@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.
@Alizter Solve in reals $$\frac{1}{\left\lfloor{a}\right\rfloor}+\frac{1}{\left\lfloor{2a}\right\rfloor}=a-\left\lfloor{a}\right\rfloor+\frac{1}{3}$$
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 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".
@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.
@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.
@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.
@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?