Theorem: Let $(X,\tau)$ be a topological space. If there is no nonempty closed subsets $F$ and $G$ in $X$ such that $F\cap G=\emptyset$ and $X=F\cup G,$ then $X$ is connected.


Let's suppose X is no connected and let's see that there exist nonempty closed subsets $F$ and $G$ in $X$ such that $F\cap G=\emptyset$ and $X=F\cup G$.

As X is no connected, there exists open sets $U,V\in\tau$ such that $U\cap V=\emptyset$ , $X=U\cup V$ and both are nonempty sets...(1)

Notice that $U^c=V,V^c=U.$

Therefore $U$ and $V$ are closed. And by (1) the proof is done.

I did this proof and I don't know if it's correct. Could you check it and tell me please?

  • 2
    $\begingroup$ It is correct. Good work. $\endgroup$ – mathguy Nov 27 '17 at 20:32
  • 1
    $\begingroup$ :) I can be at peace now. $\endgroup$ – Isabella Nov 27 '17 at 20:35
  • $\begingroup$ Nice proof by contrapositive! $\endgroup$ – Perturbative Nov 28 '17 at 10:59

If we have $X = A \cup B$ where $A \cap B = \emptyset$, the following are equivalent:

  1. $A$ and $B$ are both open.
  2. $A$ and $B$ are both closed.
  3. $A$ and $B$ are separated: $A \cap \overline{B} =\overline{A} \cap B = \emptyset$.

You did the equivalence of 1. and 2. essentially: $A$ and $B$ are each other's complements so $A$ open iff $B$ closed and vice versa.

If $A$ and $B$ are separated, $A$ is closed: $A \subseteq \overline{A} \subseteq X\setminus B =A$ and similarly $B$ is closed, so 3 implies 2. 2 implies 3 is trivial.

All 3 variants are taken (depending on the textbook) as the definition of a disconnected space (if both are non-empty too).

  • $\begingroup$ Good to know that, thanks Henno. What textbook would you recommend for self-learning topology? $\endgroup$ – Isabella Nov 27 '17 at 21:30
  • $\begingroup$ @Isabella Munkres is quite good for self-study. A lot of its excercises are solved on this site (I think all of them are somewhere) which also helps when you're stuck. Also Willard, which is a lot cheaper. $\endgroup$ – Henno Brandsma Nov 27 '17 at 21:33
  • $\begingroup$ "..all of them somewhere" , you know where they are.. :D .Ok thanks again. $\endgroup$ – Isabella Nov 27 '17 at 21:46

I think your proof it's very good.


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