i move x in $(x \cos x, x \sin x)$? @LeakyNun

I don't think it's possible to generalize the one in GH or Warner.

Strange.

@Silent desmos.com/calculator/0lkm2r7pyq

@BalarkaSen hi, are you familiar with filters?

Nope, @Leaky. I think @Akiva is though.

4:19 PM

also, I have a question in chain homotopy

thank u so much!

@BalarkaSen If you knew their proof you wouldn't find it strange!

@BalarkaSen why do we require that $g-f = dP+Pd$ even though the $Pd$ is not used?

I mean, can't we set $g-f=dP+2Pd$?

@LeakyNun Ah, by the way, I looked for that pdf but I can't find it anymore... What's your question about filters?

@AlessandroCodenotti well I was not getting the generalization of limit (as in analysis) via filters

4:22 PM

Ocelot: H(M, dM) is the homology of M with a cone attached to it's boundary. You most often use it if you want your intersection-theoretic pairing to be non singular.

Or whatever you want to call the pairing that gives you Poincare duality

I think you can use the Lefschetz duality to prove a version of the Poincare-Hopf theorem for manifold with boundaries, say. (With vector fields vanishing on the boundary)

@BalarkaSen Oh, that would be very nice.

Or at least that's one way to do it.

Lefschetz duality eg can be used to prove the half-lives-half-dies principle for 3-manifolds: if M has a surface boundary components S, then the map HS -> HM has half of the homology in the kernel

@MikeMiller Wow that looks like Lefschetz hyperplane theorem

Well, "looks like"

Semiclassical

4:25 PM

fun fact: my thesis secretly has an example of the Picard-Lefschetz formula in it

Maybe only superfically though

@MikeMiller Where is the mistake here? If there is one math.stackexchange.com/questions/2756119/…

@LeakyNun Why this is not workinf for negative t's?

@Semiclassical maybe you can help too

@Silent it is

4:26 PM

@MikeMiller I see.

is an image in my thesis

I do not draw images

@MikeMiller @BalarkaSen I'm looking for a quick application of $H^k_\text{dR}(M)\cong H^{n-k}_\text{dR}(M,\partial M)$.

Because I am an unreadable author

Yeah I saw

4:27 PM

I can confirm ^

I'm just throwing shit

@LeylaAlkan maybe you know where I'm wrong, math.stackexchange.com/questions/2756119/…

@LeakyNun or you.

I have used that duality n+1 times but it's hard to immediately name examples

Oh, Alexander duality for submanifolds of S^n.

@LeakyNun Well a filter $F$ converges to a point $x$ if every open nbhd of $x$ is in $F$

Wise @BalarkaSen, where is my mistake?

4:29 PM

@AlessandroCodenotti why?

If you think about sequences as functions $f:\Bbb N\to X$ you can recover the usual notion of convergence, $f(n)\to x$ is the same as asking that the pushforward of the Fréchet filter on $\Bbb N$ through $f$ converges to $x$

as in, what is the intuition behind?

what is the Frechet filter?

@Waiting Please don't spam pings. I am sure whoever knows can help. I certainly don't!

3

the cofinite filter, it is made of all the cofinite subsets

@Secret It's you!

4:30 PM

@MikeMiller You mean I can get a proof of Jordan curve theorem from it?

@Waiting and who told you you're wrong? (not that I'm saying you aren't)

For smooth spheres at least (lame but ok)

@Secret When I posted this answer here I though it's a brilliant one, but the no one voted it. Maybe it's wrong? math.stackexchange.com/questions/2756119/…

@AlessandroCodenotti ok more context. They defined $\displaystyle \lim_{x \to c} f(x) = L$ as "the filter around c tends to the filter around L"

@0celo7 For smooth embedded curves, since you're going to use the normal bundle. For these I think it's more or less already trivial.

4:32 PM

it's just 2 hours, just wait for it

it is natural to have new answers not get lot of votes just yet

Gabriel Romon

@Waiting where's your book ?

"The filter around" means the neighbourhood filter?

@LeakyNun Just kidding people! ;)

@MikeMiller Yeah.

@LeakyNun Hm, it's not immediately clear to me if writing $f - g = dF + 2Fd$ doesn't harm anything. What if you're in $\Bbb Z/2$?

4:33 PM

@AlessandroCodenotti well they permit other filters. For example, there's the "top" filter representing the filter around infinity

@GabrielRomon What book? You mixed the users.

and I find this unification quite intriguing

meanwhile, on that integral question, I don't know how to visualise $\Gamma_{\epsilon}, \Gamma_{\infty}$

unification as in a unified theory behind everything

@LeakyNun I guess that agrees with the neighbourhood filter of infinity in some compactification though

What's the source?

Gabriel Romon

4:34 PM

@Waiting Aren't you Chris's sis ?

I'm looking at Bredon, all of his applications require more machinery than I'd like. No point in giving an application of some analysis that requires a book of topology behind it.

By the way, just out of curiosity, does any understand my answer here? math.stackexchange.com/questions/2756119/…

@AlessandroCodenotti it's in a definition inside a library of a proof assistant

How to they define convergence of a filter to another filter? I've always seen convergence of filters to points

@GabrielRomon I'm Waiting.

