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Ted Shifrin
Ted Shifrin
12:01 AM
@robjohn If only I'd invested in ponies.
 
2132123
2132123
12:56 AM
Hi I just have quick and silly question. Why $\int_{0}^{2\pi}1/(a+bcos(x)=2\int_{0}^{\pi}1/(a+bcos(x)$ I tried two u-subs $u=x/2$ and $u=x-pi$ both not giving me what i want.
@robjohn I like your new avatar btw :)
 
Ted Shifrin
Ted Shifrin
1:07 AM
@2132123 No, substitute $-x$ or $2\pi-x$.
 
AlgebraicGeometryStudent
AlgebraicGeometryStudent
Got a non-crackpot question for AG people: math.stackexchange.com/questions/3791221/…
From Hartshorne
@mick are you, like me, another victim of prime numbers?
 
2132123
2132123
@TedShifrin Ah that did it, thank you :)
 
 
2 hours later…
Binod
Binod
3:19 AM
@robjohn Sir i was facing a problem . If f(x) has a codomain a , Is domain of $f^{-1}(x)=a$?
 
Adam L
Adam L
3:49 AM
$\mathbb U$ is reserved for the unit fractions everyone happy unanimous vote? good to hear reason has been restored and everyone is on team adam
 
robjohn
robjohn
4:07 AM
@Binod No. The range of $f$ is contained in the codomain of $f$. The range of $f$ equals the codomain of $f$ if the function is surjective. The domain of $f^{-1}$ is the range of $f$.
 
Ted Shifrin
Ted Shifrin
4:48 AM
@robjohn actually, the codomain, not the range, and assuming $f$ is injective?
 
robjohn
robjohn
@TedShifrin The way I learned, it was $f:\color{#C00}{\mathbb{R}}\to\color{#090}{\mathbb{R}}$ by $f(x)=x^2$ has domain $\color{#C00}{\mathbb{R}}$ and codomain $\color{#090}{\mathbb{R}}$, but range $\{x\in\color{#090}{\mathbb{R}}:x\ge0\}$
 
Ted Shifrin
Ted Shifrin
Oh, hell, I use range for codomain and image for range. I give up.
 
robjohn
robjohn
@TedShifrin It was always a change of notation from what I learned in school, where I never heard of codomain.
but that earlier terminology seemed to be non-uniformly defined.
 
Binod
Binod
5:13 AM
Sir If $f:[-\infty,1] \rightarrow [-\infty,1]$ f(x) =x(2-x) . I need to find it's inverse.

I am getting $y= 1 _-^+ \sqrt{1-x}$
Now how to check , which of the one is true?
 
 
3 hours later…
Knight
Knight
7:47 AM
If $$|a_2 - a_1| \gt |a_1|$$ can I conclude that $a_2| \gt |a_1|$ ?
Given that $a_1 , a_2 \neq 0$
 
Knight
Knight
8:06 AM
Given that $a_1, a_2 \in \mathbb{Z}$
 
robjohn
robjohn
8:46 AM
@Binod think of the range of $1+\sqrt{1-x}$ and the range of $1-\sqrt{1-x}$.
remember you want to end up in $(-\infty,1]$
The triangle inequality applied to $a_2=(a_2-a_1)+a_1$ gives $|a_2|\le|a_2-a_1|+|a_1|$
The triangle inequality applied to $(a_2-a_1)=a_2-a_1$ gives $|a_2-a_1|\le|a_2|+|a_1|$
The triangle inequality applied to $a_1=a_2-(a_2-a_1)$ gives $|a_1|\le|a_2|+|a_2-a_1|$
So, if you want to bound $|a_2|$ below, you get $|a_2|\ge|\,|a_2-a_1|-|a_1|\,|$
which means $|a_2|\ge|a_1|-|a_2-a_1|$, but you don't necessarily have $|a_2|\ge|a_1|$
 
Anonymous
9:13 AM
Is there an easy way (without counterexamples) to see that something like $d(x, y) = |x - y|^d$ does not satisfy the triangle inequality for $d > 1$ and thus cannot be a metric on $\mathbb R$?
 
Anonymous
Actually, I'm not sure if it is a metric for $d < 1$ either.
 
Anonymous
Oh, and I should have used some different symbol for the metric. Say $\rho$.
 
Paramanand Singh
Paramanand Singh
@S.D.: can you notice that the inequality $a^d+b^d\geq 1$ can't hold if $d>1$ and $a, b$ positive with $a+b=1$?
 
Anonymous
9:52 AM
@ParamanandSingh Uh I don't see it right away but I think the answer I got just now on the main site is similar to yours math.stackexchange.com/a/3791585
 
Anonymous
It seems easier to notice that $2^d \leq 2$ cannot hold for $d > 1$
 
Paramanand Singh
Paramanand Singh
Well we have both $0<a<1,0<b<1$ so that $a^d<a, b^d<b$ and their sum is thus less than $a+b=1$.
 
Anonymous
@ParamanandSingh Ooh makes sense!
 
Binod
Binod
@robjohn Ah, as range is a subset of the codomain ,this $1+\sqrt{1-x}$ produces value greater than codomain.
 
