Is the map, $ f:(0,1)⊂ \mathbb{R}$ → $(1,∞)⊂ \mathbb{R}$ : $x ↦ 1/x $continuous? I feel it is, but cannot prove why.
Also is it bijective, and is its inverse continuous?
 A: Elaborating on that hint:
Let $p$ be an arbitrary point in  $(1,\infty)$; you want to show the map is continuous at $p$.
Fixate on some value of $\delta > 0$.  Now look at the region near $1/p$.  For some $\epsilon_1 > 0$, which will depend on both $p$ and $\delta$, the entire interval $(x-\epsilon_1,x)$ maps into the interval $p,p+\delta$ -- find the largest such $\epsilon$ explicitly, by solving $f(x) = p$.  (This works in for this function because it is monotonic.)  
Similarly, find $\epsilon_2$ such that the interval $(x,x+\epsilon_2)$
maps into the interval $p-\delta,p$ .  Then let $\epsilon$ be the smaller of those two $\epsilon_i$; this $\epsilon (p,\delta)$, used in the  continuity definition, shows that the function is continuous at point $p$.
Also, to ask whether it is bijective:  It is clearly injective since the definition specifies one and only one value in the co-domain $(1,\infty)$ for each value in the domain
$(0,1)$.  To see if it is surjective, ask yourself: for which point $p$ in $(1,\infty)$ is there no point $x$ in $(0,1)$ such that $1/x = p$?  If you can prove there is no such point, then you have shown it is surjective as well.
Finally, I believe any bijection which is continuous has a continuous inverse.
A: Hint: Write down the definition of continuous. Rely heavily on the fact that $x\neq 0$ (in fact, if you close the interval around $1$ on both sides, it remains true).
