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12:45 AM
If you dig deep enough, every proof involves Cauchy sequences.
 
2 hours later…
2:18 AM
@EE18 you can do this fairly explicitly, it is the kind of thing lots of books will have some clever way of doing. i see no 'one obvious way' of doing it.
also the kind of thing where imvho there is absolutely nothing lost by just seeing how one or two books do it and adding that to the list of things you now know, instead of "how could i have come up with that" or "how could i make this really long thing that i came up with as elegant as something in a textbook"
how you prove it matters way less than being able to notice when you can make use of it in other, more involved / less "fundamental principles" kind of arguments
 
2 hours later…
4:06 AM
1
Q: Examine the convergence of $\int_0^1x^{n-1}\log x dx.$

Thomas FinleyExamine the convergence of $\int_0^1x^{n-1}\log x dx.$ The solution given is as follows: $0$ is the only point of infinite discontinuity of the integrand. Let us examine the convergence of $\int_0^{\frac 12}x^{n-1}\log x.$ The integrand is negative in $(0\frac 12]$. Let $f(x)=-x^{n-1}\log x,x\in ...

I need some help with understanding this proof.
 
1 hour later…
5:30 AM
Do number theorists like when someone expresses $\pi(x)$ as a sum over primes less than a given magnitude?
5:48 AM
honestly, Chebyshev, for what it's worth, stands as an exemplary model of efficiency in the context of the aforementioned example. His erudite capabilities transcended the complexities of mathematical language, making significant contributions to classical theory, despite occasional deviations from traditional norms
I truly admire Chebyshev's efforts. When he resourcefully defined his functions, he provided radical contributions and honest expressions of primal tameness, reducing the redundancy of the non-summatory prime counting function. Past number theorists and those of the classical era had deemed this function inefficient for computational and analytical purposes. Consequently, many, including me, regard Chebyshev as revitalizing the field and inspiring numerous successful analytical endeavors
Chebyshev's bright mind,
Numbers dance, theories unwind...
In primes, truth we find.
Thank you Chebyshev!
6:48 AM
> Problem: Let $X$ and $Y$ be independent r.v. such that $X\in U(0,1)$ and $Y\in U(0,\alpha)$. Find the density function of $Z=X+Y$. Remark: Note that there are two cases: $\alpha\geq 1$ and $\alpha <1$.
Let's recall the densities of $X$ and $Y$: $$ f_X(x)=\mathbf{1}_{(0,1)}(x), \quad f_Y(y)=\frac{1}{\alpha}\mathbf{1}_{(0,\alpha)}(y) $$ Let $z\in (0,1+\alpha)$. So we know that $f_Z(z)$ is given by: $$f_Z(z)=\int_\mathbb{R} f_X(t)f_Y(z-t)\,dt.$$ Both $f_X$ and $f_Y$ are zero most of the time. We check $f_X$ and $f_Y$ one by one. We start with $f_X(t)$; it is nonzero when $0<t<1$. We have that $f_Y(z-t)$ is nonzero when $0<z-t<\alpha$.
That means $t<z$ and $z-\alpha<t$. We want to satisfy all these inequality at once. So that means $\max\{z-\alpha,0\}<t<\min\{1,z\}$. Hence: $$ f_Z(z)=\int_{\mathbb{R}}f_X(t)f_Y(z-t)\,dt=\int^{\min\{1,z\}}_{\max\{z-\alpha,0\}}\frac{1}{\alpha}\,dt=\frac{\min\{1,z\}- \max\{z-\alpha,0\}}{\alpha} .$$
Now, what troubles me is the remark in the problem statement. I don't see that there are two cases to consider. For me, the density is the one given in the last equation, or?
 