Anonymous
@ParamanandSingh By any chance do you also happen to know how to show that $(a + b)^d - a^d - b^d$ is positive for all $a, b > 0$. Then I guess by taking partial differentials w.r.t. $a, b$ we can show that it is decreasing if either of $a, b$ decreases.
 
ronak jain
ronak jain
11:21 AM
Hi everyone
I have read that :- A function f(x, y) is said to be homogeneous of degree m if f(tx,ty)=tmf(x,y) for any real number t≠0. Euler’s theorem states that if f is homogeneous of degree m and has all partial derivatives of first order, then x∂f/∂x+y∂f/∂y=mf(x)
But how can if come to df = (df/dx)dx + (df/dy)dy, from the above expression
 
Thorgott
Thorgott
I don't understand what you're asking
 
ronak jain
ronak jain
@Thorgott actually when I read the Euler's theorem that it relates the actual function and it's partial derivative. But I didn't understand how to find the differential of the function as it is shown in last two lines
That is df = (df/dx)dx + (df/dy)dy
 
Thorgott
Thorgott
11:42 AM
well, that will depend on the function
 
Threnody
Threnody
12:41 PM
Am I butchering the contour deformation theorem? (Complex Analysis)

Suppose we have $\int_\psi\frac{tan(\pi z)}{z^2}dz$ where $\psi$ is the ellipse given by $\frac{x^2}{16} + y^2 = 1$.

Since $\frac{tan(\pi z)}{z^2}$ is not defined at $z \in \{0, \pi/2, -\pi/2\}$ we cannot outright use Cauchy's Integral Formula (because two of the points are due to $tan$). Then... deform the ellipse into a 'triple circle' where each loop is centered around those points. I suppose the specific radius for each loop is not important here.
 
Threnody
Threnody
12:52 PM
Sigh... no I can't do that... otherwise I won't have $0$ in the interior of all of the smaller circles.
 
Threnody
Threnody
1:03 PM
I guess the application of the deformation theorem is ok but not so useful in this case.
 
Paramanand Singh
Paramanand Singh
1:18 PM
@S.D.: well just note that $(a+b) ^d-a^d-b^d=(a+b) ^d(1-(x^d+y^d))$ where $x=a/(a+b), y=b/(a+b) $ so that $x, y$ are positive and $x+y=1$. As discussed in my last message we then have $x^d+y^d<1$ for $d>1$ and thus your expression is positive for all positive $a, b$ and $d>1$.
 
robjohn
robjohn
@TedShifrin It seems that Wikipedia says that "range" is used by different authors to mean "codomain" or to mean "image". So according to that, I guess the terms to use would be domain, codomain, and image.
it's hard to teach an old dog, new tricks; so it will be almost impossible for me to stop using range=image.
 
 
2 hours later…
Knight
Knight
3:28 PM
Is an odd degree polynomial always injective?
No
An odd degree polynomial need not to be injective, example $x^5 -x^4 -^2 -1$ just drew its graph and saw that it doesn’t qualify horizontal line test.
 
Thorgott
Thorgott
3:44 PM
for a simpler example, you can take something like $x^3-x$
on the other hand, it is an important fact that an odd degree polynomial function R->R is always surjective
 
Knight
Knight
Okay!
 
Anonymous
I'm having some trouble understanding the concept of open cover in topology. They say $(0, 1)$ is not compact in $\mathbb R$, but isn't say $(-1, +2)$ a finite open cover for $(0, 1)$?
 
Knight
Knight
If I have a sequence $$a_1, a_2 , a_3 \cdots$$ and we’re given that the first and second term cannot be zero and given that $$|a_1| = |a_2 -a_1| = |a_3 - a_2| = |a_4-a_3|= \cdots$$ then can we conclude that the sequence is an AP?
 
Anonymous
If I'm understanding correctly an open cover must be a union of open sets in $\mathbb R$ containing the set $(0, 1)$
 
Thorgott
Thorgott
yes, it is a finite open cover
 
Knight
Knight
3:58 PM
I think given that modulus equality, the sequence can never converge to zero.
 
Anonymous
@Thorgott Oh, but is the point is that there exists some open cover of $(0, 1)$ that does not have a finite subcover?
 
Thorgott
Thorgott
yes
compactness means that any open cover has a finite subcover
 
Anonymous
I see, thanks. There's another related confusion I have. Suppose $X$ is a compact set in $\mathbb R$. Then can it also be shown to be a compact set in $X$ itself, with the metric induced from $\mathbb R$? That is for some collection $\mathcal C$ of open subsets (open cover) of $X$, we can show that $\bigcup \mathcal C = X$ and for some finite subcollection $\mathcal C' \subseteq \mathcal C$ we have $\bigcup \mathcal C' = X$?
 
Thorgott
Thorgott
Yes, that is the better definition of compactness anyhow
Compactness is a property intrinsic to a topological space
defining it in the presence of an ambient space is more confusing than helpful imo
@Knight I'm pretty sure such a sequence doesn't converge at all
 
Anonymous
@Thorgott Interesting!
 
