Question : Prove that $C+C=\{x+y\mid x,y\in C\}=[0,2]$, using the following steps:
We will show that $C\subseteq [0,2]$ and $[0,2]\subseteq C$.
- a) Show that for an arbitrary $n\in\mathbb{N}$ we can always find $x_n,y_n\in C_n$, where $$C_n=\left[0,\frac1{3^n}\right]\bigcup \dots \bigcup\left[\frac{3^{n}-1}{3^n},1\right]$$such that for a given $s\in[0,2]$ we have $x_n+y_n=s$.
- b) Then we will set $x=\lim x_n$ and $y=\lim y_n$, then $x+y=s$.
My progress : Showing that $C+C\subseteq [0,2]$ is obvious, and I did part a) by showing that if $x_n,y_n$ are in different subintervals then concluding that $x_n+y_n$ covers $[0,2]$ (can be done using induction on $n$).
My difficulty is in the second part. The sequence $x_n$ is bounded so it must have a convergent subsequence $(x_{n_{k}})$. If we set $x=\lim x_{n_{k}}$, then we can conclude $\lim y_{n_{k}}=y=s-x$, thus $x+y=s$. First I thought that $x,y$ will be in $C$ as it is closed. However $(x_{n_{k}})$ may not necessarily be in $C$, so $x$ can't be in $C$ for sure.
Is this last thought correct? How do I overcome this last gap in my solution? Thanks for your help.
Best Answer
To carry out (b), note that if $\ell\le n_k$, then $x_{n_k}\in C_\ell$. Thus, for each $\ell$ there is an $m(\ell)$ such that $x_{n_k}\in C_\ell$ whenever $k\ge m(\ell)$. Let $\ell$ be arbitrary. Clearly
$$x=\lim_{k\ge m(\ell)}x_{n_k}\;,$$
and $C_\ell$ is closed, so $x\in C_\ell$. Thus, $x\in\bigcap_\ell C_\ell=C$.
(You might be interested in comparing your argument with the first proof here; they’re basically the same idea, just expressed a bit differently.)