4:35 PM

/-- `tendsto` is the generic "limit of a function" predicate.
  `tendsto f l₁ l₂` asserts that for every `l₂` neighborhood `a`,
  the `f`-preimage of `a` is an `l₁` neighborhood. -/
def tendsto (f : α → β) (l₁ : filter α) (l₂ : filter β) := l₁.map f ≤ l₂

how does the floor thing in the sum become a harmonic number?

the ≤ is reverse inclusion @AlessandroCodenotti

Q: Arc contribution in $\int_{-\infty}^\infty \mathrm{d}z \frac{e^{-z^2}}{z-1}$

5

Consider an improper integral with a pole on the integration contour at say $z=1$, $$ \tag{1} I = \int_{-\infty}^\infty \mathrm{d}z\ \frac{e^{-z^2}}{z-1+i\epsilon},~~~~~\epsilon>0. $$ Let $$f(z) = \frac{e^{-z^2}}{z-1+i\epsilon}$$ then $$ \sum_{residues~inside~\Gamma} = 0 = \oint_\Gamma f(z) ...

integration complex-analysis improper-integrals contour-integration residue-calculus

and map is defined as follows:

/-- The forward map of a filter -/
def map (m : α → β) (f : filter α) : filter β :=
{ sets            := preimage m ⁻¹' f.sets,
  exists_mem_sets := ⟨univ, univ_mem_sets⟩,
  upwards_sets    := assume s t hs st, f.upwards_sets hs (assume x h, st h),
  directed_sets   := assume s hs t ht, ⟨s ∩ t, inter_mem_sets hs ht,
    inter_subset_left _ _,  inter_subset_right _ _⟩ }

I need help to visualise the contours of this one

4:36 PM

@Secret thanks for splitting my message in three parts

four

lol

@Secret I used simple inequalities with $\lfloor x\rfloor$.

Letting the joke apart, this question should be the easiest one for getting a job in a researching institute for example. To be done under a minute.

It takes 5 seconds to guess the limit.

And at most 1 min to finish it easily not with fancy tools.

what does $\Gamma_{\infty}, \Gamma_{\epsilon}$ even look like?

4:42 PM

Uhm ok, so $\lim\limits_{x\to l_1}f(x)=l_2$ if for every filter $F$ with $F\to l_1$ we have $f^*F\to l_2$ (where $f^*F$ denotes the pushforward filter)

hello,

@Secret No one finished that question with an inspired contour integration. Still waiting. :-)

how are you?

Hm, the number of subsets of $\{1, 2, \cdots, n\}$ such that any pair of elements of such a subset differ by at least two is exactly the Fibonacci sequence.

(The recurrence is easy to derive)

I guess checking the neighbourhood filter of $l_1$ is actually enough

4:43 PM

@Waiting I don't even understand the question, what do the two other contours look like?

The integral on the big arc doesn't vanish. In fact, the integral on the big arc has 2 components that blows up at infinity, but wth opposite sides, which together leads to a finite value. So we also view that integral on the arc as a CPV integral.

(sorry I cannot edit my messages at al)

try pressing "edit" and post again, the chat sometimes does that lag thing, you modified message shoudl be in there

The proposed contour doesn't lead to the extraction of the value of the integral on the real line.

Anyway, I don't know what $\Gamma_{\infty}, \Gamma_{\epsilon}$ will look like if plotted in the diagram of the function, which is what currently stumped me

for any $r>0$ we have $card\{n\in\mathbb{N}, (\frac1n,1)\in B_r\}=+\infty$ where $B_r=\{(x,y)\in\mathbb{R}^2, (x-3)^2+y^2<r^2\}$

4:49 PM

@BalarkaSen You meant you certainly don't understand such solutions?

Not sure I got your point.

@Waiting Put it simply, I have the diagram of the function $e^{-z^2}/(z-1)$ plotted above. What shape does $\Gamma_{\infty}$ look like cause I need to be able to visualise the contour said by the OP to make sense of the question?

Anyway, that's less important.

I am bad at English, I meant I certainly can't help you.

$(\frac1n,1)\in B_r\Rightarrow (3-\frac1n)^2<r^2-1$ for $r>1$ we have $|3-\frac1n|<\sqrt{r^2-1}$ how to continue please

have you an idea @Secret?

uh what is the question?

4:58 PM

for any $r>0$ we have $card\{n\in\mathbb{N}, (\frac1n,1)\in B_r\}=+\infty$ where $B_r=\{(x,y)\in\mathbb{R}^2, (x-3)^2+y^2<r^2\}$

but what is the question?

find the adherent value of $(\frac1n,1)$ in the toplogy $\{\mathbb{R}^2,\emptyset, (B_r)_{r>0}\}$

Almost every time I come in here I have some fun. ;)

No more fun, I have to go now.