Anonymous
4:10 PM
I a few more confusions. Say we take the set $X = \{1, 2, 3, 4, 5\}$ with the metric inherited from $\mathbb R$. Then what would be an open cover for $X$ taking open sets from $X$ only. Would $\{1\} \cup \{2\} \cup \{3\} \cup \{4\} \cup \{5\}$ be an open cover?
 
Thorgott
Thorgott
Yes, it would. Can you tell me why?
 
Knight
Knight
@Thorgott But if one of the subsequent terms is zero, I’m getting no contradiction. Why?
 
Anonymous
@Thorgott Because each singleton set like $\{1\}$ is open as we can enclose it within some open ball of say radius $1/2$?
 
Anonymous
And what I wrote is a union of such open sets
 
Anonymous
I'm also trying to think of an open cover for $(0, 1)$ that does not have a finite subcover and thereby prove that $(0, 1)$ is not compact
 
Thorgott
Thorgott
4:15 PM
That's correct
For $(0,1)$, try using balls with smaller and smaller radii
@Knight I think no terms in this sequence can be zero, but I haven't worked out the details
 
Knight
Knight
@Thorgott I too feel that, but I’m unable to prove it. Can you give some hint? Something about how would you go on proving it.
 
Thorgott
Thorgott
For each $n$, you have either $a_{n+1}-a_n=a_n-a_{n-1}$ or $a_{n+1}-a_n=a_{n-1}-a_n$. This allows you to study the sequence recursively.
 
Anonymous
@Thorgott Thanks, I notice that they're using the open cover $\{(1/n, 1)\}$ for $n \geq 1$ here.I understand the proof!
 
Anonymous
The first answer suggests that we can also show it using the property of sequential compactness, i.e., a subset of a metric space is compact iff it is sequentially compact. Indeed, $\{x_n = \frac{1}{n}\}$ is a Cauchy sequence but how do we prove that it does not have a subsequence converging to a point in $(0, 1)$?
 
Anonymous
Is there any property like if $\{x_n\}$ converges to $x$ then any of its subseqence must also converge to $x$?
 
Anonymous
4:26 PM
I think I saw that somewhere but can't recall
 
Thorgott
Thorgott
That is indeed true (and intuitive). Try proving it.
 
Anonymous
4:46 PM
Thanks, I will check it. Unrelated question: Is the topology generated by $\tau = \{(a, \infty): a \in \mathbb R\}$ the same thing as $\tau$ with just $(-\infty, +\infty)$? It seems like any arbitrary union of elements in $\tau$ also results in an open set of the form $(a, \infty)$.
 
Anonymous
I guess we can express it as $\bigcup_{a \in \alpha} (a, \infty) = (\mathrm{inf}_\alpha(a), \infty)$
 
Anonymous
And $\mathrm{inf}_\alpha(a)$ must lie in $\mathbb R$
 
Anonymous
I wonder what kind of metric space subsets always contain their infimums
 
Thorgott
Thorgott
what do you mean by "the same thing as $\tau$ with just $(-\infty,\infty)$" exactly?
 
Anonymous
@Thorgott I mean $\mathcal O_\tau = \tau \cup (-\infty, +\infty)$ where $\mathcal O_{\tau}$ is the topology generated by the basis $\tau$
 
Thorgott
Thorgott
4:51 PM
That's not true, the topology generated by $\tau$ contains a lot more sets than just those in $\tau$ and $(-\infty,\infty)$
 
Anonymous
@Thorgott For example? I was thinking that any arbitrary union of elements in $\tau$ always results in something of the form $(a, \infty)$ or $(-\infty, \infty)$
 
Anonymous
$\bigcup_{a \in \alpha} (a, \infty) = (\mathrm{inf}_\alpha(a), \infty)$
 
Thorgott
Thorgott
it does, but the topology generated by $\tau$ contains more than just arbitrary unions of elements in $\tau$
for example, it also contains finite intersections of elements in $\tau$
e.g. all the half open intervals $(a,b]$
 
Anonymous
@Thorgott But isn't $\tau$ supposed to be a basis and not a subbasis?
 
Anonymous
The topology generated by a basis only contains its arbitrary unions
 
Thorgott
Thorgott
4:54 PM
evidently, it is not a basis
 
Anonymous
This is basically lemma 13.1 in Munkres I think
 
Thorgott
Thorgott
if it were a basis, then there would need to be a set of the form $(c,\infty)$ contained in $(a,b]$
but that cannot happen
 
Anonymous
@Thorgott Oh you mean it does not satisfy the conditions of a basis?
 
Thorgott
Thorgott
precisely
on the other hand, the half open-intervals $(a,b]$ will form a basis for the resulting topology
 
Anonymous
Umm, but a basis just needs to satisfy these two conditions right: (1) For each $x \in \mathbb R$ there is at least one basis element $B$ containing $x$. (2) If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$ then there is a third basis element $B_3$ such that $B_3 \subset B_1 \cap B_2$. I guess $\tau$ does not satisfy the second condition?
 
