0
$\begingroup$

I need to show that there exists a homeomorphism from $(0,1) \to(0,4)$ but I don't understand how to do it.

I would guess we could have the function $f(x)=4x$ then this function is continuous, bijective and has a continuous inverse but I don't know how to show that it has these properties. I guess the bijection is obvious but how would I show such a map is continuous in the topological sense?

My definition is that if $(X,\tau_X),(Y,\tau_Y)$ are topological spaces and $f:X \to Y$ then $f$ is continuous at $x$ if the preimage of any neighbourhood of $f(x)$ is a neighbourhood of $x$ (in the respective topological spaces).

My problem is I don't know how to prove this at all I have only ever proved continuity in metric spaces before never in topological spaces (I am new to these).

Could any show me the best way to approach a problem like this.

(If anyone could give me a full example on a similar question that would help a lot I think)

thanks!

  • 1
    polynomial maps are continuous – 2017-01-16
  • 0
    I think you can find the inverse, no? – 2017-01-16
  • 0
    The inverse is $y/4$ (with the correct domain and codomain). I suspect but don't actually know that polynomials are continuous in the topological sense hence the confusion!!!!! – 2017-01-16

1 Answers 1

0

If a topology arises from a metric space, then you can use the $\epsilon-\delta$ definition of continuity as they are the same. I.e., if $(X, d_X)$ is a metric space and $(Y,d_Y)$ is a metric space, then $f:X\to Y$ is continuous if and only if, for every $x_0\in X$ and for every $\epsilon>0$, there exists some $\delta>0$ that if $d_X(x,x_0)<\delta$, then $d_Y(f(x), f(x_0))<\epsilon$.