1 hour later…
7:57 AM
@copper.hat :-)
psie: without reading your calculations at all, i'm not sure that i like or understand the comment, but, if you wanted to "simplify" (another term i don't often like or understand) that thing you got at the end there, e.g. as a "piecewise" function of z defined in terms of "simpler" functions of z, it seems like cases involving the relation of alpha to 1 would come up
psie: i.e. "max" and "min" both implicitly contain "cases", they just aren't expressly written out
it strikes me as conceivable that whoever made that comment might not have been thinking of using functions that encode 'cases' in some non express way
if you're satisfied with your answer i would not waste even 30 seconds on attempted mind reading
8:11 AM
ok, thanks 👍
Here's the next problem :)
Let $X$ and $Y$ have joint density $$f(x,y)=\begin{cases} 1& \text{for }0\leq x\leq 2,\max(0,x-1)\leq y\leq\min(1,x) \\ 0 &\text{otherwise}.\end{cases}$$ Then I need to find the marginal density functions and the joint and marginal distribution functions. "Integrating out" the $y$-variable and $x$-variable separately, we see that $f_Y(y)=2$ and $f_X(x)=\min(1,x)-\max(0,x-1)$.
From my previous problem, we see that $X$ is the sum of two independent $U(0,1)$-distributed r.v.s. What is the distribution of $Y$ though?I'm tempted to say it is a $U(0,1/2)$-distribution, but it looks like $y$ doesn't range from $0$ to $1/2$, so I'm not sure what it is.
8:43 AM
I guess my approach is not really the right one...since the density of $Y$ just doesn't make sense
 
3 hours later…
12:03 PM
I guess $$f_Y(y)=\int_\mathbb{R} f(x,y) \,dx \neq \int_0^2 1\, dx=2,$$but I do not see why we have $\neq$...
12:15 PM
I think it is because $[1,2]$ on the $x$-axis is actually not in the support of $f$, hence the integral only goes from $0$ to $1$.
 
