Show closedness of $D:=\bigcup\limits_{k=1}^{\infty}\left[k-\frac{1}{2^{k+1}},k+\frac{1}{2^{k+1}}\right]$

Let's consider $D:=\bigcup\limits_{k=1}^{\infty}\left[k-\frac{1}{2^{k+1}},k+\frac{1}{2^{k+1}}\right]\subseteq\mathbb{R}$.

Show that $D$ is closed.

We know that if $D$ contains all its limit points then it is closed. Let be $a^*$ such a limit point, then its neigbourhood $N:=\{a\in \mathbb{R}\mid|a-a^*|<\frac{1}{2}\}$ contains a point $a\neq a^*$ with $a\in D\implies a\in \left[k_0-\frac{1}{2^{k_0+1}},k_0+\frac{1}{2^{k_0+1}}\right]$ for some $k_0\in\mathbb{N}$. If we shrink the neighbourhood of $a^*$, then again we find an $a'\neq a^*$ with $a'\in\left[k_0-\frac{1}{2^{k_0+1}},k_0+\frac{1}{2^{k_0+1}}\right]$ because any other interval would be too far away. So $a^*$ is a limit point of the interval $\left[k_0-\frac{1}{2^{k_0+1}},k_0+\frac{1}{2^{k_0+1}}\right]$. As $\left[k_0-\frac{1}{2^{k_0+1}},k_0+\frac{1}{2^{k_0+1}}\right]$ is closed $a^*$ must be an element of this interval. Hence, $D$ is closed.

Is this correct? Any suggestions are welcome :)

Solution 1:

You can express $D^c$ as a (countable) union of open sets, i.e., $$ D^c=(-\infty,3/4)\cup\bigcup_{k=1}^{\infty}\left(k+2^{-k-1},(k+1)-2^{-k-2}\right). $$

Solution 2:

For a direct approach to solve this question, you must prove that any convergent sequence $\{a_n\}\subseteq D$ has its limit in $D$. To prove this, if $a_n$ converges to $p$, prove that there is a unique subinterval $[k-\frac{1}{2^{k+1}},k+\frac{1}{2^{k+1}}]$ for which, all terms of $\{a_n\}$ after a specific term, say $a_N$, lie in (refer to the convergence of $\{a_n\}$). The conclusion will be easy from this point, as $p$ must also be in that subinterval due to its closedness.