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Math Help - topology-homomorphic

  1. #1
    MHF Contributor Amer's Avatar
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    topology-homomorphic

    Prove that  \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

    topology-homomorphic-topology.jpg
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  2. #2
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    shouldn't it be homeomorphic instead of homomorphic?
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Amer View Post
    Prove that  \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

    Click image for larger version. 

Name:	Topology.JPG 
Views:	9 
Size:	14.8 KB 
ID:	15697
    I can't understand the question. Type in LaTeX.

    Quote Originally Posted by facenian View Post
    shouldn't it be homeomorphic instead of homomorphic?
    Yes.
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  4. #4
    MHF Contributor Amer's Avatar
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    if we defined
    \

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

    as I said t_u is the usual topology

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

    since

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

     [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

    (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

    topology-homomorphic-topology2.jpg

    or instead of that prove that A_r is connected space
    Last edited by Amer; March 1st 2010 at 08:49 PM.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Amer View Post
    if we defined
    \

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

    as I said t_u is the usual topology

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

    since

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

     [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

    (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

    Click image for larger version. 

Name:	Topology2.JPG 
Views:	5 
Size:	14.2 KB 
ID:	15719

    or instead of that prove that 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?
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  6. #6
    MHF Contributor Amer's Avatar
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    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
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Amer View Post
    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, A_r is surely connected since x-a)+(y-b)^2=r^2\right\}\cap\mathbb{H}\right)" alt="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.
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  8. #8
    MHF Contributor Amer's Avatar
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    the semi-circle is connected how I can prove that ???

    I know it is connect but how I can prove that
    Last edited by Amer; March 1st 2010 at 09:27 PM.
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Amer View Post
    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 S^1_+=\left\{(x,y):x^2+y^2=1\right\} so it remains to show that this is homeomorphic.

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