Les $R$ be the ring of cauchy sequences in the field $\mathbb{R}$ of real numbers and $I=\{(x_n):(x_n) \rightarrow 0 \}$ be a subset of $R$. Then which of the following is correct:
$(1)$ For every real number $a$ their exists a $(x_n)$ in $R$ such that $x_n\rightarrow a$.
$(2)$ $I$ is prime ideal.
$(3)$ $I$ is maximal ideal.
$(4)$ $I$ is prime but not maximal.
$(5)$ $I$ is principal ideal.
$(6)$ Every real number has a unique representation in $R/I$.
I can see that for $a\in \mathbb{R}$ we have a sequence $(x_n)=(a,a,a,\ldots)$ such that $(x_n)\rightarrow a$. And if $x_n y_n\rightarrow 0$ then at least one of $x_n$ or $y_n$ must converse to $0$, so $I$ is prime ideal. But I am not able do anything for the rest options. Please help me
Questions (1) - (4) and (6) are addressed in the comment.
For (5), we can show it is false by contradiction. Suppose that $t = (t_n)$ is a generator of this ideal. Note that no $t_n$ can be zero (otherwise no element with a non-zero entry in that spot could be a multiple of $t$).
Define $s = (t_n / n)$. Since $t$ is a generator, one has $$ t = s\alpha $$ for some $\alpha$. We compute $\alpha_n = n$ and therefore $\alpha$ is not a Cauchy sequence, a contradiction.
(We could probably word this in a less sloppy way to avoid proof by contradiction; the point is that nothing can be a generator because we can construct something such that the quotient isn't Cauchy.)