:)

Hi @jasper

@JasperLoy replied, I'll go eat

@JasperLoy :)

@BrianM.Scott Hey there.

@PeterTamaroff What’s up?

The thing is: Let $G$ be an additive subgroup of $\Bbb R$. Let $G^{+}=G\cap (0,\to)$. Then $$\alpha=\inf G^{+}>0\implies G=\alpha \Bbb Z$$ and $$\alpha=0\implies \bar G=\Bbb R$$

Hi all

I have proven the first.

Now, for the second, one can prove easily that $(-\epsilon,\epsilon)\cap G\neq \varnothing$ for each $\epsilon >0$.

Then I'm just thinking about translating the group (i.e. $G+g=G$ for $g\in G$) to get any other ball centered at $x\in\Bbb R$ to instersect.

Looks okay so far.

I can't make it formal though.

@pourjour مساء الخير

@JasperLoy yes - very slowly

@PeterTamaroff Suppose that there is some $\epsilon>0$ such that the translates $(g-\epsilon,g+\epsilon)$ for $g\in G$ don’t cover $\Bbb R$. Pick an $x$ that isn’t covered. Then for any $g,h\in G$ with $h<x<g$ you have $g-h>\epsilon$. In other words, $G$ has a ‘hole’ of size at least $\epsilon$. Get a contradiction with $\alpha=0$.

@anon Did you resolve the cyclotomic field problem?

@CтарыйДжон كيف حالك؟

Hello

@pourjour أنا بخير، شكرا

I have a subset A of $\mathbb{R}^n$ which is open. What is $\partial A$ ?

(=ana bikhair, shukran?)

@BrianM.Scott I want to prove that $\forall x\in\Bbb R$; $(x-\epsilon,x+\epsilon)\cap G\neq\varnothing$.

@PeterTamaroff And the negation of that is the hypothesis in my last comment.

They denote in my problem $\partial A$. What is partial of a set ?

@CтарыйДжон جيد

@user43418 The boundary.

$\bar A\cap\overline{X-A}$, usually.

@pourjour sorry - I don't understand " جيد"

@CтарыйДжон it means "Good"

@CтарыйДжон You are good at arabic when did you start learning it?

@pourjour Hmm - is that modern standard arabic, or dialect?

Does anyone know how the boundary is called in french ?

@pourjour I started learning arabic about 35 yearsd ago :)

@BrianM.Scott Sorry, let me rephrase, see if I can get the negation straight. That $\bar G=\Bbb R$ means that for every $\epsilon >0$, for every $x\in\Bbb R$; $$B(x,\epsilon)\cap\Bbb R\neq \varnothing$$

@user43418 frontiere?

How do I negate that double $\forall$? @BrianM.Scott

@CтарыйДжон it's standard

I got something like $\forall\epsilon\forall x P(x,\epsilon)$

@pourjour darn - then I should have known it :(

@CтарыйДжон On wikipedia it is written probleme aux limites

@CтарыйДжон hmm interesting

But I don't think it is exactly boundary

@PeterTamaroff There are an $x\in\Bbb R$ and an $\epsilon>0$ such that $B(x,\epsilon)\cap G=\varnothing$.

@CтарыйДжон there many words such as "جيد"

like "ممتاز"

@pourjour aha! jayidan?

I never read or heard the term "frontiere" used

yeah - mumtaaz is one I have met

@user43418 I have seen English frontier used for the topological boundary, but it’s not common and appears to be a borrowing of French frontière, which is the normal French term.

@CтарыйДжон yeah

@pourjour I sometimes struggle when the diacritics for vowels are missing :(

@BrianM.Scott Generally frontiere just means border in french

@user43418 As a technical term in topology it means the boundary of a set.

@CтарыйДжон yeah you have to read a lot to read without those signs

@BrianM.Scott Oh, and you take $(g-\epsilon,g+\epsilon)$ but with $g\in G$ because...? If $B(x,\epsilon)\cap G=\varnothing$ then

@pourjour yep - I am OK with books for kids :)

@CтарыйДжон there are good to get started

for each $g\in G$

$|x-g|\geq \epsilon$

french wikipedia

@PeterTamaroff Actually, the argument that I originally suggested is more complicated than necessary. Just start with $x\in\Bbb R$ and $\epsilon>0$ such that $B(x,\epsilon)\cap G=\varnothing$. Show that this implies a gap in $G$ of size at least $2\epsilon$, then get a contradiction with $\alpha=0$.

or $x\notin (g-\epsilon,g+\epsilon)$

@CтарыйДжон Which is exactly what I checked a few minutes ago to verify my memory!

@CтарыйДжон Thank you

@pourjour yep - but not easy to find in the UK

@BrianM.Scott OK.

