1. ## 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?

2. 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 $\displaystyle x~\&~y$ are in the same set then $\displaystyle y~\&~x$ 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 $\displaystyle x~\&~y$ are in the same connected subset and $\displaystyle y~\&~z$ are in the same connected subset then the union is connected having $\displaystyle y$ in common. Thus $\displaystyle x~\&~z$ are in the same connected subset.

3. Thanks for your help, Plato.
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

4. 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.