0
$\begingroup$

Show that 1) $(-1,1)$ is homeomorphic to $R$

2) $(-1,3]$ is homeomorphic to $[1,3)$.

The definition of homeomorphism :- Let $X$ & $Y$ be topological space . A mapping $f$ from $X$ to $Y$ is said to be homeomorphism if f is bijective, continuous and $f^{-1}$ is continuous.

So for $(1)$ I should define a map. Can I define $f$ as $f(x)=x$ ?

And for $(2)$, $f(x )=\frac{(x+1)}{3}+1 ?$

Clearly both functions are bijective and continuous. Also inverse of f exists and continuous.

$\endgroup$
  • 1
    $\begingroup$ Surely $f(x)=x$ is not surjective. Remember that $f: (-1,1) \to \Bbb R$. $\endgroup$ – DHMO Apr 17 '17 at 12:43
  • 1
    $\begingroup$ For (1), think of a function $f:(-1,1) \to \mathbb{R}$ such that $\lim_{x\to \pm 1} f(x) = \pm \infty$. $\endgroup$ – Prahlad Vaidyanathan Apr 17 '17 at 12:45
  • $\begingroup$ Yes you are right $\endgroup$ – Kavita Apr 17 '17 at 12:46
  • $\begingroup$ What if I take f(x)=$\frac{x}{(1+x)(1-x)}$? If I want to find $y\in R$ such that f(x)=y then will you help me for this? $\endgroup$ – Kavita Apr 17 '17 at 12:48
  • $\begingroup$ I think for (2) also the choice of f is wrong $\endgroup$ – Kavita Apr 17 '17 at 12:52
1
$\begingroup$

For (2) consider a linear function $f$ with $f(-1)=3$ and $f(3)=1$, that is $f(x)=(5-x)/2.$ Restrict of $f$ to the domain $[-1,3)$.

For (1) the most commonly used example for a homeomorphism from $(-1,1)$ to $\mathbb R$ is $g(x)=\tan (\pi x/2).$

$\endgroup$
0
$\begingroup$

For any two intevrals $(a,b),(c,d)$ there is a natural homeomorphism between them is $f : (a,b) \to (c,d)$ . Note that if $x \in (a,b)$ then $x$ has form $\lambda.a + (1-\lambda)b$ then we define $f(x) = \lambda.c+(1-\lambda)d$ . The remainder of proof is $f,f^{-1}$ continuty and bijective ( you should check that ! That it is easy ) and an analogous proof for close interval .

For the case one of $c,d$ is infinite , example the problem which you suggest $(-1,1)$ is homeomorphic to $\mathbb{R}$ , define a map :

$$f : (-1,1) \to \mathbb{R}$$

$$f(x) = \frac{x}{x^{2}-1}$$

$\endgroup$
  • $\begingroup$ @BangPhanKhoa may I know what is $\lambda$? $\endgroup$ – Kavita Apr 17 '17 at 12:57
  • 1
    $\begingroup$ First $x \in (a,b)$ then $\lambda = \frac{b-x}{b-a} , 0 < \lambda < 1$ $\endgroup$ – Gankedbymom Apr 17 '17 at 12:59
  • $\begingroup$ For every y in R is there x in (-1,1) such that f(x)=y. That means can you tell me what is x in terms of y? So that I can show f is onto(surjective)? $\endgroup$ – Kavita Apr 17 '17 at 13:02
  • 1
    $\begingroup$ Please be careful with latex after each formula . Here you can see $f$ is continuous and $\lim_{x \to 1} f(x) = -\infty$ and $\lim_{x \to -1} f(x) = +\infty$ . By immediate value theorem show that $f$ is surjective $\endgroup$ – Gankedbymom Apr 17 '17 at 13:05
  • $\begingroup$ What if instead of open intervals I have semi open and semi closed intervals as I have in example 2)? How should I define f? $\endgroup$ – Kavita Apr 17 '17 at 13:10
0
$\begingroup$

consider $f:\mathbb R\to (1,1)$ defined by $$f(x)=\frac{x}{1+|x|}$$

or

$$f(x)=\frac 2\pi\tan^{-1}x$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.