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I have the set $A = \{\frac{1}{n} + \frac{1}{k} \mid n,k \in \mathbb{N}\} \subseteq \mathbb{R}$. My exercise asks me to find the closure of this set, but my question in this post is simply asking for clarification on how open sets work.

A set is closed if its compliment is open. A set is open if every point has a neighborhood lying in the set.

I want to show that $A$ is not closed. So I search for a point in $A - \mathbb{R}$ which has a...neighborhood (I need a precise definition for this) that overlaps the neighborhood of a point in $A$. This would show that the compliment $A - \mathbb{R}$ is not open. Right?

Definition of limit $$\lim_{x \to a} f(x) = L \iff \forall \epsilon >0 \; \exists \delta >0, \; |x - a| < \delta \implies |f(x) - L| <\epsilon$$

By this definition, we need $\delta = \frac{1}{\epsilon}$ so that $$\lim_{n \to \infty} \frac{1}{n} = 0$$

More specifically, we need $0 < |n| < \frac{1}{\epsilon}$ for that limit to be true.

So now I know $\frac{1}{n} + \frac{1}{k} \to 0$ as $n,k \to \infty$ if $n,k \in (0, \frac{1}{\epsilon})$

  1. I haven't shown that $0$ is not in $A$ yet. How do I do that?
  2. What is the definition of "neighborhood"? As I understand, it is an epsilon-ball centered at a point with radius $\epsilon$. So in this case, $B_{\epsilon}(p) = \{x \in A \mid |x - p| < \epsilon\}$ since $A \subseteq \mathbb{R}$.
  3. A set is open if every point in the set has an $\epsilon$-ball lying completely in the set. Is this right?

The epsilon ball about $0$ is $B_{\epsilon}(0) = \{ x \in \mathbb{R} \mid |x - 0| < \epsilon\}$

The interval for which our limit earlier is true is contained within this ball. So that means the point $0 \in A \setminus \mathbb{R}$ has a neighborhood which overlaps into the set $A$. This means the compliment is not open.

After 1,2,3 questions, is my conclusion correct?

user463756
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  • $\frac{1}{n} + \frac{1}{k} > 0$ so $0 \notin A$. – mechanodroid Sep 07 '17 at 22:29
  • A subset $S$ of $\mathbb R$ is open iff for every $x\in S$ there exists $r_x>0$ such that $(-r_x+x,r_x+x)\subset S.$.... A neighborhood of $x $ is a set $T\subset \mathbb R$ such that there exists an open set $S$ with $x\in S\subset T.$... Note that the empty set $\phi$ is open because there is no $x\in \phi$ that violates the condition of being open. – DanielWainfleet Sep 08 '17 at 18:39

2 Answers2

1

  1. Note that $\frac{1}{n}+\frac{1}{k}>0$ for all $n,k \in \mathbb N$, so by definition, $0 \notin A$.

  2. Your definition of neighborhood is correct. It is an open set containing your point, and in the euclidian case, just a ball around your point with some nonzero raadius

  3. Yes, you are correct.

To answer the overall question at hand, you have shown that $0$ is a limit point, but not in $A$, so $A$ cannot be closed.

Now, to finish you must either show that either there are no other points in the closure, or find any more.

Andres Mejia
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(I).For any sequence $S=(s_n)_{n\in \mathbb N}$ of reals there exists a strictly increasing $f:\mathbb N\to \mathbb N$ such that one of the following holds for the sequence $F=(s_{f(n)})_{n\in \mathbb N}$ :

(i). $F$ is constant. I.e. $s_{f(m)}=s_{f(n)}$ for all $m,n$

OR (ii). $F$ is strictly increasing

OR (iii). $F$ is strictly decreasing.

In other words $S$ has a sub-sequence $F$ which is constant or strictly monotonic.

(II). Suppose $y_n,z_n\in \mathbb N$ and $x=\lim_{n\to \infty}(1/y_n+1/z_n).$ In other words, $x\in \overline A.$ We show that $x=0$ or $x\in A$.

(a). Neither of the sequences $(1/y_n)_n$ or $(1/z_n)$ has a strictly increasing sub-sequence $(1/y_{f(n)})_n$ or $(1/z_{f(n)})_n$ because otherwise $(y_{f(n)})_n$ or $(z_{f(n)})_n$ would be a strictly decreasing infinite sequence of natural numbers.

(b). Suppose $(1/y_n)_n$ has a constant sub-sequence $(1/y_{f(n})_n$ with $y_{f(n)}=y\in \mathbb N$ for all $n.$ Then $$x=\lim_{n\to \infty}(1/y_{f(n)}+1/z_{f(n)})=1/y+\lim_{n\to \infty}1/z_{f(n}.$$

Now if $\lim_{n\to \infty}1/z_{f(n)}=0$ then $$x=1/y\in A.$$ Or if $\lim_{n\to \infty}1/z_{f(n)}=K\ne 0$ then $\lim_{n\to \infty}z_{f(n)}=K^{-1}.$ But a convergent sequence $(z_{f(n)})_n$ of natural numbers can only converge to a natural number, so $K^{-1}=z\in \mathbb N$ and $$x=1/y+1/z\in A.$$ (c). Suppose $(1/z_n)_n$ has a constant sub-sequence. Then by interchanging the letters $y,z$ in (b) above, we also obtain $$x\in A.$$(d). If neither $(1/y_n)_n$ nor $(1/z_n)_n$ has a constant sub-sequence then by (I) and by (II)(a) there is a strictly increasing $f:\mathbb N\to \mathbb N$ such that $(y_{f(n)})_n$ is strictly decreasing, and by (I) and by (II)(a) again, the sequence $(1/z_{f(n)})_n$ must have a strictly decreasing sub-sequence $(1/z_{g(f(n))})_n$ (with $g:\mathbb N\to \mathbb N$ strictly increasing).

Then $(y_{g(f(n)})_n$ and $(z_{g(f(n)})_n$ are strictly increasing sequences of natural numbers, so $$x=\lim_{n\to \infty}(1/y_{g(f(n))}+1/z_{g(f(n))})=0.$$

DanielWainfleet
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