Let $B = \{-n +(1/n) \mid n = 2,3,4,\ldots \}$. Prove that $B$ is closed in $\mathbb R$

Question

Good night, I'm trying to prove this theorem:

Let $B = \{-n +(1/n) \mid n = 2,3,4,\ldots \}$. Prove that $B$ is closed in $\mathbb R$.

My attempt:

Let $A = \{2,3,4,\ldots\}$. I claim that $A$ is closed in $\mathbb R$. If not, there is a sequence $(a_n)$ in $A$ such that $a_n \to a \notin A$. Clearly, $a \ge 2$. It follows that $n < a < n+1$ for some $n \in A$. Let $\epsilon = \min\{(n+1)-a, a-n\}$. Then there is $a_N \in (a-\epsilon,a+\epsilon)$, so $a_N$ is not an integer. This contradicts the fact that $a_N \in A$.

Consider $f: B \to A, \quad -n +(1/n) \mapsto n$. Next we prove that $f$ is continuous. Given $\delta >0$, if $|[-n +(1/n)] - [-m +(1/m)]| < \delta$ then $|(m-n)[1+1/(mn)]| < \delta$ and thus $|m-n| < \delta$. Hence $f$ is continuous. Since $A$ is closed in $\mathbb R$, $f^{-1}[A] = B$ is closed in $\mathbb R$.

My questions:

1. Could you please verify if my proof looks fine or contains logical gaps/errors?

2. If possible, I would like to ask for more direct or alternative proofs.

Thank you so much ^o^

Answer 1

Another possible approach for 2): $|(-n+\frac 1 n)-(-m+\frac 1 m)| \geq |n-m| -|\frac 1 n -\frac 1 m| \geq 1-\frac 1 2 =\frac 1 2$ whenever $n \neq m$. So any convergent sequence in $B$ is eventually constant which implies that the limit belongs to $B$.

Answer 2

Answer to OP's question (1):

The function $y = f(x) = \frac{1}{x} - x$ is a continuous mapping $f: (0,+\infty) \to \Bbb R$.

Solving for $x$ we get

$\tag 1 x = \frac{1}{2} (\sqrt{y^2 + 4} - y)$

So $f$ is a homeomorphism. If you know that the set

$\tag 2 C = \{n \in \Bbb N \, | \, n \ge 2\}$

is closed in $(0,+\infty)$ then the image $f(C) = B$ is closed in $\Bbb R$.

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Answer to OP's question (2):

Let the strictly increasing sequence $(a_n)_{\, n \ge 0}$ be given.

Proposition 1: If ${\displaystyle \lim _{n\to +\infty} a_n = +\infty}$ then the set

$\quad C = \{a_n \, | \, n \ge 0 \}$

is closed.

Proof
Define $U_0 = (-\infty, a_0)$ and for each $n \ge 1$ define $u_{n} = (a_{n-1}, a_n)$. The complement of the union of the open intervals $U_n$ is closed and equal to $C$. $\quad \blacksquare$

Observe that the same argument applies for a sequence starting at any offset integer. Also, proposition 1 can be 'flipped over' when a sequence is strictly decreasing.

Consider the sequence

$\quad a_n = \frac{1}{n} - n, \; \text{ for } n \ge 2$

All that remains to show that the OP's set is closed is to verify the following two assertions.

$\tag 3 a_{n+1} \lt a_n, \; \text{ for } n \ge 2$

$\tag 4 {\displaystyle \lim _{n\to +\infty} a_n = -\infty}$