1. ## topology-homomorphic

Prove that $\displaystyle \left( R^2 - \left{ (a,b) \right} , t_{u} \right)$
is a connected space, t_u is the usual topology

defined

A_r = ( (-infinity , a-r ] U [ a+r , infinity ) ) X {b} U {(x,y) in R^2 : (x-a)^2 + (y-b)^2 =r^2 y>= b }

B_r = ( (-infinity , a-r ] U [ a+r , infinity ) ) X {b} U {(x,y) in R^2 : (x-a)^2 + (y-b)^2 =r^2 y<= b }

A_r homomorphic with R with usual topology , thats my question I do not know why A_r homorophic with R^1 ???

A_r connected
B_r connected .... I get all the rest it is easy

here is a picture that shows A_r and B_r

2. shouldn't it be homeomorphic instead of homomorphic?

3. Originally Posted by Amer
Prove that $\displaystyle \left( R^2 - \left{ (a,b) \right} , t_{u} \right)$
is a connected space, t_u is the usual topology

defined

A_r = ( (-infinity , a-r ] U [ a+r , infinity ) ) X {b} U {(x,y) in R^2 : (x-a)^2 + (y-b)^2 =r^2 y>= b }

B_r = ( (-infinity , a-r ] U [ a+r , infinity ) ) X {b} U {(x,y) in R^2 : (x-a)^2 + (y-b)^2 =r^2 y<= b }

A_r homomorphic with R with usual topology , thats my question I do not know why A_r homorophic with R^1 ???

A_r connected
B_r connected .... I get all the rest it is easy

here is a picture that shows A_r and B_r

I can't understand the question. Type in LaTeX.

Originally Posted by facenian
shouldn't it be homeomorphic instead of homomorphic?
Yes.

4. if we defined
$\displaystyle \$

$\displaystyle A_r = ( (-\infty, a-r ] \cup [ a+r , \infty) \times \{b\} \cup ( (x,y) \in R^2 : (x-a)^2 + (y-b)^2 = r^2 , y\leq b )$

prove that this is homeomorphic with R I draw A_r in the picture that I attached
in latex Prove
$\displaystyle (A_r,t_u ) \cong (R, t_u)$

as I said $\displaystyle t_u$ is the usual topology

what I came up is if I can prove that $\displaystyle (a-r , a+r) \cong ((x,y) \in R^2 : (x-a)^2 + (y-b)^2 = r^2 , y\leq b )$

since

$\displaystyle ( -\infty , a-r] \times {b} \cong (-\infty , a-r]$ two spaces with usual topology

$\displaystyle [a+r , \infty) \times {b} \cong [a+r ,\infty)$ two spaces with usual topology

so the the problem is how I can prove the semi-circle (x-a)^2 + (y-b)^2 y>=b is homemorphic with the interval (a-r , a+r) or can it be, my prof said

$\displaystyle (A_r, t_u) \cong (R,t_u)$

but it is not clear how

that's my question and here is A_r after I changed something

or instead of that prove that $\displaystyle A_r$ is connected space

5. Originally Posted by Amer
if we defined
$\displaystyle \$

$\displaystyle A_r = ( (-\infty, a-r ] \cup [ a+r , \infty) \times \{b\} \cup ( (x,y) \in R^2 : (x-a)^2 + (y-b)^2 = r^2 , y\leq b )$

prove that this is homeomorphic with R I draw A_r in the picture that I attached
in latex Prove
$\displaystyle (A_r,t_u ) \cong (R, t_u)$

as I said $\displaystyle t_u$ is the usual topology

what I came up is if I can prove that $\displaystyle (a-r , a+r) \cong ((x,y) \in R^2 : (x-a)^2 + (y-b)^2 = r^2 , y\leq b )$

since

$\displaystyle ( -\infty , a-r] \times {b} \cong (-\infty , a-r]$ two spaces with usual topology

$\displaystyle [a+r , \infty) \times {b} \cong [a+r ,\infty)$ two spaces with usual topology

so the the problem is how I can prove the semi-circle (x-a)^2 + (y-b)^2 y>=b is homemorphic with the interval (a-r , a+r) or can it be, my prof said

$\displaystyle (A_r, t_u) \cong (R,t_u)$

but it is not clear how

that's my question and here is A_r after I changed something

or instead of that prove that $\displaystyle A_r$ is connected space
So do you want to prove that the semi-circle and the interval are homeomorphic. Is that what you're asking for? Or, the whole question?

6. yeah I want to prove that A_r homeomophic with R with usual topology, I want to prove that to make sure that A_r is connected since connectivity is topological property, or is there away to prove that A_r is connect directly ?

thanks

7. Originally Posted by Amer
yeah I want to prove that A_r homeomophic with R with usual topology, I want to prove that to make sure that A_r is connected since connectivity is topological property, or is there away to prove that A_r is connect directly ?

thanks
Well, $\displaystyle A_r$ is surely connected since $\displaystyle A_r=\mathbb{R}-(a-r,a+r)\cup \left(\left\{(x,y)x-a)+(y-b)^2=r^2\right\}\cap\mathbb{H}\right)$. Now, the union of two disjoint connected spaces is connected.

8. the semi-circle is connected how I can prove that ???

I know it is connect but how I can prove that

9. Originally Posted by Amer
the semi-circle is connected how I can prove that ???

I know it is connect but how I can prove that

and as I know the union of two connect set is connect if the intersect between them is not empty

example (1,2) U (4,5) is not connected although each one is connect in usual topology
Clearly your semi-circle is homeomorphic to the unit semi-circle $\displaystyle S^1_+=\left\{(x,y):x^2+y^2=1\right\}$ so it remains to show that this is homeomorphic.

Define $\displaystyle \varphi:[0,1]\mapsto S^1_+$ by $\displaystyle x\mapsto \sqrt{1-x^2}$. This is readily verified to be a continuous.