Rationals in [0,1] are $F_{\sigma}$? I have the following problem that I think I have worked out.


  Let $A$ be the set of rational numbers in $[0,1]$. Is $A$ an $F_{\sigma}$ set?


My Attempt: Yes, $A$ is an $F_{\sigma}$ set. We recall that an $F_{\sigma}$ set is one which may be expressed as the countable union of closed sets. Since the set of rationals $\mathbb{Q}$ are countable, so too is the set $A$. Hence, we may enumerate it as $A=\{q_{k}\}_{k=1}^{\infty}$. Now, each singleton set $\{q_{k}\}$ is closed in $\mathbb{R}$ (with the standard topology), since its complement $\{q_{k}\}^{c}=(-\infty,q_{k})\cup(q_{k},\infty)$ is open. Thus, we have $A=\bigcup_{k=1}^{\infty}\{q_{k}\}$, so $A$ is indeed an $F_{\sigma}$ set.

My concerns: I think my above argument is okay, but I am shaky on one part. Does it matter if I consider each $\{q_{k}\}$ as closed in $\mathbb{R}$, or do I need to show explicitly that they are closed in $[0,1]$? Thanks in advance for any help!
If $S$ is closed in $\Bbb R$, then $S\cap X$ is closed in $X$ for all $X\subseteq \Bbb R$, so showing closedness in $\Bbb R$ is sufficient.
Yes it is a set. What you did is sufficient. Good job!