Thorgott
Thorgott
4:58 PM
yeah, I just told you why it doesn't
 
Anonymous
Say $2 \in (1, \infty)$ and $(0, \infty)$ but there is no basis element of the form $(a, \infty)$ containing $2$ and lying within $(0,1]$
 
Anonymous
@Thorgott I see, makes sense!
 
Anonymous
Okay, so ummm $\tau$ is a subbasis I suppose. Arbitrary unions and finite intersections
 
Thorgott
Thorgott
to say that $\tau$ is a subbasis for the topology just means that the topology is generated by $\tau$, which is precisely how we define this topology
so it's true by definition
 
Anonymous
@Thorgott Ah, a topology can be generated by a basis too but then again every basis is a subbasis I think
 
Anonymous
5:05 PM
So in general when they say a topology generated by some random set we can consider it to be a subbasis. I get your point
 
Thorgott
Thorgott
yeah, the topology generated by a set is just the coarsest (meaning smallest with respect to inclusion) topology containing that set
 
Anonymous
Anyway, this is basically a part of a question that asks me to check whether $X = (\mathbb R, \tau)$ is $\sigma$-compact, sequentially compact, limit point compact, Lindelof, pseudocompact and so on. I haven't really covered much of Munkres yet but I'm trying to see if I can deduce the answers from just the definitions.
 
Anonymous
@Thorgott I see!
 
Anonymous
To prove $\sigma$-compactness I guess I first need to determine what the compact subsets in $X$ are. I'm not sure how to approach that
 
Anonymous
In the standard topology on $\mathbb R$ we would have the Heine-Borel theorem to help
 
Thorgott
Thorgott
5:17 PM
To get you started, you can try proving that a) any open interval $(a,b)$ is open in $X$ (that is to say, the topology on $X$ is finer than the metric topology on $\mathbb{R}$) and conclude that b) the basic open sets $[a,b)$ are not only open, but also closed
 
Anonymous
@Thorgott Minor confusion actually now that I think about it again. Isn't the intersection of $(a, \infty) \in \tau$ and $(b, \infty) \in \tau$ just $(b, \infty)$ if $a < b$, rather than $(a, b]$? (For instance, Munkres says that the sets containing the sets $(-\infty, a)$ is a basis and not just a subbasis in exercise 7 of section 13)
 
Anonymous
5:34 PM
@Thorgott See example 2.3 on page 2 here: math.toronto.edu/ivan/mat327/docs/notes/02-bases.pdf
 
Anonymous
They do say $\mathcal A = \{(a, \infty): a\in \mathbb R\}$ is a basis
 
Thorgott
Thorgott
5:45 PM
oops, you're absolutely right
for some strange reason I complemented instead of intersecting, sorry
so forget my last couple comments
good news is that this makes the exercise easier
 
user378298
user378298
Hello ,a question in linear algebra :If subspace W_1 is generated by linear combination of set of vectors ,and subspace W_2 is generated by a different set of vectors, W_1=lin([1,0,1], W_2 = lin([1,1,0]) how to find formula for linear transformation phi: W_1 -> W_2?
 
Thorgott
Thorgott
there are many such transformations
 
robjohn
robjohn
@Knight I didn't see if anyone replied, but the absolute values allow for sequences which are not arithmetic progressions.
I believe $1,2,1,0,\dots$ satisfies your criteria
 
user378298
user378298
yes, and I meant phi:R_3 -> R_3. By Steinitz's theorem we can complete to the canonical base, what's an example solution?
 
Anonymous
@Thorgott Yes, thanks for clarifying! But even now I'm not sure how to determine the compact sets in $X = (\mathbb R, \tau)$. And then to show that $X$ is $\sigma$-compact I need to show that it is a union of countably many compact subsets. Do you have some hints?
 
Thorgott
Thorgott
6:02 PM
@S.D. this is relevant
keep in mind that you were initially right, i.e. every open set is of the form $(a,\infty)$ or it's $\mathbb{R}$ or the empty set
 
Anonymous
Uh, so let's say $\bigcup_{a \leq x \leq b} (x, \infty)$ is an open cover for $(a, b)$ and it has a finite subcover $\{(a, \infty)\}$. So open intervals like $(a, b)$ are compact in $X$. Is that correct?
 
Anonymous
Oh no. I need to show that every open cover has a finite subcover.
 
Anonymous
Say, there is any open cover of $(a, b)$ indexed by some set $\alpha$. Then a finite subcover can always be given by $(\mathrm{inf}_{\alpha}(x), \infty)$.
 
Thorgott
Thorgott
but $(\inf\alpha,\infty)$ may not have been part of the original cover
 
Anonymous
@Thorgott True....
 
Anonymous
6:09 PM
So $(a, b)$ is not compact in general?
 
Anonymous
I guess right closed intervals like $[a, b)$ could be compact
 
Anonymous
Any open cover for $[a, b)$ must contain the point $a$
 
Knight
Knight
@robjohn Thank you sir. How come you’re so good at giving counter-examples?
 