@BrianM.Scott Yeah - I recall reading some topology papers in French years ago - about boundaries and stuff in topologies associated with potential theory :)

I am trying to find on the stack exchange a proof of the following: If A is an open subset of $\mathbb{R}^n$, then every continuous function $f: A \rightarrow \mathbb{R}$ is unbounded

@user43418 that doesn't sound true!

@user43418 That’s clearly false: look at any constant function.

@pourjour Prove that if $p\not\mid a $ then $p^2\not\mid a^2$.

is not closed equivalent to open ?

@user43418 Uh?

Equivalent?

@PeterTamaroff and it's not easy

Ah that is why

@user43418 No: in $\Bbb R$ the set $[0,1)$ is neither closed nor open.

Sorry

See my post again then

open sets are not like doors - closed and open are not opposites

darn - too slow :(

@user43418 Again, just look at a constant function: it’s continuous on any domain and certainly bounded.

Suppose A ⊂ Rn is not closed. Show that there is a continuous function f : A → R which is not bounded.

maybe something along the lines of "if a set is open then there exists a continuous function which is not bounded ..."?

Your examples

@BrianM.Scott So if such $x,\epsilon$ exist, there exist $g,h\in G$ for which $g-h\geq 2\epsilon$ and there is no $h'\in G$ for which $g<g'<h$?

@user43418 That at least has the virtue of being true!

On stackexchange

but I don't seem to find it anymore

@PeterTamaroff Make that $\ge2\epsilon$, but otherwise it’s okay.

brooonzyah.net/vb/t56815.html

brooonzyah.net/vb/t56815.html

st-takla.org/Learn_Languages/01_Learn_Arabic-ta3leem-3araby/…

brooonzyah.net/vb/t56815.html

st-takla.org/Learn_Languages/01_Learn_Arabic-ta3leem-3araby/…

@PeterTamaroff Now it looks okay. And now subtract $g$ from everything to get $\alpha\ge h-g$.

@JasperLoy I didn't come here to find the solution

I thought someone could help me find the post on stackexchange

@JasperLoy Yes

I was reading a book about counterexamples

and this problem came up

And then I looked for it on stacexchange

found it

and now I can't seem to find it

@user43418 It would be easier for me to produce the argument than to find the question.

The argument is short.

Ok

... well yes for the solution lol

(in a way)

@pourjour Many thanks - I will see what I can learn from those links

@JasperLoy you are annoying :)

@user43418 If $A$ is not closed, choose a point $p\in(\operatorname{cl}A)\setminus A$. Define $f:A\to\Bbb R$ by $f(x)=\frac1{\|x-p\|}$; then $f$ is continuous and unbounded.

@JasperLoy I told you

@CтарыйДжон you're welcome - I hope you can read those links :)

@pourjour I am managing with http : //st-takla.org/ - so far!

@CтарыйДжон you can use google translator I you get any difficulties

@JasperLoy I was just joking when I said you were annoying. I didn't mean it

@pourjour :)

@CтарыйДжон ok I'm here if you need any help

@BrianM.Scott Do you think there is a direct proof using that $(-\epsilon,\epsilon)\cap G^{+}\neq\varnothing$?

@pourjour Thanks!

Because we essentially used that.

@JasperLoy I wasn't trying to trick you

Time I got some sleep - goodnight all

@PeterTamaroff There might be one, but I didn’t see one, and the other argument is so easy that I didn’t spend a lot of time looking.

@JasperLoy I am new here. Therefore I can't judge what is and isn't "usual"

@JasperLoy Bye for now

Suppose i have a system of equations At=b representing two planes in the euclidian space R^3 where A is 2x3 and t is (x y z). My same system has a 1-dimension NullSpace, that means the intersection of the planes shifted at the origin is a line.If the only possible free variables are x and y for example, does it mean the line is in the x-y plane ? If the only possible free variables are x and z for example, does it mean the line is in the x-z plane?

@JasperLoy I am sure there are

@JasperLoy Yes :)

@BrianM.Scott Did you see it Brian?

@PeterTamaroff Yes, but it takes a while to write a proper response.

@BrianM.Scott Sorry, I'll be more patient next time.

@JasperLoy I know.. I didn't really know what to write

ok\

@PeterTamaroff A couple of points. First, you do have to justify the claim that there are points of $G$ on both sides of the ‘bad’ $x\in\Bbb R$. (This is not hard.) Then the second sentence is awful: there’s no reason to write $\ell'$ as $\ell-h$. If you don’t want to say simply that $G\cap(0,h-g)=\varnothing$ because $G\cap(g,h)=\varnothing$ $-$ i.e., if you want to prove that assertion $-$ do it like this. Suppose that $\ell\in G\cap(0,h-g)$; ...

... then $\ell+g\in G\cap(g,h)=\varnothing$, which is impossible.

@BrianM.Scott OUCH! Awful is a strong word! =(