2 hours later…
2:33 PM
@psie your calculation isn't quite right
this is more subtle than what Leslie was suggesting
@psie for the second equality in here to be true, you need $\min(1, z) \geq \max(z-\alpha, 0)$
this is where the cases come from
@psie if you can, please refer to them as cumulative distribution functions or CDFs instead of calling them distribution functions. This is confusing
in the definition of $f$, the inequality $0\leq x \leq 2$ is redundant. The region given by $\max(0, x-1)\leq y\leq \min(1, x)$ gives us a parallelogram already contained in the region $0\leq x \leq 2$.
although for purposes of integration this is okay
you found $f_X$ correctly. But not $f_Y$. You need to write the conditions $0\leq x \leq 2, \max(0, x-1)\leq y\leq \min(1, x)$ in terms of wanting to integrate by $x$ first
$0\leq y\leq 1, \max(0, y)\leq x\leq \min(2, y+1)$
Of course formula for $f_X(x)$ is for $0\leq x\leq 2$
the distribution of $Y$ looks to me as sum of two independent $U(0, 1/2)$ random variables, but shifted and scaled
3:17 PM
@Jakobian hmm, I don't see how the inequality is not true. I've tried $\alpha<1$ and $\alpha>1$, but I don't see that it doesn't hold. Could you give an example when it does not hold?
@psie what do you mean
you say that we need $\min(1, z) \geq \max(z-\alpha, 0)$. Doesn't this always hold?
no. In fact this is equivalent to $0\leq z\leq 1+\alpha$
well, I guess Leslie was right, the cases of $\alpha > 1$ and $\alpha\leq 1$ are irrelevant in this approach
but you still didn't calculate this fully correctly
How does the following axiom (of extensionality) assert that two classes $A,B$ are equal if they contain the same elements: $(x\in A \iff x\in B)\implies A = B$. If $x \notin A$ but $x \in B$ or vice versa, we have $A = B$ or $A \neq B$ is true. Shouldn't this be an iff?
I think this makes more sense no? $(x\in A \iff x\in B)\iff A = B$
@Jakobian is it still the second equality in $$f_Z(z)=\int_{\mathbb{R}}f_X(t)f_Y(z-t)\,dt=\int^{\min\{1,z\}}_{\max\{z-\alpha,0\}}\frac{1}{\alpha}\,dt=\frac{\min\{1,z\}- \max\{z-\alpha,0\}}{\alpha}$$ that is not correct?
3:30 PM
@psie yes, in the sense that the equality holds only for $0\leq z\leq 1+\alpha$
for other $z$, $f_Z(z) = 0$
I've read this answer but it only makes sense if the definition of membership/equality is given and in hungerford it's not. It's a primitive notion in NBG set theory
@Obliv axiom of extensionality doesn't assert when two classes are equal (in ZFC)
@psie ok, but I did write at the beginning of my calculation that $z\in (0,1+\alpha)$ ;) but I probably should have said that $f_Z(z)=0$ otherwise
In axiomatic set theory and the branches of logic, mathematics, and computer science that use it, the axiom of extensionality, axiom of extension, or axiom of extent, is one of the axioms of Zermelo–Fraenkel set theory. Informally, it says that the two sets A and B are equal if and only if A and B have the same members. == Formal statement == In the formal language of the Zermelo–Fraenkel axioms, the axiom reads: ∀ A ∀ B ( ∀ X ( X ∈ A ⟺...
3:33 PM
stop spamming links I know what this is
I'm not spamming? I am providing context to my question. This is the information I am using
I also don't understand the axiom of class formation in the 2nd paragraph either. I don't really know formal logic so the predicate notation is unfamiliar to me maybe that's why?
@psie it was in the beginning so I missed that as part of that calculation, my fault, althought it was well hidden
no worries
@Obliv for this, its actually not NBG specific and the same discussion appeared here before for axiom of extensionality in ZFC
the implication that $A = B$ implies $\forall_z (z\in A\iff z\in B)$ can be considered an axiom of logic
so there is no need to write it down
Okay, I guess that's what he means implicitly by the primitive undefined notion of equality
3:39 PM
one of the things those axioms of logic assert is that from $A = B$ it follows that we can replace $A$ by $B$ in formulas
3 mins ago, by Jakobian
the implication that $A = B$ implies $\forall_z (z\in A\iff z\in B)$ can be considered an axiom of logic
so membership is $\in$ and equality is $=$, both are equivalence relations on the universe of classes?
of course, this is wrong, and I meant that it follows from something that can be considered an axiom of logic
and the whole implication I wrote isn't an axiom of logic
@Obliv they are relations
not equivalence relations
but isn't $=$ reflexive, symmetric, transitive?
is $\in$
@Jakobian no
$x \in A$ doesn't mean $A \in x$, etc
but it's transitive at least
3:45 PM
Show that $p$ prime $\implies$ $p$ irreducible for $p\in \mathbb{Z}$. My work: let $p$ be a prime and let $x,y \in \mathbb{Z}$ such that $p=xy$. Hence, from $1\cdot p=p=xy$, we have that $p|xy$ and so, being $p$ a prime, we have that $p|x$ or $p|y$.
If $p|x$, there exists $k\in\mathbb{Z}$ such that $x=kp$ and so $p=xy=p(ky)$; since $p \ne 0$ and since the cancellation property holds in $\mathbb{Z}$, we have $1=ky$ and so $y \in U(\mathbb{Z})$. The case $p|y$ is similar. This shows that $p$ is irreducibile. Is this correct?
@Obliv that's also not true
its neither reflexive, symmetric nor transitive
no I think it is transitive, but probably not reflexive because of axiom of foundation/regularity
idk it depends on which set theory
like $x \in A, A \in B \implies x \in B$ no?
@Obliv its not transitive in NBG nor in ZFC
@Obliv no
is it a semantic difference between $\in$ and $\subset$?
3:48 PM
$\subset$ is surely transitive?
@Jakobian isn't $0\leq y\leq 1, \max(0, y)\leq x\leq \min(2, y+1)$ simply $0\leq y\leq 1, y\leq x\leq y+1$ and so $$f_Y(y)=\int_\mathbb{R}f(x,y)\, dx=\int_y^{y+1}1\, dx=1?$$
$\in$ and $\subset$ mean totally different things
in English language there might be some confusion but as symbols they mean something else (always)
$\in$ is never transitive, although $\subseteq$ is
$x\in A$ simply means "$x$ is an element of the class $A$"
if $x$ is a set, though, then doesn't $\in$ do the same job as $\subset$ in that case?
@Obliv what do you mean
doing the same job is vague and unmathematical
3:50 PM
$a \in \{a\}$, but it is not true that $\{a\} \in \{a\}$, while it is true that $\{a\} \subseteq \{a\}$.
like $\mathbb{N} \in \mathbb{Z}$ same thing as $\mathbb{N}\subset \mathbb{Z}$ is wrong?
@Obliv yes thats very wrong
@Obliv Yes, that's not right.
that's something that would offend me as a mathematician
that's how wrong it is
@XanderHenderson I didn't claim it was reflexive, but in your example if $\{a\} \in \{\{a\},b\}$ then it seems transitive
since $a \in \{\{a\},b\}$
3:51 PM
@Obliv What? No... that isn't what transitive means...
and $a \not\in \{\{a\},b\}$.
$\{a\} \in \{\{a\},b\}$.
I feel like john cleese right now in the restaurant skit.
some sets are transitive
which means that $x\in y\in A$ would imply $x\in A$
but not all of them, this is an important concept which makes birth to such concepts as ordinal numbers
ok I will think this over thank you, I will not spam chat for now.
you're not spamming
