1-1 correspondence between [0,1] and [0,1)

Question

I wonder how to build a 1-1 correspondence between [0,1] and [0,1). My professor offers an example such that 1 in the first set corresponds to 1/2 in the second set, and 1/2 in the first set corresponds to 1/4 in the second. I don't quite understand it. Does it mean every element in the first set corresponds to its half value in the second set? wouldn't that make some elements left out in the second set? Does it still count as 1-1 correspondence? Does it connect to Schroder-Bernstein theorem?

Answer 1

The complete mapping $f$ is:

If $x$ is one of $1, \frac12, \frac14, \ldots$, then $f(x) = \frac x2$. If $x$ is not one of these, then $f(x) = x$.

Now nothing is left out of the second set. (If you disagree, please say what number you think is left out.)

Schröder-Bernstein is not involved.

Answer 2

This is not what your teacher suggested, but I hope it makes sense. Let $C= [0,1] \cap \Bbb{Q}$ be all the rational numbers in $[0,1]$. $C$ is countable so we can enumerate the elements, $C=\{c_i\}_{i=1}^\infty$ but designate $c_1=1$. Now define the map $f:[0,1]\to [0,1)$ with $$f(x) = \begin{cases} c_{i+1} &\mbox{if } x=c_i \in C \\ x& \mbox{if } x \notin C\end{cases}$$ It remains to be shown that $f$ is a bijection.

Answer 3

for all x∈R(the set of all reals)to exp(x)/(1+exp(x)) is bijection from R to (0,1). So card R= card (0,1). (0,1)is proper subset of [0,1), [0,1)is proper subset of [0,1],[0,1]is proper subset of R. By Schroder-Bernstein theorem they havesame cardinality