Find a $1–1$ function that maps $(0,1)$ into, but not necessarily onto, $S$. Where $S$ is the set of points in the open unit square; that is, $S = {(x, y) : 0 < x, y < 1}.$ (This is easy.)
After thinking and rereading about an hour I give up trying answer the question. Especially, when the author states, that it is trivial. Nothing comes to my mind :(.
Can you, please, help me? In general, are there any systematic ways to answer such kind of questions, rather than guessing a function (as it is so far in the textbook: Understanding Analysis, Stephen Abbott)?
Best Answer
Expanding on some of the comments to the question itself:
For
$x \in (0, 1), \tag 1$
set
$f(x) = (x, x) \in S; \tag 2$
it is clear that $f$ is not onto, since for any
$(y, z) \in S \; \text{with} \; y \ne z \tag 3$
there is no $x \in (0, 1)$ with
$f(x) = (x, x) = (y, z), \tag 4$
since this leads to the contradiction
$x = y \ne z = x. \tag 5$
On the other hand, $f$ is injective since
$f(x_1) = f(x_2) \Longrightarrow (x_1, x_1) = (x_2, x_2) \Longrightarrow x_1 = x_2. \tag 6$