Prove that $[0,1)\cong( -\infty,a]$

Question

Prove that $[0,1)\cong( -\infty,a]$.(is homeomorphic)

I know I need to find a decreasing bijective continuous function so that the homeomorphism is possible. However I cannot think of a functions whose domain restricts to the interval $[0,1)$ and its codomain is the interval $(-\infty,a]$

Question:

Can someone provide me a function with the aforementioned desirable characteristics?

Thanks in advance!

Answer 1

Consider $x \mapsto a/(1 - x)$ which sends $[0,1) \simeq[a, \infty)$ then consider $y \mapsto 2a - y$ so that $[a, \infty) \simeq (-\infty, a]$. Take the composition of the two.

Answer 2

Hints

• find a homeomorphim $[0,1)\cong(0,1]$.
• find a homeomorphim $(0,1]\cong(-1,0]$.
• find a homeomorphim $(-1,0]\cong(-\infty,0]$.
• find a homeomorphim $(-\infty,0]\cong(-\infty,a]$.

Now take the composition.

Answer 3

I am having a blind spot on whether or not order is important here.

Consider, $[0,1)\rightarrow (-\infty,a]$, $x\mapsto a+\ln(1-x)$.

This was constructed by trying to get a map sending $a\mapsto 0$, and $``-\infty''\mapsto 1^-$, and then inverting. Something like $1-e^x$ was good for this but then one had to normalise and invert.

Answer 4

$y=\frac{1}{x-1}+a+1$ Is this work?