0
$\begingroup$

I need to find a bijective map from $A=[0,1)$ to $B=(0,1).$ Is there a standard method for coming up with such a function, or does one just try different functions until one fits the requirements?

I've considered some "variation" of the floor function, but not sure if $\left \lfloor{x}\right \rfloor$ is bijective. Now i'm thinking of maybe some trig function that takes elements of $A$ and maps them to $B$.

I know how to show that a function is injective and surjective, but how do I find such a function? Am I overthinking this or am I under thinking this?

Any help would be greatly appreciated!!!

Lastly, since we are to find a bijective map from $A$ to $B$, assuming one exists, is this equivalent to saying that these two sets have the same cardinality?

Thank you

$\endgroup$
1
$\begingroup$

Consider the function $\psi: A \rightarrow B$ given by $$ \psi(x) = \begin{cases} x & \text{ if}\; \; \; x \neq 0 \; \; \text{and}\; \; x \neq \frac{1}{n} \in \mathbb{N} \\ \frac{1}{2} & \text{if} \;\; x = 0 \\ \frac{1}{n+1} & \text{if } \; \; x= \frac{1}{n}, \; n \in \mathbb{N} \end{cases} $$ It is not difficult to verify that $\psi$ is bijective. The problem with constructing a bijection using our favorite elementary functions is that no continuous bijection between the sets exists. This is due to the differences in character (or topology) of the two sets. So unfortunately one must turn their attention to different functions.

The function above relies on the fact that we can get this bijection by taking a sequence in $(0,1)$, say $\{x_n\}$ and mapping $x_1$ to the extra point $0$, and then mapping $x_n$ to $x_{n-1}$ for all other $n$. Hopefully this trick helps with your future work.

$\endgroup$
  • $\begingroup$ I need some function to map from A to B? If I needed B to A, could I just find the inverse of the piece-wise function that you provided? $\endgroup$ – mathamphetamines Mar 6 '15 at 6:25
  • $\begingroup$ Yes, you could. Sorry about that. I did not notice it was from $A$ to $B$. $\endgroup$ – BigMathTimes Mar 6 '15 at 6:35
  • $\begingroup$ I've now edited the solution so that the map is from $A$ to $B$. @mathamphetamines $\endgroup$ – BigMathTimes Mar 6 '15 at 6:40
  • $\begingroup$ Why is $\psi (x)= \frac{1}{2}$ if $x \ne 0$? For $x \ne 0$, does this just need to be any element within $B$? Was $\frac{1}{2}$ chosen arbitrarily? $\endgroup$ – mathamphetamines Mar 6 '15 at 6:51
  • $\begingroup$ Why wouldn't $\log (x+1)$ work? I believe $\log (x+1)$ is a bijective function. $\endgroup$ – mathamphetamines Mar 6 '15 at 6:55
1
$\begingroup$

To find the bijection from $[0,1)$ to $(0,1)$, remember that you are basically looking for a function that maps a subset of $\mathbb{R}$ to another subset of $\mathbb{R}$.

(An example would be $f:R\mapsto (0,1)$ where $f(x)=1/(1+e^x)$)

.. assuming one exists, is this equivalent to saying that these two sets have the same cardinality?

Yup, that is indeed the definition of cardinality. This is how we can prove that $\mathbb{Z}$ has the same cardinality as $\mathbb{N}$.

$\endgroup$
  • $\begingroup$ $\ln x$ would not work, but how about $\ln (x+1)?$ $\endgroup$ – mathamphetamines Mar 6 '15 at 6:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.