[Solution] When $n=1$, $\phi$ and $\{1\}$ are two alternating sequences, so $A(1)=2$. When $n=2$, $\phi, \{1\}, \{1,2\}$ are three alternating sequences, so $A(2)=3$. If $\left\{a_{1}, a_{2}, a_{3}, \cdots, a_{m}\right\}$ is a non-empty alternating sequence taken from the set $\{1,2, \cdots, n\}$, then when $a_{1}=1$,
$$\left\{a_{2}-1, a_{3}-1, \cdots, a_{m}-1\right\}$$

is an alternating sequence taken from the set $\{1,2, \cdots, n-1\}$. And when $a_{1}=1$, the non-empty alternating sequence $\left\{a_{1}, a_{2}, a_{3}, \cdots, a_{m}\right\}$ taken from the set $\{1,2, \cdots, n\}$ corresponds one-to-one with the alternating sequence $\left\{a_{2}-1, a_{3}-1, \cdots, a_{m}-1\right\}$ taken from the set $\{1,2, \cdots, n-1\}$, so when $a_{1}=1$, the non-empty alternating sequence $\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$ taken from the set $\{1,2, \cdots, n\}$ has $A(n-1)$ sequences.

When $a_{1} \geqslant 3$,
$$\left\{a_{1}-2, a_{2}-2, \cdots, a_{m}-2\right\}$$

is a non-empty alternating sequence taken from the set $\{1,2, \cdots, n-2\}$. And when $a_{1} \geqslant 3$, the non-empty alternating sequence $\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$ taken from the set $\{1,2, \cdots, n\}$ corresponds one-to-one with the non-empty alternating sequence $\left\{a_{1}-2, a_{2}-2, \cdots, a_{m}-2\right\}$ taken from the set $\{1,2, \cdots, n-2\}$. So, when $a_{1} \geqslant 3$, the non-empty alternating sequence $\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$ taken from the set $\{1,2, \cdots, n\}$ has $A(n-2)-1$ sequences.

From the above discussion, the number of alternating sequences taken from the set $\{1,2, \cdots, n\}$
$$A(n)=A(n-1)+(A(n-2)-1)+1$$

i.e.,
$$A(n)=A(n-1)+A(n-2)$$

Using the above equation, we calculate
$$A(1)=2, A(2)=3, A(3)=A(1)+A(2)=5,$$

Similarly, we get
$$\begin{array}{llll} A(4)=8, & A(5)=13, & A(6)=21, & A(7)=34 \\ A(8)=55, & A(9)=89, & A(10)=144, & A(11)=233 \\ A(12)=377, & A(13)=610, & A(14)=987, & A(15)=1597 \\ A(16)=2584, & A(17)=4181, & A(18)=6765, & A(19)=10946 \\ A(20)=17711. & & \end{array}$$