INTRODUCTION TO REAL ANALYSIS. Fourth Edition. Robert G. Bartle. Donald R. Sherbert. page 83.
3.4.10 Definition : Let $X = (x_n)$ be a bounded sequence of real numbers.
(a) The limit superior of $(x_n)$ is the infimum of the set $V$ of $v \in \mathbb R$ such that $v < x_n$ for at most a finite number of $n \in\mathbb N$. It is denoted by : $\limsup (x_n)$.
3.4.11 Theorem: If $(x_n)$ is a bounded sequence of real numbers, then the following statements for a real number $x^*$ are equivalent :
(a) $x^* = \limsup(x_n)$. (b) If $\epsilon > 0$, there are at most a finite number of $n\in \mathbb N$ such that $x^* + \epsilon < x_n$, but an infinite number of $n \in \mathbb N$ such that $x^*-\epsilon < x_n$.
Proof : (a) implies (b):
If $\epsilon > 0$, then the fact that $x^*$ is an infimum implies that there exists a $v$ in $V$ such that $x^* \le v < x^* + \epsilon$. Therefore $x^*$ also belongs to $V$, so there can be at most a finite number of $n\in \mathbb N$ such that $x^* +\epsilon < x_n$.
On the other hand, $x^* -\epsilon$ is not in $V$ so there are an infinite number of $n \in\mathbb N$ such that $x^* - \epsilon < x_n$.
https://mathworld.wolfram.com/Infimum.html
The infimum is the greatest lower bound of a set S, defined as a quantity m such that no member of the set is less than m, but if ε is any positive quantity, however small, there is always one member that is less than m + ε (Jeffreys and Jeffreys 1988).
Question:
From the information given, how can we conclude that "Therefore $x^*$ also belongs to $V$"? The definition of infimum from Wolfram's guarantees the existence of a $v \in V$ smaller than $m + \epsilon$ which implies that $m +\epsilon$ belongs to $V$.
Is it a typo in the textbook in which place "Therefore $x^* + \epsilon$ also belongs to $V$" should be written or am I mistaken?