well I just don't want to push the previous questions too far out of sight
3:56 PM
eh, don't worry about it
@ZaWarudo yes, assuming the property $p|xy \implies p|x$ or $p|y$ is known for primes
(this is something that might require further diving into depending on the definition and known properties of prime numbers that we are to assume)
@Jakobian Yes, my textbook defines primes with that implication; thanks for checking my work. My proof doesn't work in a generic ring for prime element $p$, because the cancellation property does not hold in a generic ring, right?
@psie ah yes, sorry. We have $f_Y = 1_{(0, 1)}$ then so $Y$ is $U(0, 1)$ r.v.
👍
@ZaWarudo yeah, the proof holds for integral domains
4:49 PM
@nickbros123 Who is that in your profile picture?
suppose I have a set $E \subset \mathbb{R}^n$ which is bounded, and also infinite, and also not closed ie- has a limit point $p\in \mathbb{R}^n$ but $p \not\in E$. Is there a way for me to construct a subset in $E$ which is infinite, which has its limit point as "only" $p$?
@nickbros123 Yes, use the definition of limit point
@SoumikMukherjee richard dawkins, the evolutionary biologist known for his vocal criticism of Christianity (although I dont use him as my pfp for that reason)
hint: You are in a metric space, so chose open sets as open balls of radius 1/n for each n
@nickbros123 okay
@SoumikMukherjee so I choose an element in every 1/n ball around p?
4:54 PM
yes an element that lies in E also
you can construct an infinite set in this way whose limit point is p. Next you need to show that no other point can be a limit point of this set.
@nickbros123 What does it mean for $p$ to be a limit point of $E$? Every open set containing $p$ also contains an element of $E$, yes?
@XanderHenderson yea
Ill go off on what soumik said, Ill come back if I need more help. I actually just wanted to know "if" its possible
@nickbros123 Right. So in every ball around $p$, you can find some $q \in E$. In particular, because $E$ is infinite and $p$ is a limit point, there are infinitely many points of $E$ in any neighborhood of $p$ (why?).
So in $B(p,1/n)$, you can always find a point of $E$.
oh wait, uniqueness just comes from it being a convergent sequence :facepalm:
Yes. Though the fact that the space is metric helps with that (there are spaces in which the limit of a sequence needn't be unique).
5:05 PM
I see
For example, the dumb space $\{0,1\}$ with the topology $\tau = \{ \varnothing, \{0,1\}\}$ has the property that every sequence converges to both $0$ and $1$. This is not a terribly interesting space otherwise, but it gives the example.
@XanderHenderson right, cause every neighbourhood of 0 has 1 and vice versa?
Yes. In a completely stupid way, but yes.
5:38 PM
I was basically trying to prove that "if E is not closed or not bounded" implies "there is an infinite set in E that doesnt have a limit point in E", and for the unbounded case, I chose a point $x$ in E, and $y_1$ some $z$ distance away from $x$. I then choose $y_2$ that is more than $z$ farther from $x$ but in the $2z$ ball of $x$(if it doesnt exist, make it $3z$ or $4z$ and so on).
So for every $\varepsilon$ there are utmost finite $y_n$ inside $B_{\varepsilon}(x)$, thus rendering $x$ a non limit point. Now If I consider another "possible" limit point $y$, This would require that infinite $y_n$ exist in every neighbourhood of $y$, but there is a neighbourhood of $x$ that would subsume one of $y$s, thus making it impossible for any $y$ neighbourhood to have infinite points.
my question is, where have I used a property of $R^n$
@nickbros123 by contraposition, limit point compact => closed and bounded
limit point compact and compact is equivalent in a metric space, and bounded is of course only really defined there
this implication then always holds, so you certainly didn't need any property of R^n
its the reverse implication that only holds in R^n (where by only I just mean it doesn't hold in all metric spaces)
compact implies closed and bounded is true for all compact sets in metric spaces, im ok with that. I also proved that in Rn, closed+bounded=> compact (by first proving k-cells are compact), now I am tasked with the following: in R^n: E is closed+bounded <=> (call this T)every infinite subset of E has a limit point in E (Compact=>(T) is something I have proved for metric spaces), so I only have to prove (T)=> closed and bounded for which I have given the outline above.
This sketch is quite rough, I just wanted to know where ive used a properry of Rn here
(T) is called limit point compact
you actually want to prove limit point compact and compact coincide
not just for R^n but all metric spaces
I have proved compact=> limit point compact
is the back implification also true?
yes, for metric spaces it is
compact and limit point compact only start to be different concepts in topological spaces
perhaps they wanted you to give a different proof of the equivalence, proving something weaker result because just for R^n, by going to show limit point compact => closed and bounded => compact
this doesn't mean you have used that this is R^n anywhere in the first implication, but that you need it for the second
5:53 PM
this is just me trying to prove things. Rudin tells me that limit point compact is equive to compact in metric spaces, but closed+bounded is not in general
(just saw that remark at the bottom of the page)
my question was, I have a direct proof of closed+bounded <=> limit point compact (in R^n) outlined like 4 messages above, I wanted to know where I would have used a properry of R^n
bounded is not just not a topological property
you can easily switch your metric from $d$ to $\min(1, d)$ and get that every set is bounded
while topology is still the same
@nickbros123 you need to use it to claim closed + bounded implies limit point compact
I'm not exactly sure where you have this since the argument is quite scattered
@nickbros123 okay so you use that you are in R^n by saying closed + bounded => compact
so nowhere in any explicit arguments (probably)
I guess ill have to write it down explicitly and see
thanks for the time
6:56 PM
In mathematics, a topological space X {\displaystyle X} is said to be limit point compact or weakly countably compact if every infinite subset of X {\displaystyle X} has a limit point in X . {\displaystyle X.} This property generalizes a property of compact spaces. In a metric space, limit point compactness, compactness, and sequential compactness are all equivalent. For general topological spaces, however, these three notions of compactness are not equivalent...
as you can see on this wikipedia page, $$\text{compact}\implies \text{countably compact}\implies\text{limit point compact}$$
and $$\text{limit point compact} + T_1\implies \text{countably compact}$$ where $T_1$ is a certain very weak separation axiom (that metric spaces definitely satisfy)
the real issue for topological spaces is whetever countably compact and compact are equivalent
$$\text{countably compact} + \text{Lindelof} \iff \text{compact}$$
 
