DAILY MATHS CHALLENGE #2

that's right ! got a proof ?

Provide a bijection of (0, 1) into [0, 1].

It's possible.

Let
[math] N=\{ \frac{1-3^{-k}}{2}, 3^{-k}+\frac{1-3^{-k}}{2} \mid k\in \mathbb{N}\cup \{0\} \} [/math]
[math] M=[0,1]-N [/math]

and so we get the bijection
[math] \varphi: [0,1] \to (0,1)[/math] defined by
[math] \varphi(x)= \begin{cases} (x+1)/3 & x \in N \\ x & x \in M \end{cases} [/math]

Send 0 to 1/2, 1 to 1/3, 1/n to 1/(n+2) for each n >= 2 and leave everyone else fixed. That gives you a bijection from [0,1] to (0,1).
An interesting follow-up question might be to prove that you can't find a continuous bijection, one way or the other

How did you come up with such a convoluted formulas? Did you know the answer already?

>How did you come up with such a convoluted formulas?
an algorithmic proof of Cantor–Schröder–Bernstein

>Did you know the answer already?
yes i posted the same bijection a week or so ago

its not really that convoluted, you just contract the endpoints along an infinite sequence similarly to the way did and leave the rest fixed

What's a continuous bijection?
> an algorithmic proof of Cantor–Schröder–Bernstein
So basically picrel?

Well a bijection [math]f: [0,1] \to (0,1)[/math] (or the other way around) that is also a continuous function in the usual sense

yes basically

1 to 0.9 to 0.99 to 0.99 etc
0 to 0.1 to 0.11 to 0.11 etc
Everything else the same

Easy.