Is the set of points whose even binary digits are zero closed in $[0,1]$? Let $A$ be the set of points $x$ in the unit interval $[0,1]$ that can be expressed as a binary series 
$$x = \sum_{k=1}^\infty 2^{-k}d_k, \qquad d_i \in \{0,1\},$$
Such that $d_i = 0$ for each even index $i$.

I would like to determine whether $A$ is a closed subset of $[0,1]$.

I’ve tried to show this by supposing we have a sequence $x_n$ from $A$ converging to a point $y$, and then showing all even digits of $y$ must also be zero as well. But I haven’t been able to see how closeness of $x_n$ and $y$ should force all even digits to match. 
Hints or solutions greatly appreciated.
 A: One way to do it: $A$ is the image of the compact space $\{0,1\}^{\mathbb N}$ under the map
$t \mapsto \sum_{k=0}^\infty 2^{-2k-1} t_k$, therefore is compact. 
A: Let $P= \{0,{1 \over2} \}$, $I=[0,1]$. Define the set operation $f(B) = P + {1 \over 4} B$. Note that if $B$ is closed, then so is $f(B)$.
Note that $x$ has an expansion $0.00***\cdots$ or $0.10***\cdots$ iff $x \in f(I)$.
Note that $x$ has an expansion $\sum_{k=1}^\infty p_k {1 \over 4^k}$, with $p_1,...,p_n \in \{0,2\}$ and $p_{n+1},... \in \{0,1,2,3\}$ iff
$x \in f^n(I)$.
Note that $x \in A$ iff $x \in f^n(I)$ for all $n$ iff $x \in \cap_n f^n(I)$.
Note that each $f^n(I)$ is closed, hence $\cap_n f^n(I) = A$ is closed.
(It doesn't matter here, but we also have $f^{n+1}(I) \subset f^n(I)$.)
A: Hint:
If $0<x<1$ and $x\notin A$, let $j$ the smallest even index such that $d_j=1$. Can you find an open neighbourhood of $x$? Note that the biggest number in $A$ lesser than $x$ have the same $j-1$ first digits as $x$ and then it follows $\ldots01010101\ldots$.
A: Followings hints given in other answers, we will show $A$ is closed by showing $B = [0,1] \setminus A$ is open. That is, for each $x \in B$, we will find an $\varepsilon_x > 0$ such that
$$(x - \varepsilon_x, x + \varepsilon_x) \cap [0,1] \subseteq B.$$
First consider $x = 1 \in B$. Any $y \in A$ may be written as 
$$y = \frac{1}{2} \sum_{k =0}^\infty 4^{-k} y_k, \quad y_k \in \{0,1\} \implies y \le \frac{2}{3}.$$
Therefore  we may take $\varepsilon_1 = 1/3$.
Now suppose $x \in B \cap  (0,1)$. Write
$$x = \sum_{k =1}^\infty 2^{-k} x_k,$$ 
such that there exists some smallest even index $j$ such that $x_j = 1$. 
We claim that, without loss of generality, we can assume the sequence $\{x_k\}_{k=1}^\infty$ is such that there exists an in index $J > j$ so that $x_J = 0$. To see this, suppose
$$ 1 = x_{j+1} = x_{j+2} = \cdots.$$
Then we must have $x_{j-1} = 1$, otherwise $\sum_{k =1}^\infty 2^{-k} x_k$ may be re-expressed to give $x \in A$. Moreover, we cannot have $j = 2$, because then $x = 1$. So, in the scenario that $1 = d_{j +1} = d_{j+2} = \cdots$, we can rewrite the series for $x$ so that $d_{j-2}$ is now one and all subsequent binary digits are zero. 
To proceed, suppose $y \in A$ and  $y = \sum_{k = 1}^\infty 2^
{-k} y_k < x$. Let $N \le j$ be the first index such that $y_N \neq x_N$. Then it must be that $y_N = 0$ and $x_N = 1$ and so
$$ \sum_{k = N+1}^\infty \frac{y_k}{2^k} \le \frac{1}{2^N}\left(1 - \sum_{\substack{k \ge N :\\ k \text{ even}}}   \frac{1}{2^{k-N}}  \right) \le  \left(\frac{1}{2^N} - \sum_{\substack{k \ge j :\\ k \text{ even}}}   \frac{1}{2^{k}}  \right) \le \sum_{k = N} \frac{x_k}{2^k} - \sum_{\substack{k \ge j :\\ k \text{ even}}}   \frac{1}{2^{k}}. $$
On the other hand, suppose $y \in A$ and $x < y$. Again, take $N \le j$ to be the first index such that $y_N \neq x_N$, and as before $x_N = 0$ and $y_N =1 $. This time we have the estimate
$$\sum_{k = N+1} \frac{x_k}{2^k} \le \frac{1}{2^N} - \frac{1}{2^J} \le \sum_{k = N}^\infty \frac{y_k}{2^k} - \frac{1}{2^J}. $$ 
So we can take 
$$\varepsilon_x = \frac{1}{2} \min\left\{ \sum_{\substack{k \ge j :\\ k \text{ even}}}   \frac{1}{2^{k}} ,\frac{1}{2^J}\right\} > 0.   $$