1 hour later…
8:20 PM
@leslietownes The "explicitly" you mention I assume is the usual "divide into bounded and unbounded sequences". The former is the hard case, apply Bolzano Weierstrass and go from there...?
i.e. no use of Cauchy sequences?
oh no i meant more that the hard case (that of bounded sequences) can be handled fairly explicitly even if you don't have big theorems
maybe it was confusing, i don't really know the context in which that exercise was coming up. the details may obviously depend on your construction of R (or your axiom set if you did not construct R) and whether you have any theorems you can use as black boxes
all of which maybe make it less interesting, say, than almost anything else in analysis
8:50 PM
@EE18 most of the times this is used to prove the bolzano weierstrass theorem--> a possible reason why this is given in the chapter of cauchy sequences may be that, to prove cauchy sequences are convergent in R, we first prove it is bounded=> it has a convergent subsequence.
9:06 PM
one arguable lesson of some of this is that one could spend one's life writing and rewriting chapter 0 of an analysis book without ever getting anywhere
9:18 PM
@nickbros123 so this sequence I have seen. But this exercise is asking me about showing that a monotone subsequence exists always for sequences in R. But to Leslie’s point maybe I shouldn’t worry about how or why this is here in this section
@leslietownes I think there is a good chance we are seeing this in practice, and we might see this in practice when replacing "real analysis" with other subjects
10:17 PM
I never learned if Bolzano Weierstraß and Heine Borel refer to distinct statements or if its just a matter of taste which wowrds you use
there isn't enough consistency in how people use those names for me to be comfy with using them
@EE18 maybe. the proof is pretty cool tho. IMHO the thing this proof ties neatly with is not cauchy sequences, but the concept of lim sup and lim inf
@leslietownes I think the development of Bartle and Sherbert and Rudin are rock solid, wouldnt even dare trying to rewrite it

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