bijection between [0,1) and [0,infinity)

[Carl R.]

Hi, I'm trying to find a bijective map between [0,1) and [0,infinity).

So I define f: [0,1) -> [0,infinity) by f(x) = - Log ( 1/x).
Then f( 1) = 0 and if x->0 then f(x) -> infinity. Then f ([0,1) =
[0,infinity) and the map is injective.

Is this correct? What would be another example?

I'm also trying to find a bijection between [0,1] and [0,infinity),
wouldn't the same example work?

[Virgil]

In article
<9820935c-996c-408f...@k30g2000hse.googlegroups.com>,
"Carl R." <solr...@hotmail.com> wrote:

> Hi, I'm trying to find a bijective map between [0,1) and [0,infinity).
>
> So I define f: [0,1) -> [0,infinity) by f(x) = - Log ( 1/x).
> Then f( 1) = 0 and if x->0 then f(x) -> infinity. Then f ([0,1) =
> [0,infinity) and the map is injective.
>
> Is this correct? What would be another example?

g(x) = x/sqrt(1-x^2)

>
> I'm also trying to find a bijection between [0,1] and [0,infinity),
> wouldn't the same example work?

There cannot be any continuous bijection.

However, one can map [0,1] to [0,1) discontinuously and then map [0,1)
to [0,oo) by a variety of mappings.

For instance, for n in {1,2,3,...} = the set of positive naturals,
map 1/n to 1/(n+1) and for all other reals,, map x to x.

This will map [0,1] to [0,1)

[The World Wide Wade]

In article
<9820935c-996c-408f...@k30g2000hse.googlegroups.com>,
"Carl R." <solr...@hotmail.com> wrote:

> Hi, I'm trying to find a bijective map between [0,1) and [0,infinity).
>
> So I define f: [0,1) -> [0,infinity) by f(x) = - Log ( 1/x).
> Then f( 1) = 0 and if x->0 then f(x) -> infinity. Then f ([0,1) =
> [0,infinity) and the map is injective.

Plenty wrong here: f is not defined at 0 and f is negative on (0,1)!

> Is this correct?

No.

> What would be another example?

You've seen plenty of examples like this before in your studies. Hint:
What's the easiest example of a function on [0,1) with a vertical
asymptote at 1?

> I'm also trying to find a bijection between [0,1] and [0,infinity),
> wouldn't the same example work?

This is a different animal altogether. There will be no continuous
example this time. By the exercise above, it's enough to find a
bijection between [0,1] and [0,1). You are familiar with countable
sets?

[Frederick Williams]

"Carl R." wrote:
>
> ... What would be another example?

x |-> tan(x*pi/2)

--
He is not here; but far away
The noise of life begins again
And ghastly thro' the drizzling rain
On the bald street breaks the blank day.

[JEMebius]

I guess the simplest pair of functions for a bijection between [0, 1) and [0, infinity)
is

F: x -> y = x/(1 - x) for 0 <= x < 1;

G: y -> x = y/(1 + y) for 0 <= y.

Johan E. Mebius