Find a 1−1 function that maps ( 0,1 ) into, but not necessarily onto, S.

Question

I am self-studying real analysis from Understanding Analysis by Stephen Abbott. I want to say that some of the exercises in the book are really challenging. Am I expected to solve these exercises as I don't have any significant background in mathematics other than high school Algebra and Calculus 1 & 2? Now to the main point. I actually don't understand the exercise. And what is meant by open unit square? What am I asked to find or prove? Thanks!

Consider the open interval (0,1), and let S be the set of points in the open unit square; that is, S = { ( x , y ) : 0 < x , y < 1 }. (a) Find a 1−1 function that maps ( 0,1 ) into, but not necessarily onto, S. (b) Use the fact that every real number has a decimal expansion to produce a 1−1 function that maps S into ( 0,1 ). Discuss whether the formulated function is onto. (Keep in mind that any terminating decimal expansion such as .235 represents the same real number as .234999… ).

Answer 1

(a) is simple: $x\mapsto(x,0)$. This is injective because $(x,0)=(y,0)$ clearly implies $x=y$. For (b) use the digit-interleave function (for a fuller discussion see here) where $0.999\dots$ is not used in favour of $1.000\dots$: $$f(0.2568,0.0376)=0.20536786$$ This is not surjective, since the preimage of $0.909090\dots$ would be $(0.999\dots,0.000\dots)$ which is not in $S$.