I have a question about item (f) in a very long problem in Spivak's Calculus.
Here is the problem statement
- This problem investigates for which $a>0$ the symbol
$$a^{a^{a^{a^{...}}}}$$
makes sense. In other words, if we define a sequence $\{b_n\}$ by
$$b_1=a$$
$$b_{n+1}=a^{b_n}$$
when does
$$\lim\limits_{n\to\infty} b_n=b$$
exist?
(a) If $b$ exists, then $a$ can be written in the form $y^{1/y}$ for some $y$. Describe the graph of $g(y)=y^{1/y}$ and conclude that $0<a\leq e^{1/e}$.
(b) Suppose that $1\leq a \leq e^{1/e}$. Show that $\{b_n\}$ is increasing and also that $b_n\leq e$. This proves that $b$ exists (and also that $b\leq e$)
The analysis for $a<1$ is more difficult.
(c) Using Problem 25, show that if $b$ exists, then $e^{-1}\leq b\leq e$. Then show that $e^{-e}\leq a \leq e^{1/e}$.
From now on we will suppose that $e^{-e}\leq a \leq 1$.
(d) Show that the function
$$f(x)=\frac{a^x}{\log{x}}$$
is decreasing on the interval $(0,1)$.
(e) Let $b$ be the unique number such that $a^b=b$. Show that $$a<b<1\tag{1}$$ Using part $(e)$ [sic], show that if $0<x<b$, then $$x<a^{a^{x}}<b\tag{2}$$ Conclude that $$l=\lim\limits_{n\to\infty} a_{2n+1}\tag{3}$$ exists and that $$a^{a^l}=l\tag{4}$$
(f) Using part $(e)$ again [sic], show that $l=b$.
I've asked about item (c) before, and there I show the solution up to item (c).
Let me continue almost where I left off in that other question, skipping item (d) and starting from item (e) to reach my question in (f).
Item e
Note the graph of $f(x)=x^{1/x}$ below. Since $e^{-e}\leq a<1$ then we are on a portion of the graph where it is increasing and thus one-one.
Note that on this interval, $x^{1/x}$ lies below $x$ because
$$0<x<1\implies \frac{1}{x}>1\implies \frac{1}{x}\log{x}<\log{x}\implies x^{1/x}<x$$
Thus, since $a$ is in this interval, we have
$$a=b^{1/b}<b<1\tag{1}$$
Now, assume that $0<x<b$.
Then $a^x>a^b$ since
$$f(x)=a^x, f'(x)=a^x\log{a}\leq 0\text{ since } 0\leq e^{-e}<a<1$$
Then,
$$a^{a^x}<a^b=b\tag{5}$$
by the same argument. That is, $a^x$ is decreasing for $e^{-e}\leq a<1$. Hence, given $a^x>a^b$ we have $a^{a^x}<a^b=b$.
In part (d) we showed that $\frac{a^x}{\log{x}}$ is decreasing on $(0,1)$. Thus, since $x<b$, then $\frac{a^x}{\log{x}}>\frac{a^b}{\log{b}}$ and
$$a^x\log{b}>a^b\log{x}$$
$$a^x\log{(a^b)}>b\log{x}$$
$$a^xb\log{a}>b\log{x}$$
$$a^x\log{a}>\log{x}$$
$$\log{a^{a^x}}>\log{x}$$
$$a^{a^x}>x\tag{6}$$
(5) and (6) give us (2)
$$x<a^{a^x}<b\tag{2}$$
In particular, we have
$$0<a=b_1<b\implies a<a^{a^a}=b_3<b\implies a^{a^a}<a^{a^{a^a}}=b_5<b$$
Thus, the subsequence $\{b_{2n+1}\}$ is increasing and bounded, hence convergent. That is, for some $l$ we have
$$\lim\limits_{n\to\infty} b_{2n+1}=l\tag{3}$$
Now, $\{b_{2n+1}\}$ is the sequence $a,f(a),f(f(a)),...$ with $f(x)=a^{a^x}$.
By a previous result, since this sequence converges, it must converge to a fixed point of $f$.
Thus, $$f(l)=a^{a^l}=l\tag{4}$$
Now for the question.
Item f
We want to prove that $l=b$. The solution manual simply says
Since $a^b=b$ we have $a^{a^b}=a^b=b$. Since we also have $a^{a^l}=l$ we must have $l=b$ by part $(e)$.
What exactly is happening in this solution?