Prove that the set $\Big\{ 1/(n+1): n \in \mathbb{N} \Big\} \cup \big\{ 0 \big\} $ is closed. 
Prove that the set $\Big\{ 1/(n+1): n \in \mathbb{N} \Big\} \cup \big\{ 0 \big\} $ is closed

By definition of closed, I know that you have to show that the complement of the set is open. But I don't know how to take its complement. How should I do it? Or are there other ways to show that a set is open? 
 A: Let $A = \{ \frac{1}{n+1}: n\in\mathbb{N} \} \cup \{0\}$. Then 
$$A^{c} = \left(\bigcup_{n=1}^{\infty}\left(\frac{1}{n+1},\frac{1}{n}\right)\right) \cup (-\infty,0) \cup (1,\infty)$$ 
is a countable union of open intervals (which are open sets), hence it is open. Therefore $A$ is closed.
$\textbf{Edit}$: my answer assumed that $\mathbb{N}$ includes $0$. If your convention for $\mathbb{N}$ does not include $0$, then we would have
 $$A^{c} = \left(\bigcup_{n=2}^{\infty}\left(\frac{1}{n+1},\frac{1}{n}\right)\right) \cup (-\infty,0) \cup \left(\frac{1}{2},\infty\right)$$ 
and the conclusion is the same.
A: Thanks for clarifying that $\mathbb{N}$ does not contain $0$ in your world.  Here's your answer.  Let's clean it up a bit. 
Let $A = \{\frac{1}{n} \mid n \geq 2\} \cup \{0\}$.  This is the same set as yours but maybe presented a little clearer (all symbols $n$ are integers).  By writing it out, your complement is everything EXCEPT
$$
\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, 0.
$$
Thus the complement of $A$ is 
$$
(-\infty, 0) \cup \left(\frac{1}{2}, \infty\right) \cup \bigcup_{k=2}^\infty \left(\frac{1}{k+1}, \frac{1}{k}\right).
$$ 
This is a union of open sets, so it's open. Hence $A$ is closed.
Of my three "terms" the first gets rid of the negatives, the second gets you everything above $\frac{1}{2}$, and the last gets all those teeny intervals between. 
A: You can also show it's closed directly using sequences.
Namely, you must show that for all $(x_n) \in A^\mathbb{N}$ such that $x_n$ converges, its limit is in $A$.
A: Take any sequence $s$ of points from your set $A$ (they can repeat). What are its accumulation points?


*

*If $s$ has infinitely many copies of $0$, then $0$ is an accumulation point. Otherwise, $s$ is eventually $0$-free, so assume for the rest that $s$ has no $0$.

*If the reciprocals of elements of $s$ are bounded (i.e., $s$ contains values $1/(n+1)$, but no $n$ is larger than $M$), then there is a finite number of distinct elements in $s$, and the accumulation points will be among those.

*If $s$ contains $1/(n+1)$ terms with arbitrarily large $n$, then $0$ is again an accumulation point.


ALL the accumulation points of $s$ belong to $A$, so $A$ is closed.
(Disclaimer: you still have to show that in the last case, no point outside $A$ can be an accumulation point. This is in a sense, the point of the exercise, so I leave it to you to complete the argument)
A: You can prove that this set is compact if your are familiar with compactness.
Let $\{A_i|i \in I\}$ be an open cover of $B=\{\frac{1}{n+1}|n \in \mathbb{N}\} \cup\{0\}$
The sequence $x_n=\frac{1}{n+1} \rightarrow 0$ .
We have that $\exists i_0 \in I$ such that $0 \in A_{i_0}$
Because of the fact that $A_{i_0}$ is open exists $\epsilon >0$ such that $0 \in (-\epsilon,\epsilon) \subseteq A_{i_0}$.
Because of the fact that $x_n \rightarrow 0$ we have that $\exists m \in \mathbb{N}$ such that $x_n \in (-\epsilon,\epsilon) \subseteq A_{i_0}, \forall n \geq m$
Now for the terms $x_1,x_2....x_{m-1}$  we have that $\exists i_1,i_2...i_{m-1} \in I$
such that $x_1 \in A_{i_1},x_2 \in A_{i_2}.....x_{m-1} \in A_{m-1}$
So $B \subseteq A_{i_0} \cup A_{i_1} \cup.....\cup A_{m-1}$ 
So we proved that every open cover of $B$ has a finite subcover thus $B$ is a compact  subset of the real line.
From Heine-Borel theorem we have that $B$ is closed and bounded.
