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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.

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    $\begingroup$ Surely $f(x)=x$ is not surjective. Remember that $f: (-1,1) \to \Bbb R$. $\endgroup$
    – DHMO
    Apr 17, 2017 at 12:43
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    $\begingroup$ For (1), think of a function $f:(-1,1) \to \mathbb{R}$ such that $\lim_{x\to \pm 1} f(x) = \pm \infty$. $\endgroup$ Apr 17, 2017 at 12:45
  • $\begingroup$ Yes you are right $\endgroup$
    – Kavita
    Apr 17, 2017 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, 2017 at 12:48
  • $\begingroup$ I think for (2) also the choice of f is wrong $\endgroup$
    – Kavita
    Apr 17, 2017 at 12:52

3 Answers 3

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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 $g(x)=\tan (\pi x/2).$

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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 :

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

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

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  • $\begingroup$ @BangPhanKhoa may I know what is $\lambda$? $\endgroup$
    – Kavita
    Apr 17, 2017 at 12:57
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    $\begingroup$ First $x \in (a,b)$ then $\lambda = \frac{b-x}{b-a} , 0 < \lambda < 1$ $\endgroup$ Apr 17, 2017 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, 2017 at 13:02
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    $\begingroup$ 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 $\endgroup$ Apr 17, 2017 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, 2017 at 13:10
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consider $f:\mathbb R\to (1,1)$ defined by $$f(x)=\frac{x}{1+|x|}$$

or

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

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