Thorgott
Thorgott
$(a,b)$ is in fact never compact (for $a<b$). Here is an open cover that does not admit a finite subcover: $\{(a+1/n,\infty)\colon n\in\mathbb{N}\}$. This example is instructive enough to figure out half of the general case.
@S.D. great observation, this is the other half of the general case
 
robjohn
robjohn
@Knight experience, I guess.
 
Knight
Knight
6:13 PM
:)
 
Anonymous
@Thorgott Ah, so $(a, b)$ is never compact, makes sense. But $[a, b)$ is compact and also $[a, b]$. I am thinking about $(a, b]$...
 
Anonymous
Oh $(a, b]$ suffers from the same problem you stated (your counterexample open cover works)
 
Thorgott
Thorgott
yeah, so what's the crucial difference between $(a,b)$ and $[a,b)$ that determines this behavior?
 
Anonymous
@Thorgott That for the latter there is at least one element of the open cover that contains the point $a$
 
Anonymous
So we can just pick that (or one such) element of the open cover as the finite subcover
 
Anonymous
6:21 PM
This has an axiom of choice vibe lol
 
Thorgott
Thorgott
right, so in the general case, you want to think about the infimum of your set
 
Anonymous
@Thorgott Ummm, the infimum of the elements of the open cover containing $[a, b)$ might still not lie in the open cover I think
 
Anonymous
It seems more sensible to just choose one such element from the open cover that contains $a$
 
Anonymous
Nevertheless, since open intervals $(a, b)$ are compact, I think we have that $X$ is $\sigma$-compact. Say we can write $X = \bigcup_{n \in \mathbb N} (n-1, n+1)$ which is a countable union.
 
Thorgott
Thorgott
we just concluded that $(a,b)$ is not compact, did we not
 
Mary Star
Mary Star
6:30 PM
Hello!! I have a question about uniform convergence. My approach is at "Edit 2" part of my question:
-2
Q: Check pointwise and uniform convergence

Mary StarCheck for the below the pointwise and uniform convergence. $\displaystyle{f_n(x)=\frac{x\sqrt{n}}{1+nx^2}}, \ x\in \mathbb{R}$ $\displaystyle{g_n(x)=n^2xe^{-nx}}, \ x\in \mathbb{R}^+$ $\displaystyle{h_n(x)=\frac{\sin nx}{1+nx^2}}, \ x\in \mathbb{R}$ For the pointwise convergence we have to c...

analysis uniform-convergence pointwise-convergence
Does someone of you have an idea?
 
Anonymous
@Thorgott Oops, I meant $[a, b)$
 
Thorgott
Thorgott
yeah, that does the job
 
Anonymous
$X = \bigcup_{n \in \mathbb N} [n-1, n+1)$
 
Anonymous
@Thorgott Yay
 
Thorgott
Thorgott
if you have an arbitrary subpace $A\subseteq\mathbb{R}$, my suggestion is to look at $\inf A$ and make a distinction based on whether $\inf A\in A$ or $\inf A\not\in A$
 
Anonymous
6:33 PM
@Thorgott Umm, but do you agree with me that for both [a, b) and (a, b), infimum of all the elements of some arbitrary open cover may not be an element of that open cover?
 
Anonymous
So I'm not sure if making the distinction you mention makes sense
 
Thorgott
Thorgott
yes, but you don't need $(\inf A,\infty)$ to be part of the cover, an $(a,\infty)$ with $a<\inf A$ does the job just as well
and this is precisely what happened in the case for $[a,b)$!
 
Anonymous
Oh, you're taking about $\mathrm{inf} \ A$. I was talking about $(\mathrm{inf} \ \alpha, \infty) \in \mathcal C$ where $\mathcal C$ is an open cover of $A$.
 
Thorgott
Thorgott
yeah, that's why I clarified
 
Anonymous
Right, looking at $\mathrm{inf} \ A$ does make sense for some arbitrary subspace $A$. Gotcha!
 
Anonymous
6:39 PM
Well, there are 4 more parts to this question :P The next one asks whether $X$ is sequentially compact
 
Knight
Knight
@robjohn Your example has helped me to solve a problem. Thank you so much.
 
Anonymous
I'll try thinking about it. Anyway thanks a lot for your help so far
 
Knight
Knight
I feel good when I say “thank you” to someone who has helped me.
Sam Harris once said “when you say ‘thank you’ try to mean it”
And yes I try to mean it.
 
Thorgott
Thorgott
You can try figuring out what convergence in this space looks like
 
Hawk
Hawk
Suppose $g(z) = f(z)/z$ and $\sup_{z} |f(z)/z| < \infty$ (so this implies $f(0) = 0$). I want to show $z = 0$ is a removable sing of $g(z).$ Since the sup is bounded it means $|f(z)| \leq |z|M$ for some number $M$. Therefore, $\lim |z g(z) | = \lim |f(z)| \leq \lim |z| M = 0$. Is this right?
 
Anonymous
6:45 PM
So first let's start with the definition of convergence in a general topological space I guess
 
Anonymous
A sequence $\{x_n\}$ in $X$ converges if there exists some $x \in X$ such that for any open set containing $x$, beyond some $N \in \mathbb N$, the points of the sequence $\{x_n\}$ lie within that open set
 
Mary Star
Mary Star
Hello @robjohn ! Do you maybe have an idea about my question about the uniform convergence?
 
