Topological Homeomorphisms

Question

This problem seems to be quite hard for me to understand, I dont even understand the solution on the notes:

Problem: $U = \mathbb{R}\cup\{-\infty,+\infty\}$. Let $U$ have the topology induced by the following topological basis:

$S^*=\{(c,d)\}\cup\{(c,+\infty])\}\cup\{[-\infty,d)\}$.

We have to show that $U$ is homeomorphic to $[0,1].$

The solution writes something along the line of considering $G$ as some strictly increasing function from $\mathbb{R}$ to $(0,1),$ then $G*$ is from $\mathbb{R}\cup\{-\infty,+\infty\}$ to [0,1]. Then it says $G^*$ is invertible and sends the set $A$ to the set corresponding to $\{(c, d): 0 < c < d < 1\} ∪ \{[0, d): 0 < d < 1\} ∪ \{(c, 1]: 0 < c < 1\}.$ Hence this set induces the usual topology on $[0,1]$ , so $G^*$ is a homeomorphism.

This solution is very confusing and I can make little sense of it, can anyone help me provide a better solution?

Answer 1

Considering on $[0,1]$ the topology induced by $\Bbb R$ :

You can obtain an homeomorphism as you requested defining $ F: [0,1] \to \Bbb U$ and putting $F(x) = tan(\pi(x-1/2))$ if $x \in]0,1[$ , $F(0) = -\infty$ and $F(1) = +\infty$.

F is clearly bijective. In order to prove that it is continuous it is sufficient to prove that F is continuous at 0 and 1.

For example, pick $b\in \Bbb R$ and notice that $]\frac{1}{2}+\frac{1}{\pi}arctan(b), 1]$ is an open neighborhood whose image through F is in $]b,+\infty]$; thus F is continuous at 1. Similarly you can prove that F is continuous at 0 in the same way.

In an analogous way you can prove that $F^{-1}$ is continuous at $-\infty$ and $+\infty$ (or you can prove that F is an open function, i.e. F maps open sets to open sets) thus showing that F is really an homeomorphism.

Answer 2

Consider the function $f(x)=\frac1{\pi}\tan^{-1}(x)+\frac12$, with accepting $\tan^{-1}(\pm\infty) =\pm\frac{\pi}2$

Note that:
(1): $f$ is increasing, bijective, continuous
(2): image set of open sets are open

So $f$ is a homeomorphism.

Answer 3

Some steps that can help:

• For $a,b,c\in\mathbb R$ prove that $[a,b]$ and $[a+c,b+c]$ are homeomorphic.

• prove that for positive $a$ the spaces $[-a,a]$ and $[-1,1]$ are homeomorphic.

• with former steps prove that $[0,1]$ and $[-\frac{\pi}2,\frac{\pi}2]$ are homeomorphic.

• Prove that $\tan:[-\frac{\pi}2,\frac{\pi}2]\to[-\infty,+\infty]$ is a homeomorphism.

Answer 4

Comment.-Helping you understand the demonstration you are dating.

1) "some strictly increasing function from $\mathbb R$ to $(0,1)$". There are infinitely many of these such functions: all strictly increasing continuous graph you can draw is an example.

2) You don't need an explicit analytical example to see that your function (name it $G$) is invertible: you can see that for all $x\in\mathbb R$ there is a unique $y\in (0,1)$ such that $G(x)=y$ and reciprocally.

3)It is not difficult to see that $G^*$ satisfies 2).

4) What remains is simple.

The induced topology is that of the usual order.