4. Given a sequence of positive terms $\left\{a_{n}\right\}$ satisfying $a_{n}^{n}+a_{n}-1=$ $0\left(n \in \mathbf{N}_{+}\right)$. Below are 5 statements:
$$
\text { (1) } 0<a_{n} \leqslant \frac{n}{n+1} \text {; }
$$
(2) $\frac{n}{n+1} \leqslant a_{n}<1$;
(3) $\left\{a_{n}\right\}$ is an increasing sequence;
(4) $\left\{a_{n}\right\}$ is a decreasing sequence;
(5) $\left\{a_{n}\right\}$ is not a monotonic sequence.

Among them, the correct statements are ( ).
(A)(1)(3)
(B)(1)(4)
(C)(2)(3)
(D)(2)(5)
4. A.

Obviously, $0n a_{n}-(n-1)$
$\Rightarrow a_{n}^{n}+a_{n}-1>(n+1) a_{n}-n$.
Thus, $0<a_{n}<\frac{n}{n+1}$.
Also, $a_{1}=\frac{1}{2}$, so, $0<a_{n} \leqslant \frac{n}{n+1}\left(n \in \mathbf{N}_{+}\right)$.
If for some positive integer $n$, $a_{n} \geqslant a_{n+1}$, then $a_{n+1}^{n+1}=a_{n+1}^{n} a_{n+1}<a_{n+1}^{n} \leqslant a_{n}^{n}$.
Therefore, $a_{n+1}^{n+1}+a_{n+1}<a_{n}^{n}+a_{n+1} \leqslant a_{n}^{n}+a_{n}$.
But $a_{n+1}^{n+1}+a_{n+1}=1=a_{n}^{n}+a_{n}$, which is a contradiction.
Hence, for any $n$, $a_{n}<a_{n+1}$.