Anonymous
Hmm, not sure how to proceed
 
Thorgott
Thorgott
you know what the open sets look like, use that
 
Anonymous
Let's say we take the sequence $\{x_n = n | n \in \mathbb N\}$
 
robjohn
robjohn
6:54 PM
@Knight great!
 
Anonymous
Beyond any $N \in \mathbb N$ we'd still have points of the sequence like $N + 1, N + 2, N + 3$ and so on
 
robjohn
robjohn
@MaryStar what question is that?
Oh, I see it
 
Anonymous
If a subsequence of $\{x_n = n\}$ actually converges to some $x \in \mathbb R$ then for any open set of the form $(a, \infty)$ (containing $x$) we'd have that beyond some $N \in \mathbb N$ those natural numbers in the subsequence lie within that open set
 
Anonymous
Does this look plausible
 
Thorgott
Thorgott
well, do they?
 
Anonymous
7:01 PM
@Thorgott I can't really imagine any subsequence of $\{n\}$ that converges in the first place so I'm confused :P
 
Thorgott
Thorgott
you just spelled out the question you have to answer above
 
Anonymous
I don't think it's even possible. An infinite subsequence wouldn't even be bounded above
 
Thorgott
Thorgott
does there exist an $N\in\mathbb{N}$ such that $n\in(a,\infty)$ for $n\ge N$?
 
Anonymous
@Thorgott Yes, I guess?
 
Anonymous
Any such subsequence would be strictly increasing
 
Thorgott
Thorgott
7:04 PM
yeah, so the sequence does converge to $x$
 
Anonymous
There must be such an $N$
 
Thorgott
Thorgott
and since $x$ was arbitrary, the sequence $(x_n)$ converges to every real number
fabulous
 
Anonymous
@Thorgott But $x$ may not even exist
 
Thorgott
Thorgott
huh?
 
robjohn
robjohn
@MaryStar $\frac{x\sqrt{n}}{1+nx^2}\le\frac1{x\sqrt{n}}$ so pointwise the limit is $0$. For any $n$, $x=\frac1{\sqrt{n}}$ gives $\frac{x\sqrt{n}}{1+nx^2}=\frac12$, so the convergence is not uniforn.
 
Anonymous
7:08 PM
Oh, you're correct. It converges to every real number $x$. As for every real number $x$ and for any neighborhood/open set containing $x$ we have that beyond some $N > \mathbb N$ the elements of the subsequence being unbounded above must lie within with that open set. I think I get your point
 
Thorgott
Thorgott
yeah, but you don't need to think about subsequences
the sequence itself already converges to every real number
 
Mary Star
Mary Star
@robjohn So for this sequence is it correct what I have done? I mean is my formulation correct and complete?
 
Anonymous
Right, lol. This is awesome
 
Anonymous
I happened to pick this sequence just by uneducated guesswork :P
 
Thorgott
Thorgott
now let's put a slight twist on this
can you tell me what happens with the sequence $(-n)_n$
 
Anonymous
7:12 PM
@Thorgott It doesn't have any convergent subsequence?
 
Thorgott
Thorgott
bingo
 
Anonymous
Beyond some $N$ the elements of the sequence will walk out of the open set
 
Anonymous
Cool!
 
Anonymous
So $X$ is not sequentially compact
 
Thorgott
Thorgott
yup
 
Anonymous
7:14 PM
Yay! Topology seems fun lol
 
Anonymous
The next part asks to check if $X$ is limit point compact (whether every infinite subset has a limit point in $X$)
 
Anonymous
A limit point $x$ IIRC is one to which we can find a convergent subsequence $\{x_n\}$ in $X$ converging to that point
 
user378298
user378298
Hello, can you answer my question https://math.stackexchange.com/questions/3792050/finding-formula-for-linear-transformation
Thanks
 
Anonymous
Limit point
In mathematics, a limit point (or cluster point or accumulation point) of a set S {\displaystyle S} in a topological space X {\displaystyle X} is a point x {\displaystyle x} that can be "approximated" by points of S {\displaystyle S} in the sense that every neighbourhood of x {\displaystyle x} with respect to the topology on X {\displaystyle X...
 
Anonymous
Ummm, no Wikipedia seems to define limit point differently
 
anakhro
anakhro
7:18 PM
Some words in topology have various equivalent definitions.
Here this one is equivalent to yours
 
Anonymous
Umm, Wiki doesn't state my definition. I wonder how to show the equivalence
 
Hawk
Hawk
excuse for reposting but can someone tel me if i m right?Suppose $g(z) = f(z)/z$ and $\sup_{z} |f(z)/z| < \infty$ (so this implies $f(0) = 0$). I want to show $z = 0$ is a removable sing of $g(z).$ Since the sup is bounded it means $|f(z)| \leq |z|M$ for some number $M$. Therefore, $\lim |z g(z) | = \lim |f(z)| \leq \lim |z| M = 0$. Is this right?
 
