Q:

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

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:

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.:

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

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?

A:

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 $A$.

