3
$\begingroup$

Let A be a set such $$A=(1,2]\cup{((3,4)\cap{\mathbb{Q}})}$$ and let B be a set such $$B=(0,1]\cup{(2,3]}\cup{((3,5)\cap{\mathbb{Q}})}$$

I need to find if B is isomorphic to A. I know that a bijection from A to B is needed, but I can't find a proper bijection.

It is easy to find a bijection between $(1,2]$ and $(0,1]$ - $f(x)=x-1$, but I can't define a bijection from $(3,4)$ to $(2,5)$.

How may I continue? is it even good start defining $f(x)=x-1$ for $x\in{(1,2]}$?

Please help, thank you!


Edit:

After reading the comments, I realized that I need to construct 3 bijections:

  1. $f: \ (3,4)\cap{\mathbb{Q}} \rightarrow (3,5)\cap{\mathbb{Q}}$ - I constructed $f(x)=2x-3 \ , \ x\in{\mathbb{Q}}$.
  2. $g: \ \left(1,1 \frac{1}{2} \right] \rightarrow \left(0,1 \right]$ - I constructed $g(x)=2(x-1) \ , \ x\in{\mathbb{R}}$.
  3. $h: \ \left(1 \frac{1}{2},2 \right] \rightarrow \left(2,3 \right]$ - I constructed $h(x)=2\left(x-\frac{1}{2} \right)$

Finally, I defined $\phi(x)$ to be the proper function (f,g,h) in each range of x.

If there any corrections needed, please show them. Thanks again!

$\endgroup$
  • $\begingroup$ I don't think defining $f(x) = x-1$ will help. That covers all of $(0,1]$, but leaves you with nothing left to cover $(2,3]$. $\endgroup$ – MJD Dec 30 '13 at 17:17
3
$\begingroup$

Hints

To map $(3,4)\cap\Bbb Q$ to $(3,5)\cap\Bbb Q$ try to construct a linear map from $(3,4)$ to $(3,5)$. A linear map will take rationals to rationals in both directions.

Then you need to map $(1,2]$ to $(0,1]\cup (2,3]$. Try breaking $(1,2]$ into two disjoint parts. Map one part to $(0,1]$ and the other part to $(2,3]$.

$\endgroup$
  • $\begingroup$ Great answer :) tnx. $\endgroup$ – Galc127 Dec 30 '13 at 17:24
3
$\begingroup$

$$\text{Construct the following bijections and put them together:}$$

$$\text{$(3,4) \cap \mathbb{Q}$ and $(3,5) \cap \mathbb{Q}$}$$ $$\text{$(0,1]$ and $(1,1.5]$}$$ $$\text{$(2,3]$ and $(1.5,2]$}$$

$\endgroup$
  • $\begingroup$ Great answer :) tnx. $\endgroup$ – Galc127 Dec 30 '13 at 17:25

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.