How is 0 a limit point of $\{1/n\}_{n=1}^{\infty}$?

Question

I'm working through Principles of Topology and can't wrap my head around how $0$ is a limit point of $\{1/n\}_{n=1}^{\infty}$. For reference, this is on page 44 of the 2016 edition.

The book provides the following definition/theorem & proof:

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Definition: A point x in $\mathbb{R}$ is a limit point or accumulation point of a subset A of $\mathbb{R}$ provided that every open set containing x contains a point of A distinct from x.

Theorem 2.9: A real number x is a limit point of a subset A of $\mathbb{R}$ if and only if $d(x, A\setminus\{x\}) = 0$.

Proof: Suppose first that x is a limit point of A and let $\epsilon$ be a positive number. Then the open set $(x-\epsilon, x+\epsilon)$ contains a point y of A distinct from x. Since $y\in(x-\epsilon,x+\epsilon)$, then $d(x,y) < \epsilon$ so $d(x, A\setminus\{x\}) < \epsilon$. Since the latter inequality holds for all $\epsilon>0$, then $d(x, A\setminus\{x\}) = 0$.

Suppose now that $d(x, A\setminus\{x\}) = 0$ and consider an open set O containing x. Then O contains an open interval $(x-\delta, x+\delta)$ for some positive number $\delta$. Since $d(x, A\setminus\{x\}) < \delta$ and the interval $(x-\delta, x+\delta)$ consists precisely of all points at a distance less than $\delta$ from x, then $(x-\delta, x+\delta)$ must contain a point z in $A\setminus\{x\}$. Thus

$$z \in (x-\delta, x+\delta) \subset O$$

and $z \neq x$ since $z\in$ $A\setminus\{x\}$. Hence every open set containing x contains a point of A distinct from x, and x is a limit point of A. $$\tag*{$\Box$}$$

Additionally, the book defines distance between a real number and a non-empty subset of $\mathbb{R}$ as: $$d(a, B) = glb\{|a-b|:b \in B\}$$

and provides a few examples, e.g.:

• $d(0, [1,2]) = d(0, (1,2)) = 1$
• $d(1, [1,2]) = d(1, (1,2)) = 0$

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So far, so good. The book then notes:

$0$ is the only limit point of $\{1/n\}_{n=1}^{\infty}$. Note in this case that the limit point is outside the set.

My question is... how? Say $x = 0$ and you choose $\epsilon = 3$, and you follow along the proof: $$(x-\epsilon, x+\epsilon) = \{x \in \mathbb{R}: (x-\epsilon) < x < (x+\epsilon)\}$$

which certainly does contain a point $y = 1$ where $y \in \{1/n\}_{n=1}^{\infty}$, with $y$ being distinct from $x$.

What I don't understand: if $A = \{1/n\}_{n=1}^{\infty}$, how does $d(x, A\setminus\{x\}) = 0$?

As far as I can tell, since $0 \notin A$, then $A\setminus\{x\}$ is equivalent to $\{1/n\}_{n=1}^{\infty}$, and the distance between $0$ and $A$ would be $1$:

$$d(0, [1, \frac 12, \frac 13, \frac 14 ...)) = 1$$

What might I be missing here?

Answer 1

Say $x = 0$ and you choose $\epsilon = 3$, and you follow along the proof: $$ (x-\epsilon, x+\epsilon) = \{-3,-2,-1,0,1,2,3\} $$ which certainly does contain a point $y = 1$ where $y \in \{1/n\}_{n=1}^{\infty}$, with $y$ being distinct from $x$.

You made a mistake of understanding the interval $(x-\epsilon, x+\epsilon)$ here. It is not equal to the set $\{-3,-2,-1,0,1,2,3\}$ but equal the set of real numbers $y$ such that $-3<y<3$. Note that the set $(x-\epsilon, x+\epsilon)$ has infinitely many elements! Moreover, it contains not only the point $1\in A:=\{1,\frac12,\frac13,\cdots\}$: it contains every element in $A$.

As far as I can tell, since $0 \notin A$, then $A\setminus\{x\}$ is equivalent to $\{1/n\}_{n=1}^{\infty}$, and the distance between $0$ and $A$ would be $1$: $$ d(0, [1, \frac 12, \frac 13, \frac 14 ...)) = 1 $$

What might I be missing here?

First of all, $[1, \frac 12, \frac 13, \frac 14 ...)$ is not a valid set notation. you could at most write $A=\{1,\frac12,\frac13,\frac14,\cdots\}$. In order to understand why $$ d(0,A)=0, $$ you need to read the definition of the notation $d(0,A)$ word by word slowly: $$ d(0,A)=\text{glb}\{|0-a|:a\in A\}. $$ Be very careful that "glb" means the "greatest lower bound". Now, without the "glb" symbol, look at the set $\{|0-a|:a\in A\}$. It happens to be exactly $A$: $$ \{|0-a|:a\in A\}=\{a:a\in A\}=A. $$ Now your question becomes:

why is the greatest lower bound of $A$ is $0$ not $1$?

Note that $1$ is not even a lower bound of the set $A$.

Answer 2

The greatest lower bound on $\{|0-\frac{1}{n}|:n\in\mathbb{N}\}$ is $0$, because any $\delta>0$ exceeds $\frac{1}{n}$ provided $n>\frac{1}{\delta}$.

Answer 3

An equally valid way to define the limit point of a set is: “A point $p$ of a set $E$ is a limit point if every neighborhood of $p$ contains a point $q \neq p$ such that $q \in E$.” Rudin, Principles of Mathematical Analysis

Let’s use this definition. Let $E$ be the set described in the problem. Consider a arbitrary neighborhood of $0$. For any $\delta > 0$, we can find a positive $\frac{1}{n}$ such that $\frac{1}{n} < \delta$. This is clear. Thus, all neighborhoods of $0$ will have at least one point (in fact they will have infinite points) that is not $0$ and in $E$ within the region $(-\delta, \delta)$. So, it follows that $0$ is a limit point.

Answer 4

You calculate this wrongly:

$$d(0, {1, \frac 12, \frac 13, \frac 14 ...}) = 1$$

It should be $0$. And as it has been stated in Theorem 2.9, $0$ is the limit point.