Anonymous
I can show that my definition implies Wiki's definition but I don't think I can show the converse
 
anakhro
anakhro
In a metric space you can use balls of radius 1/n.
 
Thorgott
Thorgott
they're not equivalent
 
Anonymous
7:21 PM
@anakhro Well, in a metric space I think they're equivalent
 
Thorgott
Thorgott
but they're equivalent in first-countable spaces
 
Anonymous
But not in general topological spaces
 
Anonymous
Ah, I dunno what first-countability means though. Next chapter in Munkres!
 
Anonymous
Anyway, let's take Wiki's definition for now
 
anakhro
anakhro
For some reason I thought your question was about a metric space.
 
Anonymous
7:22 PM
A point $x$ in $X$ is a limit point if every open set containing $x$ has at least one other point apart from $x$
 
Anonymous
@anakhro Heh, nevermind. But I guess I'll go through the proof of what you're saying too, thanks anyway
 
anakhro
anakhro
Apologies. :P
 
Anonymous
Np!
 
Anonymous
Returning to the problem, let's take any infinite subset in $X = (\mathbb R, \tau)$
 
Anonymous
Is there necessarily a point $x$ in the infinite subset such that satisfies the condition of limit point
 
Thorgott
Thorgott
7:26 PM
that's not what you need
 
Anonymous
@Thorgott I need to check if every infinite subset has a limit point in $X$, right?
 
robjohn
robjohn
@MaryStar for pointwise convergence, it looks okay
 
Thorgott
Thorgott
yes, but that's not what you just said
 
Anonymous
Am I missing something? I think for this we need to show that for any infinite subset $A \subseteq X$ there exists a point $x \in A$ such that $x$ is a limit point. That is, any open set $(a, \infty)$ containing $x$ must also contain another point in $A$ apart from $x$.
 
Thorgott
Thorgott
you need a limit point in $X$
not necessarily in $A$
 
Anonymous
7:34 PM
Oh! So we rather need that for any infinite subset $A \subseteq X$ there exists a point $x \in X$ such that $x$ is a limit point. That is, any open set $(a, \infty)$ containing $x$ must also contain a point in $A$. Is that right?
 
Thorgott
Thorgott
yes
 
Anonymous
Well, intuitively it seems true but I can't frame it rigorously
 
Anonymous
There can be weird subsets of $\mathbb R$
 
Anonymous
But I think we can always choose some point somewhere in the middle of a subset such that any $(a, \infty)$ will contain some other points of the subset
 
Anonymous
Maybe there can be a proof by contradiction
 
Anonymous
7:40 PM
Suppose there is no limit point for some subset $A \subseteq X$
 
Anonymous
Hmm, not sure how to proceed
 
Thorgott
Thorgott
you're making this too complicated, this can be done straightforwardly
there is an obvious sufficient condition for being contained in any $(a,\infty)$ which contains $x$
 
Mary Star
Mary Star
@robjohn Great!! And the way I formulated the proof for the uniform convergence for $f_n$ and $g_n$ ?
 
Anonymous
@Thorgott That there must be points in $A$ greater than $x$?
 
Anonymous
What if the subset $A$ has $x$ as a least upper bound
 
Anonymous
7:45 PM
Oh, but then we could perhaps have an element in between $a$ and $x$ where $(a, \infty)$ is the open set
 
Thorgott
Thorgott
keep in mind what you are trying to do
 
Artificial Stupidity
Artificial Stupidity
I think the following site is wrong.
My answer is 120.
 
Anonymous
I guess if a subset $A$ has at least two elements say $\{m, n\}$ we can always pick the non-maximum element as a limit point as it has some element in $A$ to the right of it
 
Anonymous
But $A$ is said to be infinite so this can always be done. Any non-maximum element of $A$ is a limit point I guess
 
Anonymous
Plus all elements in $X$ that are lesser than the supremum of $A$
 
Thorgott
Thorgott
7:50 PM
yup
 
Anonymous
Cool, phew. I was indeed making it too complicated :D
 
Anonymous
So $X$ is definitely limit point compact
 
Anonymous
The next questions asks if $X$ is Lindelof, that is whether every open cover of $X$ has a countable subcover
 
Anonymous
This seems like asking whether the whole space $X$ is compact
 
Thorgott
Thorgott
no, compact means every open cover has a finite subcover, Lindelöf means every open cover has a countable subcover
 
Anonymous
7:53 PM
We can always cover $\mathbb R$ using $(n-1, n+1) \ \forall n \in \mathbb N$ balls I guess
 
Anonymous
@Thorgott Ah, right
 
Thorgott
Thorgott
might as well use
general fact: $\sigma$-compact implies Lindelöf
 
Anonymous
Oh, since every compact subset of $X$ has a finite subcover and because $X$ is a countable union of compact subsets
 
Anonymous
I'm not sure if I'm missing some subtlety
 
Anonymous
Well, any open cover of $X$ is also an open cover of a subset of $X$
 
Anonymous
7:59 PM
And any such subset has a finite subcover
 
Anonymous
We can just take the union of such subcovers
 
Anonymous
It will always be countable
 
Anonymous
The converse doesn't seem to be true though. Lindelof may not mean $\sigma$-compactness
 
Anonymous
The last and final part of the question asks whether $X$ is pseudocompact :D
 
Anonymous
Where pseudocompactness means every continuous function $f: X \to \mathbb R$ is bounded
 
Anonymous
8:04 PM
Hmm, in general metric spaces I know the proof that continuous functions over a compact interval are bounded
 
Anonymous
Well, here $X$ is not compact but is close. It is Lindelof. Maybe I can frame a proof along the same lines.
 
