1) Let $M=\{\frac{1}{n}:n\in\mathbb{N}\}$. How do I show $M$ is not open in $\mathbb{R}$?

2) Let $K=\{\frac{1}{n}:n\in\mathbb{N}\} \cup \{0\}$. How do i show $K$ is closed in $\mathbb{R}$?


1) I've taken the point $1 \in M$ and placed an open ball around it such that $B_r(1)$.

I want to prove that I can find an $x \in B_r(1)$ such that $x \not\in M$ because then, by definition, $M$ is not open.

I understand that it isnt open, by drawing and looking at the interval. But how do i prove it mathematically?

Could i simply write something like this:

Let $r>0$ and $x=1+\frac{1}{2}r$ then $x \in B_r(1)$ while $x \not\in M$ therefore $B_r(1) \not\subset M$ and therefore $M$ is not open.

I'm not sure how to argue how/why I defined x as such. Is the above an okay proof and how could I proof/argue how x is defined?

2) To show that $K$ is closed, I will show that $K^c$ is open.

I know $K^c$ is an open union of sets but I'm not certain how to make a proper proof. Here is what I've tried

$K$ will never be negative and will be greater than 1, therefore it will never be within $(-∞;0)$ & $(1;∞)$.

Now I'm not certain what to do.

  • $\begingroup$ Your first proof is sufficient, but you are right that your chosen $x$ is highly arbitrary. (For example, $x=1+r/2017$ would work too.) All that matters is that any open interval around $1$ contains reals larger than 1, but all the points in $M$ are at most $1$. Hence no open interval around $1$ can lie in $M$. (Of course, there's nothing special about the point $1$, either; it's simply the easiest case to deal with. You could run a similar but ever so slightly more complex argument for any other point of $M$.) $\endgroup$ Mar 10 '17 at 5:26
  • $\begingroup$ To clarify, you should emphasize that "let $r>0$ and $x=...$" means "for any $r>0$, let $x=...$" (That is how I interpreted you, but you should make it clear because it is possible to read it as an existential statement rather than a universal statement.) The key step is that no matter what radius you pick, you can find an element in your ball that isn't in $M$. $\endgroup$ Mar 10 '17 at 5:33
  • $\begingroup$ @symplectomorphic Thank you for the confirmation. So rather "for any $r$ let $r>0$ be given" and then let $x=...$. Do you have any hints for (2)? I've seen something alike: $K^c=(-∞;0) \cup\bigcap_{n=1}^\infty (\frac{1}{1+n};\frac{1}{n})\cup (1;∞)$. But I'm not sure why the big union symbol should be included. Can you make sense of it? $\endgroup$
    – Sirmimer
    Mar 10 '17 at 5:40
  • $\begingroup$ Have you already tried to use the definition of closed sets in terms of sequences? $\endgroup$
    – user42912
    Mar 10 '17 at 5:48
  • $\begingroup$ I would just write in English: "given any open ball $B$ of radius $r>0$ centered at $1$, there are points in $B$ that aren't in $M$, for example $1+r/2$. Hence no open ball centered at $1$ lies in $M$. Therefore $M$ isn't open." The virtue of writing in English is that you can acknowledge that your choice $1+r/2$ is arbitrary; that is what the phrase "for example" suggests here. This is wordier than the symbolism you use, but I prefer it; of course, this is a matter of taste. If you prefer the symbolism, I think it would be slightly clearer to say "given any $r>0$, let $x=...$." $\endgroup$ Mar 10 '17 at 5:57

Here is another proof.

A set $K \subseteq X$ where $X$ is a metric space is closed iff for every sequence $x_n \in K$ such that $x_n \longrightarrow x$ we have that $x \in K$.(It is an easy proposition to prove)

In $A= \{1/n|n \in \mathbb{N} \} \cup \{0\}$ what type of sequences do you have?

Also if you know about compactness,every compact subset of $\mathbb{R}^n$ is closed.

$A$ is compact because ,let $\{A_i|i \in I \}$ be an open cover of $A$.

For some $i_0 \in I$ we have that $0 \in A_{i_0}$ and the sequence $x_n=1/n \in A$ converges to $0$ .We know that $A_{i_0}$ is open thus there exist $\epsilon >0$ such that $0 \in (-\epsilon,\epsilon) \subseteq A_{i_0}$

From the convergence of $x_n$ exist a $N \in \mathbb{N}$ such that $x_N \in (-\epsilon,\epsilon), \forall n \geqslant N$.

For $n<N$ every term of we have that $1 \in A_{s_1}.......\frac{1}{N-1} \in A_{s_{N-1}}$.

Take the union of these sets with $(-\epsilon,\epsilon)$ and ypou found a finite subcover of $A$.

I hope this proof helps a little.

It will help if you have encounter compactenness and some theory in metric spaces.

  • $\begingroup$ I am not sure i follow 100%. I've seen an answer floating around as: $K^c=(-∞;0) \cup\bigcap_{n=1}^\infty (\frac{1}{1+n};\frac{1}{n})\cup (1;∞)$ But I do not understand why we have the large union symbol in there. Can you explain the reasoning behind this result? $\endgroup$
    – Sirmimer
    Mar 10 '17 at 6:09
  • $\begingroup$ Could i possible use morgans law to solve it? $\endgroup$
    – Sirmimer
    Mar 10 '17 at 6:13

This solution is wrong. The second member of the union is not an intersection but an union. In fact we have:


Since by definition the arbitrary union of open sets is open, we have $K^c$ is open. Therefore $K$ is closed.

  • $\begingroup$ That is my bad, obviously a typo. What I don't understand, which you may care to explain, is why you use the notation? Why is $K^c=(-∞;0)U(1;∞)$ incorrect but what is written in your post correct? How can I explain/prove that it is a set of open unions? $\endgroup$
    – Sirmimer
    Mar 10 '17 at 6:29
  • $\begingroup$ For example $2/3$ is in $K^c$ but not in $(-\infty,0)\cup(1,\infty)$, that's why $K^c\neq(-\infty,0)\cup(1,\infty)$. $\endgroup$
    – user42912
    Mar 10 '17 at 6:33
  • $\begingroup$ I clearly misunderstand it, as it looks like all $K^c=\matbb{R}$ to me. But subtracting the middle union: (1/(n+1);1/n) would be the correct result. $\endgroup$
    – Sirmimer
    Mar 10 '17 at 6:34
  • $\begingroup$ @Sirmimer you need the second union as I explained above. $\endgroup$
    – user42912
    Mar 10 '17 at 6:37
  • $\begingroup$ I understand it now! Thank you. I'm still not 100% sure how you arrive to the specific (1/(n+1);1/n) but I understand the need for the union between the two other open intervals $\endgroup$
    – Sirmimer
    Mar 10 '17 at 6:43

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