Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Connected component

  1. Feb 25, 2010 #1
    I'm not quite clear about this notion. Could anyone explain a little bit for me?
    Here is the definition:
    Let a be an arbitrary point in X . Then there exists a largest connected subset of X
    containing a, i.e. a set Ca such that:
    • a ∈ Ca and Ca is connected;
    • for any connected subset S of X containing a, S ⊆ Ca .
    We call such a set Ca the connected component of X containing a, or simply a connected
    component of X .
    I can see that the connected components of R, Q etc are the singletons.
    What about X = {(x, y) ∈ R2 ; x not equal to y} with the topology induced from R2 ?
    Btw, if the interior of a subset A of R is the empty set, what are the different possibilities? Obviously, A could be Q, C, R etc.
    Your input is greatly appreciated!
     
  2. jcsd
  3. Feb 25, 2010 #2

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Ca is the largest/maximal connected subset containing a, meaning that if S contains a, and

    S contains C, then S is not connected.

    "I can see that the connected components of R, Q etc are the singletons. "

    Q, as a subset of R, is totally disconnected, so that the singletons q_n are the components. Any larger (than {q_n}, for any n ) subset containing q_n is disconnected.

    (R , Std. Metric Topology)is connected, so the maximal connected subset containing
    R is R itself.

    "What about X = {(x, y) ∈ R2 ; x not equal to y} with the topology induced from R2 ?."

    Then check to see if X is disconnected: what is the closure of X1={(x,y) in R^2 : x>y}

    union X2={(x,y) in R^2: x<y}?. X is disconnected if ClX1/\X2 is empty,

    and so is X1/\ClX2 (/\ is intersection, Cl is closure).

    In the use I know, components are closed, since , if a subset A is connected, so is its
    closure.


    "Btw, if the interior of a subset A of R is the empty set, what are the different possibilities? Obviously, A could be Q, C, R etc."

    R does not have empty interior. The Baire Category theorem shows this.

    Q has empty interior --the irrationals are dense in the reals.

    The Cantor set has empty interior
    !
     
  4. Feb 25, 2010 #3
    Thank you very much for your reply. I have a much clearer picture in my mind.
    Just one more question, to find the connected component for X = {(z, w) ∈ C2 ; z not equal to w} with the topology induced from C2, we still need to check to see if X is disconnected right? But it's hard to find two open and disjoint sets whose union is X. Since z not equal to w only means that the real parts and the complex parts are not equal respectively. How do we treat it then? Again, I really appreciate all your input here.
     
  5. Feb 27, 2010 #4
    I can't see, right away, a straightforward way to show that [tex]X=\{(z,w)\in\mathbb{C}^2\mid z\ne w\}[/tex] can't be separated. However, it's not hard to see that it's path connected, implying that it's connected.

    First notice that for any u,v,w in C, there's a path from u to v that avoids w (if this is not obvious, draw a picture). Use that to see that there's a path from (u,v) to (z,w) in X, for any choice of (u,v) and (z,w):

    First, there's a path from (u,v) to (z,v) lying entirely in C x v which avoids (v,v), by the lemma above. Then, there's a path from (z,v) to (z,w) lying entirely in z x C which avoids (z,z) by the same reasoning. Concatenate the two and you've shown X is path connected, therefore connected.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Connected component
  1. Ndependent components (Replies: 1)

  2. Connected Sets (Replies: 7)

  3. Connections and forms (Replies: 12)

Loading...