# Connectedness

• Sep 26th 2009, 03:19 AM
bramkierkels
Connectedness
Hi, i need some help with problem;

Let (X,d) be a metric space.
I define a relation x~y on X by declaring x~y iff there exist a connected subset of X which contains both x en y.
How can i show that this is a equivalence relation and that the equivalence classes of this relation are all closed and connected?

• Sep 26th 2009, 03:52 AM
Plato
Quote:

Originally Posted by bramkierkels
Let (X,d) be a metric space.
I define a relation x~y on X by declaring x~y iff there exist a connected subset of X which contains both x and y. How can I show that this is a equivalence relation and that the equivalence classes of this relation are all closed and connected?

Because any singleton set is clearly connected, the relation is reflexive.

If $x~\&~y$ are in the same set then $y~\&~x$ are in the same set, the relation is symmetric.

Thus far the relation is no different than most equivalence relations.
But proving it is transitive requires a bit more.
You need to have proved this theorem: If two connected subsets have a point in common then their union is connected.
So if $x~\&~y$ are in the same connected subset and $y~\&~z$ are in the same connected subset then the union is connected having $y$ in common. Thus $x~\&~z$ are in the same connected subset.
• Sep 28th 2009, 04:51 AM
bram kierkels
Indeed the problem for me was proving the transitivity. Reflexivity and symmetry were quite clear.
I start the proof of transitivy with if the union were disconneted, it would split into parts X' and X'' which do not adhere to one another. At the end I get a contradiction. I think this is the right way.
Can you help me proving the equivalence classes are closed and connected?
Thanks
• Sep 28th 2009, 07:00 AM
Plato
Quote:

Originally Posted by bram kierkels
Can you help me proving the equivalence classes are closed and connected?

The equivalence classes in this relation are just the components of the space- i.e. the maximal connected subsets.
Recall this theorem: The closure of a connected set is connected.
That theorem proves the question.