$\alpha$ monotonically increasing on [a, b] $\implies$ finite at $a$ and $b$, why?

Question

The following statement is part of the definition 6.2 in Baby Rudin:

Let $\alpha$ be a monotonically increasing function on $[a,b]$ (since $\alpha (a)$ and $\alpha (b)$ are finite, it follows that $\alpha$ is bounded on $[a,b]$).

Why is the part in bold necessarily true? Isn't $\tan (x)$ on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ monotonically increasing but infinite at the endpoints?

Answer 1

If $f$ is a real-valued function on $[a,b]$, then $f(a)$ and $f(b)$ are real numbers and hence finite.

The function $\tan{x}$ isn't defined at $\pm\pi/2$, so its domain isn't on $[-\pi/2, +\pi/2]$.

Answer 2

I think the key point here is the fact that $[a,b]$ is a closed interval, in contrast with open intervals $(a,b)$.

Since $\alpha$ is defined in the closed interval $[a,b]$, both $\alpha(a)$ and $\alpha(b)$ must be defined. Probably the author assume $\alpha$ a function form $[a,b]$ to $\mathbb{R}$. In this case, $\alpha(a),\alpha(b)\in\mathbb{R}$, hence are real values (and finite values, since every real number is finite).

Note that in your example, the $\tan$ function is not defined in the closed interval, but it is in the open interval.

As a conclusion: every monotonically real function defined in a closed interval is bounded; but monotonically real function defined in an open interval is not necessarily bounded ($\tan$ is a counter-example).

Answer 3

No.

$\tan(x)$ is not monotonically increasing on $[-\frac {\pi}2,\frac {\pi}2]$.

$\tan (x)$ is monotonically increasing on $(-\frac {\pi}2,\frac {\pi}2)$.

A function can not be monotonically increasing at a point it is not defined.