It has to be r=$\sqrt{1^2+3^2}$?

yes i see it with a picture $\sqrt{10}$

i want to see it with algebra

5:14 PM

In your first step, you let $n \to \infty$ and then you will be at the limit point of the set $(\frac{1}{n},1)$

which gives $3^2 +1 < r^2$

i don't understand

why let $n\to \infty$ ?

we are not in $(\mathbb{R},|.|)$

any smaller than that and your open set $B_r$ is not a neighbourhood of $(\frac{1}{n},1)$?

The circle $r = \sqrt{10}$ contains the point $(0,1)$ and for any $r \geq \sqrt{10}$, they are neighbourhoods of $(0,1)$ and since $\lim_{n \in \Bbb{N}}\frac{1}{n} \to 0$ it must contain countably many points in $(\frac{1}{n},1)$. Thus $(0,1)$ is an adherent point of the set $(\frac{1}{n},1)$

but you speak about limits in $(\mathbb{R},|.|)$

Isn't the set $(\frac{1}{n},1)$ by definition: $(1,1), (\frac{1}{2},1), (\frac{1}{3},1) ...$?

So $n \to \infty $is only along one dimension and the limit will took place in $\Bbb{R}$

Is it correct to say that 1/x converges 'logarithmically' for x to infty?

5:28 PM

looks nothing like logarithmic, since $\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}$

jaslibra

Hey, sorry never used chat before, so not too sure about the etiquette. I thought my question was pretty simple, but I got pretty weird responses from member with reputation 63.1k. math.stackexchange.com/questions/2756359/…

Its just about really understanding the terms in the infinite series solution to $y' = y$

@Secret Yet: en.wikipedia.org/wiki/Rate_of_convergence#Basic_definition

ah ok, so logarithmically has a specific meaning, not just it behave like log

it is definitely sublinear since the stuff in the limit are both polynomials of deg 1 and hence $\mu =1 $. Meanwhile the other limit gives 3 as the limit thus it is not logarithmic (computed by plugging 1/x into the given limit in the wikipedia page)

The other limit gives 3??

ah sorry typo

so it gives 1

hence by definition, logarithmic

5:39 PM

Thanks for the sanity check.

is there a relation with my problem, because i don't understand

My question was unrelated.

ok thank you

Q: Understanding solution to $y' = y$ and exponential distribution

2

My Understanding: I would derive the exponential random variable as follows: I consider an experiment which consists of a continuum of trials on an interval $[0,t)$. The result of the experiment takes the form of an ordered $n$-tuple $\forall n \in \mathbb{N}$ containing distinct points on the...

probability differential-equations exponential-function

Why will equations that describe exponential growth will have something to do with bernoulli trials?

so if this connection is not just a mathematical coincidence, then it will mean that the dynamics of y'=y (hence exponential growth) can be approximated by a bernoulli process of adding things into the set?

I wonder if anyone have checked whether a monte carlo solution to y'=y actually give a bernoulli distribution...

5:56 PM

Is the relation "is a dense subset of" transitive?

6:07 PM

Bonsoir @TedShifrin

@feynhat what do you think?

@TedShifrin j'ai besoins de votre aide pour ca math.stackexchange.com/questions/2757794/…

s'il vous plait

@AlessandroCodenotti I think yes.

Why?

@AlessandroCodenotti Say $A$ is dense in $B$, and $B$ is dense in $C$. Then any open subset of $B$ intersects with $A$, now this open set is intersection of $B$ and an open set of $C$. So, any open set of $C$ which intersects with $B$ also intersects with $A$.

^ Any comments on this proof?

jaslibra

6:16 PM

@Secret, perhaps the question reduced to understanding each term in the series solution to $y' = y$

Nicholas Roberts

@Vrouvrou what is adherent value?

Moving sofa problem

The moving sofa problem or sofa problem is a two-dimensional idealisation of real-life furniture-moving problems and asks for the rigid two-dimensional shape of largest area A that can be maneuvered through an L-shaped planar region with legs of unit width. The area A thus obtained is referred to as the sofa constant. The exact value of the sofa constant is an open problem. == History == The first formal publication was by the Austrian-Canadian mathematician Leo Moser in 1966, although there had been many informal mentions before that date. == Lower and upper bounds == === Lower bound...

Can somebody please solve this? Thanks

I needed this result to show that $\mathcal{C}[a, b]$ (the space of all bounded, continuous, real valued functions defined on $[a, b]$) is separable. By Weierstrass approximation, set of polynomials is dense in $\mathcal{C}[a, b]$, and the set of polynomials with rational coeff. is dense in set of polynomials. So, set of polynomials with rational coeff. is dense in $\mathcal{C}[a, b]$ (this is where we use the result). And the set of polynomials with rational coeff. is countable.

@NicholasRoberts @NicholasRoberts we say that $(x,y)\in \mathb{R}^2$ is an adherent value of $u_n$ iff $\forall V\in \mathcal{V}_x, card\{n\in\mathbb{N}, u_n\in V\}=+\infty$

Please help me with this: 'Put a metric $d$ on $\Bbb R$ such that $|x_ n − x| → 0$ if and only if $d(x_ n , x) → 0$, but that ${x_ n }$ is a Cauchy
sequence in $(\Bbb R,d)$ when $|x_ n | → ∞$.'