{a.}
Prove that any open interval $(a, b)$ is homeomorphic to the interval $(0, 1)$.
Define $f:(a, b) \to (0, 1)$ by $f(x)=(x-a)/(b-a)$, which is one-to-one and onto. Consider $f^{-1}:(0, 1) \to (a, b) := f^{-1}=a(-x)+a+bx$. We can see that $f$ and $f^{-1}$ are continuous functions. Therefore, the open intervals are homeomorphic.
{b.}
Prove that the ray $(a, \infty)$ is homeomorphic to $(1, \infty)$.
Define $f:(a, \infty) \to (1, \infty)$ by $f(x)=x/a$, which is one-to-one and onto. Consider $f^{-1}:(1, \infty) \to (a, \infty) := f^{-1}=ax$. We can see that $f$ and $f^{-1}$ are continuous functions. Therefore, the defined rays are homeomorphic.
{c.}
Prove that $(a, \infty)$ is homeomorphic to $(-\infty, -a)$.
Define $f:(a, \infty) \to (-\infty, -a)$ by $f(x)=-x$, which is one-to-one and onto. Consider $f^{-1}:(-\infty, -a) \to (a, \infty) := f^{-1}=x$. We can see that $f$ and $f^{-1}$ are continuous functions. Therefore, these open intervals are homeomorphic.
{d.}
Prove that R is homeomorphic to the interval $(-\pi/2, \pi/2)$.
Define $f:R \to (-\pi/2, \pi/2)$ by $f(x)=tan^{-1}x$, which is one-to-one and onto. Consider $f^{-1}:(-\pi/2, \pi/2) \to R := f^{-1}=tanx$. We can see that $f$ and $f^{-1}$ are continuous functions. Therefore, R is homeomorphic to the interval $(-\pi/2, \pi/2)$.
{e.}
Prove that $(1, \infty)$ is homeomorphic to $(0, 1)$.
Define $f:(1, \infty) \to (0, 1)$ by $f(x)=1/x$, which is one-to-one and onto. Consider $f^{-1}:(0, 1) \to (1, \infty) := f^{-1}=1/x$. We can see that $f$ and $f^{-1}$ are continuous functions. Therefore, these open intervals of the real line are homeomorphic.
We thus conclude that any two open intervals of the real line are homeomorphic.
{f.}
Prove that any two closed intervals of the real line are homeomorphic.
Define $f:[a, b] \to [c, d]$ by $f(x)=[(d-c)(x-a)/(b-a)]+c$, which is one-to-one and onto. Consider $f^{-1}:[c, d] \to [a, b] := f^{-1}=[(b-a)(x-a)/(d-c)]+a$. We can see that $f$ and $f^{-1}$ are continuous functions. Therefore, the closed, bounded intervals $[a, b]$ and $[c, d]$ of the real line are homeomorphic.
Best Answer
Either there's a mistake in the book or you've misunderstood the book's definition of "closed interval." Certainly you're right that the line itself is not homeomorphic to a finite closed interval. When this question says "closed interval" it means the interval [a,b] for real numbers a and b (not infinity).
(It's also worth thinking about whether you can prove that [0,1] and $(-\infty, +\infty)$ are not homeomorphic.)