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

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?

• Hint: Try something with an asymptote at $1$. Commented Oct 24, 2018 at 14:01

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.

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.

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.

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