i know these is one equivalent statement to that cosnequent, but i was looking for other one

@BalarkaSen $\lfloor 2\times 0.75 \rfloor = 1$

So 0.75 is a solution

@Alizter No. $\lfloor 0.75 \rfloor = 0$ which is not invertible.

@nerdy I didn't know the name, sorry. But I think the $\sin(1/x)$ example may help you notice that a accumulation point is most probably not related to inf or sup.

@BalarkaSen $\lfloor 2\times 0.75\rfloor=2\lfloor 0.75 \rfloor +1$

That's not the equation darn it.

oh well

fi

For a sequence x_n and a real number a in R, is saying that for every positive epsilon, there exists infinitely many n in N with |x_n - a| < epsilon the same as saying that for every positive epsilon, x_n has terms in the interval (a - epsilon , a + epsilon ) ?

Ian Mateus, for example the last part would be another equivalent statement ?

@nerdy yes, that's precisely what the definition says.

@BalarkaSen indeed there are no solutions.

@Vrouvrou It would seem so. Do you have a definition of an inclusion that you are working with?

Ian Mateus, wouldn't i have to say "for every positive epsilon, x_n has terms in the interval (a - epsilon , a + epsilon ) for infinitely many n " ?

or is the last " for infinitely many n " unnecessary ?

@nerdy as long as you say $(a-\varepsilon, a+\varepsilon)\setminus\{a\}$, there is no problem, it will have infinitely many terms automatically. If you don't restrict $a$, you must say "infinitely many".

@IanMateus you mean for $\{a\} \geq 1/2$, don't you?

@BalarkaSen yes, and there are exactly three solutions otherwise. I restricted $a>0$, though

yes

well, there are no solution if $a < 0$

Why?

LHS is negative while RHS is positive.

Yeah yeah, indeed.

@Ian how did you prove that it has no solution if $\{a\} \geq 1/2$?

@BalarkaSen If all else fails, you can exhaust the few possibilities. Since the RHS is $\geqslant \frac{1}{3}$, you know that $2 < a < 5$.

@BalarkaSen calling $\lfloor a\rfloor =x$ we know $\frac 12\leq\{a\}=\frac{1}{x}+\frac{1}{2x+1}-\frac 13 \lt 1$. The leftmost inequality is equivalent to $10x^2-13x-6\lt 0$, which has $0$ and $1$ as only integer solutions. Testing the $1$ for the rightmost inequality, we get $4/3\lt 4/3$, absurd.

@DanielFischer nope. you mean $2 < \lfloor a \rfloor < 5$

@IanMateus haha, cool that's exactly what i did.

@BalarkaSen Nope, $\left\lfloor \frac{29}{12}\right\rfloor = 2$.

@DanielFischer i don't even ...

from where are you deriving $2 < a < 5$?

also, it doesn't help either. $a$ is not an integer.

@BalarkaSen $\frac{1}{3} \leqslant RHS < \frac{4}{3}$. If $a \geqslant 5$, then the left hand side is $\leqslant \frac{1}{5} + \frac{1}{10} = \frac{3}{10} < \frac{1}{3}$, so $a < 5$. If $a < 2$, then the left hand side is $\geqslant \frac{4}{3}$, and $a = 2$ doesn't work.

Aha aha

So we need to check $a = k + \{a\}$ for $k\in \{2,3,4\}$.

yes, that I already did.

the $\lfloor 2a \rfloor = 2 \lfloor a \rfloor + 1$ case was devilish.

Yes. And "if all else fails" to see that $\{a\} < \frac{1}{2}$ for all solutions, check the three possibilities manually.

@Alizter Are you replacing all the $dx$ by $\text{d}x$ ? There's like a flood of old modified questions :c

$$\frac{1}{2} + \frac{1}{5} - \frac{1}{3} = \frac{7}{10} - \frac{1}{3} = \frac{11}{30} < \frac{1}{2},$$ so no solution $2.5 \leqslant a < 3$. $$\frac{1}{3} + \frac{1}{7} - \frac{1}{3} = \frac{1}{7} < \frac{1}{2},$$ so no solution $3.5 \leqslant a < 4$. $$\frac{1}{4} + \frac{1}{9} - \frac{1}{3} = \frac{1}{36} < \frac{1}{2},$$ so no solution $4.5 \leqslant a < 5$. Done.

@BalarkaSen Only mildly tedious.

Hehe, right.

Thanks.

@DanielFischer Daniel.

@PedroTamaroff What's up? How are you?

@DanielFischer Doing fine, you?

@PedroTamaroff So-so. My aunt won't live much longer, and that makes me sad.

@DanielFischer How old is she?

92.

@DanielFischer Ah. Well, might sound silly but I've got the same issue with my dog. I guess you cannot compare, still...

Being younger, I guess I'm still a rookie with all this stuff. =/

@DanielFischer Does she live close to you?

Thanks Ian Mateus

@PedroTamaroff She does. A couple of minutes to walk.

@DanielFischer Well, that's nice.

@Hippalectryon Sorry about that. I thought I was helping with that. Turns out nope.

@Alizter There should be an option 'edit without pinging' :/

@Hippalectryon You could always click on questions then newest

personally, i think $\mathrm{d} x$ looks cool

Meh

@Alizter I know, but I usually choose changed because there are more :)

why @Ian?

@BalarkaSen I stopped using it because $\mathrm d$ is ugly.

@BalarkaSen Not everyone agrees apparently. So I read about and after @DanielFischer stopped me I think it should be left. I do agree with respecting the OP's original typographical choice.

yeah, each to his own.

I feel like a neighbor who painted both sides of the fence.

@Alizter I understand why you are making the edits, but it was a bit much at a time xD

This is what redbull does to you

uh ?

@BalarkaSen it gets worse when $\frac{\mathrm dy}{\mathrm dx}$ comes in.

yeahh, it looks ugly in differentiations.

but i use it on integrals time to time

The ISO-good-looking $\mathrm e^{\mathrm i \theta}$ is simply a joke.

The best typesetting is ones own handwriting.

@IanMateus Is that what ISO recommend?

Now I feel bad for using mathrm

@Alizter I read that somewhere some time ago. $\mathrm dx$ is also recommended, if I'm not wrong.

Great now Asaf told me off

Screw editing

I am confused

@Alizter please don't even think of undoing it! :P

@DanielFischer I was meaning to ask if you knew about a book that deals with simplices, triangulations, coloring of simplices, Sperner's lemma, &c. Is this algebraic topology?

@PedroTamaroff Simplices, triangulations, and Sperner's lemma are algebraic topology indeed. Colouring of simplices might be art, however ;)

(Sorry, no book recommendation. The books on algebraic topology I've looked in were in German.)