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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!

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    $\begingroup$ Hint: Try something with an asymptote at $1$. $\endgroup$
    – Michael Burr
    Commented Oct 24, 2018 at 14:01

4 Answers 4

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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.

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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.

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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.

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

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