# Simple quesion about metric on extended real line

1. Apr 13, 2012

### sunjin09

I was told the extended real $\hat{R}=R\cup\{-\infty,\infty\}$ is homeomorphic to [0,1], I was wondering if the mapping
$$h: [0,1]\rightarrow\hat{R}, h(x)=\cot^{-1}(\pi x), 0<x<1, h(0)=\infty, h(1)=-\infty$$
is a valid homeomorphism, so that a metric may be defined by the metric on [0,1]? Thank you.

2. Apr 15, 2012

### quasar987

Yes, of course!

The topology on the extended real line is the topology generated by the usual open sets of R plus the "neigborhoods of infinity"; i.e. the sets of the form $(a,+\infty]$ and $[-\infty,b)$.

Similarly, the topology on [0,1] is generated by the usual open sets of (0,1) plus the sets of the form $(a,1]$ and $[0,b)$

It suffices to observe that your h sends a generator of the topology of [0,1] to a generator of the topology of the extended real line, and similarly for h-1.

Last edited: Apr 15, 2012
3. Apr 16, 2012

### sunjin09

Thank you! When you say generator of the topology, is it referring to a base of the topology? So if I had a bijective (no need to be continuous, which need not be defined) mapping h from one base (collections of open sets) to another, then h would also be a homeomorphism between the two spaces generated by the two bases?

4. Apr 16, 2012

### quasar987

Yes and yes.

More generally, to check continuity of a map f:X-->Y, it is not necessary to check that f-1(U) is open in X for all open sets in Y. It is sufficient to check it for the elements U of a basis of the topology on Y. (Easy exercice in "set theory")

5. Apr 17, 2012

### sunjin09

That makes a lot of sense. Thank you.