$\sup$ and $\inf$ of this set These are exercises from my textbook, and I am not sure if the solutions are correct or not.
Given a set $B = \{\frac{n}{2n+1} : n \in \mathbb{N} \}$
Find the $\sup$ and $\inf$ of $B$, and maxima and minima if they exist.
The solutions give $\sup B = \frac{1}{2}$ and $\inf B = \frac{1}{3}$
I am confused by this result. The $\sup$ has to be the least upper bound, $\frac{1}{2} \in B$ and $\frac{1}{3}$ is also in $B$. So how can we claim that $\sup B = \frac{1}{2}$ when $\frac{1}{3} > \frac{1}{2}$
Is this a typo?
Additionally, shouldn't $\inf B = \frac{1}{2}$ since as $n \to \infty$, $\frac{n}{2n+1} \to \frac{1}{2}$

First $\frac{1}{3} < \frac{1}{2}$. 
Second $B$ is bounded above by $\lim \frac{n}{2n+1} = \frac{1}{2}$. Then $\sup B \leq \frac{1}{2}$. Take any $\epsilon > 0$ then there exists $N \in \mathbb N$ such that 
$$n > N \implies |x_n - \frac{1}{2}| < \epsilon $$ 
then there exists  $x_k = \frac{k}{2k + 1} \in B$, such that $$\frac{1}{2} - \epsilon < x_k \leq \epsilon$$ then $\sup B = \frac{1}{2}$. 
Now $\inf B = \frac{1}{3}$ because when $n = 1$ you have $\frac{1}{2 \dot \ 1 + 1} = \frac{1}{3}$. Then $B$ is bounded below by $\frac{1}{3}$, try to show that in fact $\inf B = \frac{1}{3} $.  