Thorgott
Thorgott
yeah, that argument for $\sigma$-compact implies Lindelöf works
 
Anonymous
Thanks for the confirmation!
 
Anonymous
If $f$ is continuous, then the sets $U_n = \{x \in X : f(x) > n\}$ for $n \in \mathbb N$ are open as they're preimages of the open sets $(n, \infty)$ in $X$.
 
Anonymous
$\{U_n\}_{n \in \mathbb N}$ is an open cover for $X$
 
Anonymous
8:14 PM
Hmm, not sure this helps. We only have that it has a countable subcover. It is already countable though.
 
Anonymous
Do you have any ideas?
 
Thorgott
Thorgott
try looking at the preimages of two disjoint intervals
 
Anonymous
Pre-image the union of two disjoint intervals in $\mathbb R$ will also be of the form $(a, \infty)$
 
Anonymous
Can we deduce anything from here
 
Thorgott
Thorgott
yes, we can, try thinking about why I want them to be disjoint
also not any open set in $X$ is of the form $(a,\infty)$, you're forgetting two special cases (one of which will turn out to be relevant)
 
Anonymous
8:22 PM
Ah, the empty set and the entirety of $X$
 
Anonymous
I don't know how they're relevant tho and I'm also not sure why you want them to be disjoint :P
 
Thorgott
Thorgott
what can you say about the preimages of disjoint sets under any mapping
 
Anonymous
@Thorgott I don't really know any result related to this. Does the pre-image have to be disjoint or something?
 
Anonymous
3
A: Inverse image of disjoint is disjoint?

DonAntonio$$x\in \phi^{-1}(A)\cap\phi^{-1}(B)\Longrightarrow \phi(x)\in A\cap B...\text{contradiction...?}$$

 
Anonymous
Oh, it does have to be disjoint. Interesting
 
Thorgott
Thorgott
8:28 PM
$f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$
This is a basic property of mappings. You definitely need to know this stuff.
 
Anonymous
Right, right. It makes sense and seems trivial now that I think of it
 
Anonymous
Well, which open sets in $X$ are disjoint
 
Anonymous
The empty set and any of the other open sets
 
Anonymous
So the pre-image of at least one of the disjoint components under the continuous map must be the empty set
 
Thorgott
Thorgott
now use this to conclude the map is constant
 
Anonymous
8:35 PM
Umm, you mean we can conclude from here that every element of $X$ is sent to a constant value $k \in \mathbb R$ by a continuous map $f$?
 
Anonymous
That seems strange....
 
robjohn
robjohn
@ArtificialStupidity From their instructions all the $3$s and $4$s are used, so the only optional digits are $5$ and $6$. There are $\frac{6!}{3!\,2!\,1!}=60$ ways with the $5$ and $\frac{6!}{3!\,2!\,1!}=60$ ways with the $6$.
So $120$ sounds correct
 
Anonymous
8:56 PM
Hmm, I thought of something. For any $x \in \mathbb R$, we have two disjoint open sets $A, B$ such that $A > x$ and $B < x$ and $A \cup x \cup B = \mathbb R$.
 
Anonymous
Then either $f^{-1}(A)$ is empty or $f^{-1}(B)$ is empty or both are empty.
 
Anonymous
9:10 PM
I guess I need to show that both are empty...not sure.
 
robjohn
robjohn
11:28 PM
@ArtificialStupidity Their mistake was that $_7P_6$ counts permutations where one of the repeated digits does not appear.
 
Ted Shifrin
Ted Shifrin
Howdy, @robjohn. Warm enough for you? I finally gave in and closed windows and turned on AC.
 
robjohn
robjohn
@TedShifrin It was 111°F here with 20% humidity (dewpoint 61°F) today. It was 105°F with 24% humidity (dewpoint 61°F) yesterday. It is supposed to be similarly hellish tomorrow.
 
Ted Shifrin
Ted Shifrin
I guess being over the hills in the valley has its disadvantages.
 
robjohn
robjohn
@TedShifrin yeah, it might be cooler near the ocean, but it will be more humid.
However, a dewpoint of 61° is pretty humid.
 
Ted Shifrin
Ted Shifrin
My little temperature gauge with relative humidity has disappeared ... who knows where.
 
Hawk
Hawk
11:49 PM
Suppose measurable func $(f_n) \to f$ a.e. I want to show $f$ is measurable. I want to say that since limit exists $\lim f_n = \lim sup f_n = \inf sup f_n.$ Now my problem here is I am not sure how to "say" that the null set $\{ f_n \neq\to f\}$ does not matter. I am doing things differently from my book.
 
EM4
EM4
hello guys
 

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