# Connected component

rainwyz0706
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.

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
!

rainwyz0706
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.

Tinyboss
I can't see, right away, a straightforward way to show that $$X=\{(z,w)\in\mathbb{C}^2\mid z\ne w\}$$ 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.