Prove $\{ n + \frac{1}{2n} : n \in \mathbb{N} \}$ is closed. I learned (1) the definition of a closed set: A set is closed if every limit point is a point of the set; (2) the definition of an open set: A set is open if every point is an interior point; and (3) the theorem: "A set is closed if and only if its complement is open".
I am hoping to prove that the set $E = \{ n + \frac{1}{2n} : n \in \mathbb{N} \}$ is closed. My thought is to show that its complement is open. To do so, I need to show that every $x \in E^{c}$ is an interior point of $E^{c}$, meaning that every $x$ has a neighborhood contained in $E^{c}$. I am thinking that for every $x \in E^{c}$, I can construct a neighborhood with a radius $r = \min (\lceil x \rceil - x, x - \lfloor x \rfloor, \frac{1}{2 \lfloor x \rfloor} - x, x - \frac{1}{2 \lceil x \rceil})$.
Then, is every such neighborhood is contained in $E^{c}$? I am not sure how to show this is true/untrue.
 A: Consider the set $S = \{x | \forall n \in \mathbb{N}, x \neq n + \frac{1}{2n}\}$. I claim that this set is open.
For consider some $x \in S$. Then by the Archimedean property, we can find some $n$ such that $n > x$. Then consider $\delta = \min\limits_{1 \leq m \leq n} |x - (m + \frac{1}{2m})|$. Then we see that $B_{\delta}(x) \subseteq S$. For suppose $y \in B_{\delta}(x)$. And suppose $y = k + \frac{1}{2k}$. Then $|x - y| = |k - (k + \frac{1}{2k}| < \delta$, so it must be the case that $k > n$. But then $k + \frac{1}{2k} > k \geq n + 1 > n + \frac{1}{2n} > x$, so $\delta > |x - (k + \frac{1}{2k})| = k + \frac{1}{2k} - x > n + \frac{1}{2n} - x = |x - (n + \frac{1}{2n})| \geq \delta$. This is a contradiction.
And clearly, the complement of $S$ is $\{n + \frac{1}{2n} | n \in \mathbb{N}\}$.
A: A slight modification of your attempt will work. Take $r = \min(\lceil x \rceil - x, x - \lfloor x \rfloor, \frac{1}{2}|x - \lfloor x \rfloor - \frac{1}{2 \lfloor x \rfloor}|)$. We will show all elements of $E$ are outside of the open ball of radius $r$ centered at $x$.

*

*For $n \le \lfloor x \rfloor - 1$, we have $|x - (n + \frac{1}{2n})| > (x - \lfloor x \rfloor  ) + (\lfloor x \rfloor - (n+1)) \ge x - \lfloor x \rfloor \ge r$.

*For $n \ge \lceil x \rceil $, we have $|x - (n+\frac{1}{2n})| > (n - \lceil x \rceil) + (\lceil x \rceil - x) \ge \lceil x \rceil - x \ge r$.

*It remains to consider $n=\lfloor x \rfloor$. We have $|x - (n + \frac{1}{2n})| > \frac{1}{2} |x - (n + \frac{1}{2n})| \ge r$.

A: If you know open intervals are open sets, and arbitrary unions of open sets are open, then $$E = \{ 1.5, 2.25, 3.167, 4.125, ... \}$$ has a complement which is easily seen to be a union of open intervals: $$E^c = (-\infty, 1.5) \cup (1.5, 2.25) \cup (2.25, 3.167) \cup (3.167, 4.125) \cup ....$$
A: Let $a_n = n + \frac{1}{2n}$, so that $E=\{a_n\mid n \in \Bbb N\}$.
Suppose that there were a sequence $(x_n)$ from $E$ that converges to some $x \in \Bbb R$. As all convergent sequences are bounded and $|a_n| > n$ for all $n$, we have that only finitely many different $a_n$ are used in this sequence and so there is one $a_m$ that occurs infinitely any times and that subsequence of $(x_n)$ is convergent to $a_m$ and as limits are unique $x= a_m \in E$.
So $E$ is closed as it’s closed under sequential limits and we’re working in a metric space.
