I need to show that any open interval is homeomorphic to the real line.

I know that $f(x)=a+e^x$ will work for the mapping $f:R \to (a,\infty)$ and $f(x)=b-e^{-x}$ will work for the mapping $f:R \to (-\infty,b).$

Without using two functions, how can I prove the result in general?

  • $\begingroup$ Why not use these two functions? $\endgroup$
    – user14972
    Nov 23, 2012 at 2:21

3 Answers 3


Let $$ f(x) = \frac{x}{x^2-1}. $$ This is a homeomorphism from $(-1,1)$ to $\mathbb R$.

Let $$ g(x) = \frac{1}{1+2^{-x}}. $$ That is a homeopmorphism from $\mathbb R$ to $(0,1)$.

For any other bounded interval $(a,b)$, just rescale and relocate.

  • $\begingroup$ To show that the real line is homeomorphic to any inteval,I need to show that (0,1) is homeomorphic to R and any two intervals are homeomorphic.so that I can find a composition function then that is a homeomorphism.So can you please help me with this idea? $\endgroup$
    – ccc
    Nov 24, 2012 at 8:57
  • 5
    $\begingroup$ @ccc : Let $\ell(x)=a(1-x)+bx$. Then $\ell$ is a homeomorphism from $(0,1)$ to $(a,b)$. (That is what I meant by "rescale and relocate".) Thus every open interval is homeomorphic to $(0,1)$. And since the function $g$ in the answer is a homeomorphism from $\mathbb R$ to $(0,1)$, the composition gives a homeomorphism from $\mathbb R$ to $(a,b)$. $\endgroup$ Nov 24, 2012 at 15:55
  • $\begingroup$ thanks Michael,I understand the method,but do I need to define a function for the composition,i.e from R to (a,b)? $\endgroup$
    – ccc
    Nov 25, 2012 at 8:24
  • 6
    $\begingroup$ You have $g(x)=\dfrac{1}{1+2^{-x}}$ going from $\mathbb R$ to $(0,1)$ and $\ell(x)=a(1-x)+bx$ going from $(0,1)$ to $(a,b)$, so $\ell(g(x)) = a\left(1-\dfrac{1}{1+2^{-x}}\right)+b\left(\dfrac{1}{1+2^{-x}}\right)$ goes from $\mathbb R$ to $(a,b)$. With a bit of algebra you can simplify that last expression somewhat. $\endgroup$ Nov 25, 2012 at 19:33
  • 1
    $\begingroup$ The first one is not even injective $\endgroup$
    – quantum
    Mar 22, 2022 at 10:16

You've already been given two possible homeomorphisms, but how about another one?

Say that you have two maps $\varphi : A \to B$ and $\psi : B \to C$, both of which are homeomorphisms. It's clear that $\psi \circ \varphi : A \to C$ is again a homeomorphism.

Using this fact, choose your favorite finite open interval $(a,b)$, and prove it is homeomorphic to $\mathbb{R}$.

Next up, take an arbitrary open interval $(c,d)$, and construct a homeomorphism between this an $(a,b)$, and voila, you are done.

In particular, look at the interval $(0,1)$, and its image under the function $\tan(\pi(x-\frac{1}{2}))$. This is pretty clearly a homeomorphism. Now just map an open interval to $(0,1)$ (homeomorphically), and call it a day.


Consider the function $\log(b-x)-\log(x-a)$

  • $\begingroup$ how can we show that the inverse is continuous here? $\endgroup$
    – ccc
    Nov 23, 2012 at 0:21
  • 1
    $\begingroup$ @ccc: Solving $y = \log(b-x) - \log(x-a)$ for x would work. $\endgroup$
    – user14972
    Nov 23, 2012 at 2:18

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