# Equivalence relations and connected components(Please look at my calculations)

#### Cauchy1789

1. Homework Statement

Hi I have justed switched to a new subject and have some question.

1) Show that if X is a topology space then there exist an equivalence relation if and only if there exist a connected subset which contains both x and y.
2) Show that the connected components are a closed space.
3)Show that if X has a finite amount of connected components, then these are open.
4) Find the connected components in in the following(The set $$\mathbb{Q}$$ is the set of rational numbers, which can be partioned into none-empty open subsets of
$$\mathbb{Q}$$.)
3. The Attempt at a Solution

Definition: A topology space X is to have a seperation if for a pair of non-empty open subsets where $$x_1 \cup x_2$$ A space which is connection does not a have seperation.

Proof(1)
Let T be a subspace of x, and where x and y represents a seperation i T.
Then x is both open and closed in T. Then x is a seperation in T is defined as $$\overline{x} \cap T$$ x assumed to be closed i T, then $$x = \overline{x} \cap T \rightarrow \overline{x} \cap T = \emptyset.$$, thus $$\overline{x}$$ is union of x and its limitpoints, such that it contains, but no limit points for y.
Next assume that x and y are disjoined sets who union is T. Where neither x or y contains limit points for each other, then

$$\overline{x} \cap y = \emptyset$$ and $$x \cap \overline{y} = \emptyset$$ then

$$\overline{x} \cap T = x$$ and $$\overline{y} \cap T = y$$

This implies that x = T - y and y = T - x, then x and y are both closed in T.

Since there is by the proof above doesn't not exist a seperation on X, this implies that X is connected space which again implies that x is a connected subset which connects x and y and thusly the equivalence relation exists. q.e.d.

proof(2)

Let A be a subset of X. Then A = [x] is a connected component of X. Next assume that cl(A) is connected. Finally assume that $$z \in cl(A)$$ then cl(A) contains both x and z thus $$x \thilde z$$. Then $$z \in [x] = A.$$ Thusly A = cl(A). Which implies A is closed. q.e.d.

proof(3)

How the devil do I does this?

proof(4)
Since Q is open then its subsets are also open. q.e.d.

How is this ??

I need help with 3 :(

Sincerely
Cauchy

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