Is the set $A=\{n+\frac{1}{n}:n\in\mathbb{N}\}$ closed in $\mathbb{R}$? Let $A=\{n+\frac{1}{n}:n\in\mathbb{N}\}$. Is the set $A$ closed in $\mathbb{R}$?
I think that if $n$ goes to infinity, every deleted neighborhoods of $n$ would contain the points of $A.$ So, infinity is the only limit point of $A$ and infinity is contained in the set $A.$ Hence, the set $A$ is closed? Is that right?
 A: There is no such thing as infinity in $\mathbb{R}$. You can take the limit at infinity and the limit can be infinity, but they're very different things and have a very precise meaning where infinity does not appear.
Suppose $(a_n)$ is a sequence in $\mathbb{R}$ such that $\lim_{n\to\infty}a_n=\infty$, meaning

for every $K$, there exists $\bar{n}$ such that, for every $n>\bar{n}$, $a_n>K$.

I contend that the set $A=\{a_n:n\in\mathbb{N}\}$ has no limit point (which implies it is closed). 
Let $p\in\mathbb{R}$. There is $\bar{n}$ such that, for $n>\bar{n}$, $a_n>p+1$. In particular, for $n>\bar{n}$, $a_n\notin(p-1,p+1)$. Thus the neighborhood $(p-1,p+1)$ of $p$ contains at most finitely many elements of $A$, so $p$ is not a limit point of $A$.
Can you prove that $\displaystyle\lim_{n\to\infty}\Bigl(n+\frac{1}{n}\Bigr)=\infty$?
A: We will show $A$ is closed in the classic way, based on the definition of a closed set--by showing that the complement of $A$, $A^c = \Bbb R \setminus A$, is open.
For $n \in \Bbb N$, let
$a_n = n + \dfrac{1}{n} \in A; \tag 1$
then
$a_1 = 2 < a_2 = 2 \frac{1}{2}, \tag 2$
and for $n \ge 2$,
$a_n = n + \dfrac{1}{n} < n + 1 < (n + 1) + \dfrac{1}{n +1} = a_{n + 1}; \tag 3$
thus the sequence of values $a_n$ is monotonically increasing.  Now let
$r \in A^c = \Bbb R \setminus A; \tag 4$
suppose $r < a_1 = 2$; then
$r \in (-\infty, a_1) = (-\infty, 2); \tag 5$
we note that the interval $(-\infty, a_1)$ is open.  Next, assume
$r > a_1: \tag 6$
in this case we let $L \in \Bbb N$ be the largest index with
$a_L < r; \tag 7$
we know that such $L$ exists since the set of naturals $M = \{m \mid a_m < r \}$ is non-empty by (6); furthermore the $a_m$ are bounded above by $r$, so $M$ is in fact finite and $L$ is its largest element.  Likewise there is a smallest $S \in \Bbb N$ with 
$r < a_S; \tag 8$ 
this follows since the set $K = \{l \in \Bbb N \mid r < a_l \} \ne \emptyset$ because $r$ is bounded above by some element of $A$; since $K \subset \Bbb N$ it is bounded below hence $S$ exists.  Thus
$a_L < r < a_S, \tag 9$
and we claim that in fact
$S = L + 1; \tag{10}$
for if (10) were false, then we would have
$a_L  < a_{L + 1} < a_S; \tag{11}$
and by (9), either
$r \in (a_L, a_{L + 1}), \tag{12}$
or
$r \in (a_{L + 1}, a_S); \tag{13}$
we may rule out (13) since it implies $a_{L + 1} < r$, in contradiction to the defiition of $L$; then (12) contradicts the definition of $S$; thus $S = L + 1$ and
$r \in (a_L, a_{L + 1}). \tag{14}$
The above argument shows that every $r \in \Bbb R \setminus A$ lies either lies in $(-\infty, a_1)$ or $(a_n, a_{n + 1})$ for some $n \in \Bbb N$; 
thus
$A^c = \Bbb R \setminus A = \displaystyle (-\infty, a_1) \cup \bigcup_1^\infty (a_n, a_{n + 1}); \tag{15}$
this expresses $A^c$ as the union of a countable number of open intervals; hence $A^c = \Bbb R \setminus A$ is open and thus $A$ itself is closed.
A: Note:  Let $a_n \in A$ be denoted as $n + \frac 1n$.  Then   $a_1 < a_2 < ... < a_n < a_{n+1} < ....$.  This can be proven via induction as $\frac 1n \le 1$ so $a_n = n + \frac 1n \le n + 1 < (n+1) + \frac {1}{n+1}= a_{n+1}$.
Let $r \in \mathbb R$.
So there is a unique $a_n \in A$ so that $a_n \le r < a_{n+1}$.
If $r > a_n$ then there exist an $\epsilon > 0$ so that $a_n < r-\epsilon < r < r+ \epsilon < a_{n+1}$ so $(r-\epsilon, r + \epsilon)$ contain no points of $A$.
If $r = a_n$ then there exists an $\epsilon > 0$ so that $a_{n-1} < r-\epsilon < a_n < r + \epsilon < a_{n+1}$ so that $(r-\epsilon, r + \epsilon)$ contain no points of $A$ other than $r = a_n$.
So $r$ is no a limit point of $A$.
So $A$ has no limit points.  So vacuously $A$ contains all its limit points.  So $A$ is closed.
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Note:  $\infty \not \in \mathbb R$ and even if we were using the extended real numbers to allow $\infty$ then $\infty + \frac 1{\infty} \not \in \mathbb N$ and $\infty \not \in A$.  For any finite $\epsilon > 0$ then $B_\epsilon(\infty) = \{x \in \mathbb R \cup \{\infty\}| |\infty - x| < \epsilon\} = \{\infty\}$.  And $B_\epsilon(\infty) \cap A = \{\infty\}\cap A = \emptyset$.  So $\infty$ is not a limit point of $A$.
