# homeomorphism proof

• Mar 5th 2011, 11:33 AM
MathSucker
homeomorphism proof
I have:

$\displaystyle X= \{ (x,y) \in \Re^2 : 0 < x^2+y^2<1 \}$

$\displaystyle Y= \{ (x,y) \in \Re^2 : 1 < x^2+y^2<2 \}$

both have their usual topologies as subsbaces of $\displaystyle \Re^2$

I want to show that $\displaystyle X \equiv Y$

My work:

Let $\displaystyle S= \{ x:2<x<4 \}$

Let $\displaystyle f$ map $\displaystyle \{x \in \Re^2 : 0 < ||x|| < 1 \}$ to $\displaystyle \{x \in \Re^2: 1< ||x||<2 \}$ by:

$\displaystyle f(x) = (1 + \frac{1}{||x||})x$

Now, $\displaystyle =(1 + \frac{1}{||y||})y$

so $\displaystyle f$ is one-to-one and onto.

$\displaystyle f(f^{-1}(y))=(1+ \frac{1}{||f^{-1}(y)||})f^{-1}(y)$

$\displaystyle =(1+\frac{1}{||f^{-1}(y)||})(1- \frac{1}{||y||})y$

$\displaystyle = ( 1 + \frac{1}{||(1-\frac{1}{||y||})y||})(1-\frac{1}{||y||})y$

$\displaystyle = ( 1 + \frac{1}{(1-\frac{1}{||y||})||y||})(1-\frac{1}{||y||})y$

$\displaystyle = \frac{1}{1-\frac{1}{||y||}}( 1 + \frac{1}{||y||}-\frac{1}{||y||})(1-\frac{1}{||y||})y=y$

and

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

$\displaystyle =(1-\frac{1}{||f(x)||})(1+ \frac{1}{||x||})x$

$\displaystyle = ( 1 - \frac{1}{||(1+\frac{1}{||x||})x||})(1+\frac{1}{||x ||})x$

$\displaystyle = ( 1 - \frac{1}{(1+\frac{1}{||x||})||x||})(1+\frac{1}{||x ||})x$

$\displaystyle = \frac{1}{1-\frac{1}{||x||}}( 1 + \frac{1}{||x||}-\frac{1}{||x||})(1+\frac{1}{||x||})x=x$

$\displaystyle f$ and $\displaystyle f^{-1}$ are continuous in the unduced topologies on both sets, so... that's it? (Headbang)
• Mar 6th 2011, 10:10 AM
Drexel28
Quote:

Originally Posted by MathSucker
I have:

$\displaystyle X= \{ (x,y) \in \Re^2 : 0 < x^2+y^2<1 \}$

$\displaystyle Y= \{ (x,y) \in \Re^2 : 1 < x^2+y^2<2 \}$

both have their usual topologies as subsbaces of $\displaystyle \Re^2$

I want to show that $\displaystyle X \equiv Y$

My work:

Let $\displaystyle S= \{ x:2<x<4 \}$

Let $\displaystyle f$ map $\displaystyle \{x \in \Re^2 : 0 < ||x|| < 1 \}$ to $\displaystyle \{x \in \Re^2: 1< ||x||<2 \}$ by:

$\displaystyle f(x) = (1 + \frac{1}{||x||})x$

Now, $\displaystyle =(1 + \frac{1}{||y||})y$

so $\displaystyle f$ is one-to-one and onto.

$\displaystyle f(f^{-1}(y))=(1+ \frac{1}{||f^{-1}(y)||})f^{-1}(y)$

$\displaystyle =(1+\frac{1}{||f^{-1}(y)||})(1- \frac{1}{||y||})y$

$\displaystyle = ( 1 + \frac{1}{||(1-\frac{1}{||y||})y||})(1-\frac{1}{||y||})y$

$\displaystyle = ( 1 + \frac{1}{(1-\frac{1}{||y||})||y||})(1-\frac{1}{||y||})y$

$\displaystyle = \frac{1}{1-\frac{1}{||y||}}( 1 + \frac{1}{||y||}-\frac{1}{||y||})(1-\frac{1}{||y||})y=y$

and

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

$\displaystyle =(1-\frac{1}{||f(x)||})(1+ \frac{1}{||x||})x$

$\displaystyle = ( 1 - \frac{1}{||(1+\frac{1}{||x||})x||})(1+\frac{1}{||x ||})x$

$\displaystyle = ( 1 - \frac{1}{(1+\frac{1}{||x||})||x||})(1+\frac{1}{||x ||})x$

$\displaystyle = \frac{1}{1-\frac{1}{||x||}}( 1 + \frac{1}{||x||}-\frac{1}{||x||})(1+\frac{1}{||x||})x=x$

$\displaystyle f$ and $\displaystyle f^{-1}$ are continuous in the unduced topologies on both sets, so... that's it? (Headbang)

I didn't check it 100 %, but I looked at most and it looked correct. Good job!