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?

  • $\begingroup$ I'm assuming you're taking this as a subset of the reals. In that case, the stuff between $1/3$ and $1/2$ is missing, so it's in the complement. You can write this as $(1/3, 1/2)$, which is open. Now get the rest of the stuff. $\endgroup$
    – Randall
    Aug 25, 2017 at 14:39
  • $\begingroup$ So its complement is like this: $ (-\infty, 0 ) \cup ... (1/5, 1/4)\cup (1/4, 1/3)\cup (1/3, 1/2) \cup (1/2, \infty)$? $\endgroup$ Aug 25, 2017 at 14:46
  • $\begingroup$ Yep, basically. You might need a $(1/2,1)$ before you hit that $\infty$, depending on what "natural number" means to you. Got to clean it up a bit, but you got the idea (and learning to clean up is a good exercise). $\endgroup$
    – Randall
    Aug 25, 2017 at 14:48
  • $\begingroup$ What if for the OP the natural numbers begin with 1, not 0? Has nothing to do with including 0 as a limit point. $\endgroup$
    – Randall
    Aug 25, 2017 at 14:53
  • 1
    $\begingroup$ It depends. How much detail do you want? If 0 is a natural number to you then your set (call it $A$) has 1 in it, so the complement should not. If 0 is not natural to you then your set $A$ does NOT have 1 in it and so your complement SHOULD. $\endgroup$
    – Randall
    Aug 25, 2017 at 14:57

5 Answers 5


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.


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.


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 $A$.


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)


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.


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