Show that $\{ \sum_{n \in \mathbb{N}} u_n b_n \textrm{ | } (b_n) \in \{ 0, 1 \}^\mathbb{N} \}$ is a closed subset of $\mathbb{R}$ The problem
Consider a real sequence $(u_n)$ with non-negative terms. We assume that $\sum u_n$ converges.
I would like to show that $\{ \sum_{n = 0}^{+ \infty} u_n b_n \textrm{ | } (b_n) \in \{ 0, 1 \}^\mathbb{N} \}$ is a closed subset of $\mathbb{R}$.
What I have tried
I have tried to use the sequential characterisation of closed sets: let $(b^k)_{k \in \mathbb{N}}$ be a sequence of elements of $\{ 0, 1 \}^\mathbb{N}$ and assume  $\sum_{n = 0}^{+ \infty} u_n b_n^k \to x \in \mathbb{R}$ when $k \to +\infty$. The goal is to show there exists $(c_n) \in \{ 0, 1 \}^\mathbb{N}$ such that $x = \sum_{n = 0}^{+ \infty} u_n c_n$.
I also assumed in the first instance that the $u_n$ are positive.
But I do not know how to expose such a sequence $(c_n)$.
 A: Let $(b^k)$ be a sequence in $\{0,1\}^{\mathbb{N}}$, such that there exists an $x \in \mathbb{R}$ such that
$\sum_{n\in \mathbb{N}}u_nb_n^k \rightarrow x \text{ for } k \rightarrow \infty$
Notice first that for any  $N \in \mathbb{N}$, there exists a $K \in \mathbb{N}$ such that $b_1^k,...,b_N^k$ are constant for each $k > N$ (this is because $u_n \ge 0)$. Therefore we can construct a $b \in \{0,1\}^{\mathbb{N}}$ corresponding to such constants. 
Now all you need to prove is that $x = \sum_{n \in \mathbb{N}}u_nb_n$, which should be relatively simple. For example, show that the differences of the partial sums converge to 0 by the tail lemma.
A: Let $(b^k)_{k \in \mathbb{N}}$ be a sequence in $\{0,1\}^{\mathbb{N}}$, such that there exists an $x \in \mathbb{R}$ such that $\sum_{n\in \mathbb{N}}u_nb_n^k \rightarrow x \text{ when } k \rightarrow +\infty$.
Let's study the sequence $(b_0^k)_{k \in \mathbb{N}}$. It is bounded so we can find an increasing function $\varphi_0 : \mathbb{N} \rightarrow \mathbb{N} $ such that $(b_0^{\varphi_0(k)})_{k \in \mathbb{N}}$ converges to a value $c_0 \in \{0,1\}$, according to Bolzano-Weierstrass theorem.
Now, study $(b_1^{\varphi_0(k)})_{k \in \mathbb{N}}$. Again, you can find $\varphi_1 : \mathbb{N} \rightarrow \mathbb{N} $ increasing such that $(b_1^{\varphi_0(\varphi_1(k))})_{k \in \mathbb{N}}$ converges to a value $c_1 \in \{0,1\}$.
And so on, assuming $\varphi_0, ..., \varphi_{n-1}$ are built for some $n \in \mathbb{N}$, we build $\varphi_{n}$ increasing such that $(b_n^{(\varphi_0 \circ \cdots \circ \varphi_{n})(k))})_{k \in \mathbb{N}}$ converges towards $c_n \in \{0,1\}$ (actually, it is stationary to it).
We finally set $\phi(n) = (\varphi_0 \circ \cdots \circ \varphi_{n})(n)$ for all $n$. $\phi$ is an increasing function. So, the limit of $\sum_{n \in \mathbb{N}}u_nb_n^{\phi(k)}$ when $k \to +\infty$ is still $x$.
But we can show that the limit is also $\sum_{n \in \mathbb{N}}u_nc_n$. Indeed, let $\epsilon > 0$. There exists $N \in \mathbb{N}$ such that $\sum_{n = N+1}^{+\infty}u_n \leq \epsilon$. As $b_0^{\phi(k)}, ..., b_N^{\phi(k)}$ tend towards $c_0, ..., c_N$ when $k \to +\infty$, there exists $K$ such that for $k \geq K$, these quantities are equal to their limits (because they alternate between discrete values).
So, we have, for all $k \geq K$:
$$
\begin{align*}
\left| \sum_{n = 0}^{+\infty}u_n b_n^{\phi(k)} - \sum_{n = 0}^{+\infty}u_n c_n \right| 
&= \left| \sum_{n = 0}^{+\infty}u_n \left( b_n^{\phi(k)} -c_n \right) \right| \\
&= \left| \sum_{n = N+1}^{+\infty}u_n \left( b_n^{\phi(k)} -c_n \right) \right| \\
&\leq \sum_{n = N+1}^{+\infty}u_n \left| b_n^{\phi(k)} -c_n \right| \\
&\leq \sum_{n = N+1}^{+\infty}u_n \leq \epsilon \\
\end{align*}
$$
because $\left| b_n^{\phi(k)} -c_n \right| \leq 1$ as $b_n^{\phi(k)}, c_n \in \{ 0, 1 \}$.
To conclude, as the limit is unique, $x = \sum_{n = 0}^{+\infty}u_n c_n$ and the set is closed.
$\phi$ is well-defined and increasing
Let us call an increasing function from $\mathbb{N}$ to $\mathbb{N}$ an extraction. We will prove $\phi$ is one. Clearly, $\phi$ is well-defined. A property that can be easily showed by induction is that for all extraction $s$, $s(n) \geq n$ for all $n \in \mathbb{N}$.  Moreover, composing an arbitrary number of extractions still gives you an extraction.
Let $n \in \mathbb{N}$. We will show $\phi(n+1) > \phi(n)$. 
$$
\begin{align*}
\phi(n+1) 
&= (\varphi_0 \circ \cdots \circ \varphi_{n+1})(n+1) & \\
&> (\varphi_0 \circ \cdots \circ \varphi_{n+1})(n) & \textrm{$\varphi_0 \circ \cdots \circ \varphi_{n+1}$ is increasing}\\
&= (\varphi_0 \circ \cdots \circ \varphi_{n})(\varphi_{n+1}(n)) \\
&\geq (\varphi_0 \circ \cdots \circ \varphi_{n})(n) = \phi(n) & \textrm{$\varphi_0 \circ \cdots \circ \varphi_{n}$ is increasing and $\varphi_{n+1}(n) \geq n$}\\
\end{align*}
$$
So at the end you have $\phi(n+1) > \phi(n)$: $\phi$ is an extraction.
$\forall n \in \mathbb{N}, b_n^{\phi(k)} \underset{k \to +\infty} {\longrightarrow} c_n$
Let $n \in \mathbb{N}$. Recall $\varphi_n$ has been constructed so that $b_n^{\varphi_0 \circ \cdots \circ \varphi_n(k)} \underset{k \to +\infty} {\longrightarrow} c_n$. Let $k \geq n$ and set $\psi(k) = (\varphi_{n+1} \circ \cdots \circ \varphi_k)(k)$. In the same manner as above, one can show that $\psi(\cdot + k)$ is an extraction and then $b_n^{\phi(k)} = b_n^{( \varphi_0 \circ \cdots \circ \varphi_n ) (\psi(k))} \underset{k \to +\infty} {\longrightarrow} c_n$.
