# Subspace topology is homeomorphic to $(X,T)$

Let $a$ and $b$ be real numbers such that $a<b$ and $(\mathbb R, \tau)$ a topological space generated by the Euclidean metric. I need to show that the open interval $(a,b)$ with the subspace topology is homeomorphic to $(\mathbb R, \tau)$.

Homeomorphic means that there is a function Continuous, one-to-one, in surjection, and having a continuous inverse between the two spaces.

I found the function $$f(x)=tan(\pi/(b-a)x-pi(a+b)/(2(b-a))$$ that is $f:(a,b)->\mathbb R$

How do I show that this function is continuous?

I think that since its derivative exists and is continuous then my $f(X)$ is continuous. I think it is it discontinuous only at $a$ and $b$?

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Hint: $f(x)=x/(1+|x|)$ or $\arctan$ – wj32 Nov 25 '12 at 21:07

HINT: The function $f(x)=\arctan x$ is a homeomorphism from $\Bbb R$ to $\left(-\frac{\pi}2,\frac{\pi}2\right)$; it's not hard to adjust the range to be some other open interval $(a,b)$ by scaling and translating.