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

  • Thread starter Cauchy1789
  • Start date
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 [tex]\mathbb{Q}[/tex] is the set of rational numbers, which can be partioned into none-empty open subsets of
[tex]\mathbb{Q}[/tex].)
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 [tex]x_1 \cup x_2[/tex] 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 [tex]\overline{x} \cap T[/tex] x assumed to be closed i T, then [tex]x = \overline{x} \cap T \rightarrow \overline{x} \cap T = \emptyset.[/tex], thus [tex]\overline{x}[/tex] 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

[tex]\overline{x} \cap y = \emptyset[/tex] and [tex]x \cap \overline{y} = \emptyset [/tex] then

[tex]\overline{x} \cap T = x[/tex] and [tex]\overline{y} \cap T = y[/tex]

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 [tex]z \in cl(A)[/tex] then cl(A) contains both x and z thus [tex]x \thilde z[/tex]. Then [tex]z \in [x] = A.[/tex] 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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