[a,b) and (a,b) are equinumerous

[evinda]

Hi! (Smile)

I want to show that the interval $[a,b)$ is equinumerous with this one: $(a,b)$.
How could we find a bijective function from $[a,b)$ to $(a,b)$? (Thinking)

 

[Fantini]

Well, $(a,b) \subset [a,b)$. Wouldn't this solve everything?

 

[Euge]

Well, $(a,b) \subset [a,b)$. Wouldn't this solve everything?

Not exactly. The inclusion mapping $i : (a,b)\to [a,b)$, although injective, is not surjective: there is no $x \in (a,b)$ such that $i(x) = a$.

Evinda, I suggest first considering the case $a = 0$ and $b = 1$. Construct a bijection $f : [0,1) \to (0,1)$ using a piecewise-defined function.

 

[evinda]

Evinda, I suggest first considering the case $a = 0$ and $b = 1$. Construct a bijection $f : [0,1) \to (0,1)$ using a piecewise-defined function.

What piecewise-defined function could we use?

 

[Evgeny.Makarov]

Map $0$ to $1/2$ and $1/n$ to $1/(n+1)$ for $n\ge2$. All other numbers are mapped to themselves.