1) Show that the set $\{\frac{1}{n}:n\in\mathbb{N}\}$ is neither open nor closed in $\mathbb{R}$.

2) Show the set $\{\frac{1}{n}:n\in\mathbb{N}\} \cup \{0\}$ is closed in $\mathbb R$




A set is open M is open, if for every $x_0 \in M$ exists an $r>0$ such that the open ball $B_r(x_0) \subset M$.

Show the set is not open by contraction.

Define $M=\{\frac{1}{n}:n\in\mathbb{N}\}$

Assume M is open then we have an $r>0$. Now let's evalutater the ball with center in 1 with some r>0. $B_r(1)$. Given that $M≤1$ and $r>0$ we can conclude that $B_r(1) \not\subset M$ and therefore we have a contradiction.


Show m is not closed. We assume M is closed, meaning every convergent sequence in $M$ also has a limit contained in $M$

We have the sequence $\frac{1}{n}\in M$ with $\lim\limits_{n \to\infty} \frac{1}{n}=0 \not\in M$. Therefore we have a contradiction and $M$ is not closed.

2) I'm uncertain about (2). But here is what I tried.

A set is closed if the complement is open.

I know the complement is the interval of $(-∞;0) \cup (1;∞)$ which is open. But I'm uncertain of how to prove that that is the complement and whether it is open or not.

I'm uncertain if the proofs hold up or if they are too informal. I'm also not 100% certain how to prove 2). Any advice would be greatly appreciated!

  • $\begingroup$ For $2$, one method is to show that its complement is open. $\endgroup$ – Elliot G Mar 8 '17 at 23:41
  • $\begingroup$ @ElliotG, that is what I was thinking. However I'm uncertain how to do it. Normally I let $z \in M^c$ but what would the radius in this case be? I'm not sure how to do it. $\endgroup$ – user422982 Mar 8 '17 at 23:57
  • $\begingroup$ @Bungo do you have any advise on what an appropriate radius would be? $\endgroup$ – user422982 Mar 8 '17 at 23:59

Part 1 is fine, apart from some details: $M\le1$ is incorrect notation and is not relevant anyway: the ball $B_1(1)$ certainly contains points not in $M$, take for example $9/10$.

In order to show that $M'=M\cup\{0\}$ is closed, note that if $x$ is a limit point of $M'$ there would be a subsequence of $(1/n)$ converging to $x$.

  • $\begingroup$ So i need to prove what the limit point of $M'$ is and then prove that a subsequence of $\frac{1}{n}$ converges to that limit point? I'm not sure I understand how to prove it, do you have any advise? $\endgroup$ – user422982 Mar 9 '17 at 0:14
  • $\begingroup$ @user422982 Any subsequence of a convergent sequence converges at the same limit. $\endgroup$ – egreg Mar 9 '17 at 0:18
  • $\begingroup$ I see no errors in the OP's proof of (2), although it can be done more succintly: There is a sequence in $M$ converging to $0$ ,and $0\not \in M.$ Therefore $M$ is not closed. $\endgroup$ – DanielWainfleet Mar 9 '17 at 7:02
  • $\begingroup$ @user254665 Did I say it was incorrect? Part 2 is proving that $M\cup\{0\}$ is closed. $\endgroup$ – egreg Mar 9 '17 at 7:04
  • $\begingroup$ @egreg. No you didn't say it was incorrect $\endgroup$ – DanielWainfleet Mar 9 '17 at 7:30

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