@Koro what are n_1,...,n_N

Correction: we divide the open interval into N parts and consider $N+1$ nos. of fractional parts of non-square non-negative integers that is $\{\sqrt n_1\},\{\sqrt n_2\},..., \{\sqrt n_{N+1}\}$, where $n_i$'s are non square non-negative integers (all distinct). By non-square n, I mean that $\sqrt n$ is not an integer.

ok, and then?

I want to prove or disprove that 0 is a limit point of the set S=$\{\sqrt n: n\in \mathbb Z^+\cup \{0\}\}$.

So first I want to show that for distinct $n_1, n_2$ (both non square), we can't have $\{\sqrt n_1\}=\{\sqrt n_2\}$.

@Koro: have you seen this identity, or something like it: $\sqrt{n^2+1}-n=\frac1{\sqrt{n^2+1}+n}$

I assume that is a typo

I am stuck here. Intuitively it seems true as easy examples show. $n_1=2, n_2=3$ and fractional parts of their square roots are not equal (0.414... and 0.73... respectively)

you mean $\{\sqrt{n_1}\}=\{\sqrt{n_2}\}$, presumably

@robjohn Mr. Rob: yes.

anyway, this is true, but I don't yet see how it implies the result

@Koro how big is the right side? Compare it to say $\frac1{2n}$

@robjohn 0 is a limit point of this.

the left side is $\left\{\sqrt{n^2+1}\right\}$ (fractional part)

@Thorgott It doesn't (yet) but after confirming that this is indeed true, I'll go on to apply pigeon hole principle. From there, the result will follow.

how so?

@robjohn How? That seems amazing.

Ahh.. please wait Mr. Rob. I think I got it.

okay

@Thorgott I was thinking for some i, j we'll have $0\lt \sqrt n_i -\sqrt n_j \lt \frac 1N$ but I see now that this will give not help my cause as unfortunately in this case the set S doesn't form an additive group unlike in case of the set {m+n$\pi:m,n \in \mathbb Z$}

$\{\sqrt{n^2+1}\}=\sqrt{n^2+1}-[\sqrt{n^2+1}]=\sqrt{n^2+1}-[n\sqrt{1+\frac 1{n^2}}]$ and the expression inside floor in the second term is $n(1+\frac 1{2n^2}+o(\frac 1{n^2}))=n+\frac 1{2n}+o(\frac 1{n})$ as $n\to \infty$ and so for large $n$ we'll have $\sqrt{n^2+1}-[n\sqrt {1+\frac 1{n^2}}]=\sqrt{n^2+1}-n$

Is my understanding correct, Mr Rob?

@robjohn

That is too complicated. $n=\left\lfloor\sqrt{n^2+1}\right\rfloor$, right?

so the fractional part of $\sqrt{n^2+1}$ is $\sqrt{n^2+1}-n$.

@robjohn I'm afraid Mr. Rob, I am seeing this for the first time.

you've worked with square roots?

what is the integer part of $\sqrt{17}$?

yes i have; 4

why do you say it is $4$?

i see it is true

I say it because $4^2\lt17\lt5^2$

yeah right.

@robjohn i somehow missed this message

If you look at $n^2$ and $(n+1)^2$, we have $n^2\lt n^2+1\lt(n+1)^2$, right?

so $n^2\lt n^2+1\lt (n+1)^2\implies n \lt \left\lfloor\sqrt{n^2+1}\right\rfloor\lt (n+1) $ and so $\left\lfloor\sqrt{n^2+1}\right\rfloor=n$

yes!

Wow. This is so fantastic!! I never noticed before. :-)

Have you used this identity before?

Or just invented/thought about it?

the fractional part of $x$ always differs from $x$ by an integer and is in $[0,1)$.

@Koro I've used it in many answers. or something close to it.

Either way, you are fantastic. Thanks a lot. Many thanks!! My question is answered now. :-) @robjohn

@robjohn I see :-)

@Koro That's great, as long as you understand what is going on.

@robjohn Coming up with this in the middle of a question asking 0 is a limit point or not of a given set, amazes me, fascinates me, inspires me to re-work the exercises with new methods created by me :-)

When you do, let us see those methods!

One day, I was solving exercises from chapter 2 (Topology) of Baby Rudin's book and I got stuck at one exercise so everywhere I saw contradiction was being used in solution and the methods were same. Then I came across one answer by Professor Brian M Scott and he solved it by contrapositive arguments and it was so amazing!! I thanked him and then I tried solving all the exercises using contrapositive arguments and the results were fantastic :-)

@robjohn Sure!

Different questions require different methods. Sometimes using a particular tool too much can lead to answers that are too complicated.

@copper.hat thanks for the clarification, however, you wrote “…collection of all linear combinations of elements of the set $S$ (alternatively the smallest linear subspace containing $S$).” Are these equivalent? Is not the set containing the collection of all linear combinations a set of sets whereas the smallest linear subspace containing $S$ is simply a set?

Might be down for a bit. Going to update Firefox. It's been bugging me for months now.

That was quick

@Koro: you should try to show that $\left\{\left\{\sqrt{n}\right\}:n\in\mathbb{Z}^+\right\}$ is dense in $[0,1]$.

Done! @Robjohn

How did you do that?

The previous question was part of me trying to do that :-)

@robjohn It's not complete yet though :-)

I'll share once it's finished.

That showed that $0$ is a point of density, what about the other points?

okay

@Koro Consider it done! is what I wanted to say.

@copper.hat never mind, what you wrote makes sense

@robjohn Here’s my proposed solution: I choose any open interval (a,b) in [0,1]. Now I choose N so large that $\frac 1N\lt (b-a)$ (possible due to Archimedes property). There exists some $p:=\{\sqrt n_1\}\in (0,\frac 1N)$ (we proved earlier that 0 is a limit point of the set). Now there exists minimum integer m such that $mp \le a$ hence $(m+1) p \gt a$. I claim that $(m+1) p \lt b$ for if $(m+1) p \ge b$ then $(m+1)p-mp =p \ge (b-a) \gt \frac 1N$ @robjohn

And since (a,b) is arbitrary open interval in [0,1], we are done.

No no. I’m still not done. Please ignore this.

ok

helli

I have $g: A\cup B \to \{0,1\}$ defined by f(x) if $x\in A$ and 0 whene $x\in B$

what is equal to g^{-1}(\{0;1\}$

@robjohn Tried a lot but nothing seemed to work. I thought it would be very easy as in the case of {m+n $\pi$: m,n$\in \mathbb Z$}

I said that $\frac 1{\sqrt {N^2+1}+N}\lt \frac 1 N$

and so tried to show that $\frac 1{\sqrt{(N-1)^2 +1}+N-1}$ is in (a,b).

But then realised that it may not be true that $\frac 1N\lt a$

I tried to fix it by choosing minimal N such that $\frac 1N\lt \min \{a,b-a\}$ but then I got stuck

@Koro this also doesn’t help as the set of $\{\sqrt n\}$ is not closed under scaler multiplication.

@Koro consider that $\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}}$ is a decreasing sequence.

@Vrouvrou I assume that is $f(x)=1$ if $x\in A$. Are $A$ and $B$ disjoint?