# Can a disconnected set be simply classified by it is a union of two disjoint sets (ignoring the topology)?

In a lot of proofs I am doing, I ended up proving that a set $$E$$ is disconnected by showing it is a union of two disjoint sets, without considering their topology such as saying disjoint open (equiv closed) subsets. Is this enough to say $$E$$ is disconnected? Or do I have to follow the definition it is?

Here is an example. I want to show the closure of a connected set is connected, denote such set $$S$$.

Now if $$\bar{S}$$ is disconnected, then it is separated by $$\bar{S} = A \cup B$$. Now $$\bar{S} = S \cup S'$$. Thus $$S = (A \cup B) - S' = (A - S') \cup (B - S').$$ Notice $$(A - S') \cap (B - S') = (A \cap B) - S' = \emptyset- S' = \emptyset$$.

Note $$-$$ is for relative complement, I don't know how to do the reverse slash thing, sorry or what the symbol is called.

• $\setminus$ formats as $\setminus$ – saulspatz Jul 29 at 12:40
• Every set is a sum of two disjoint ones. Connectedness is a topological property, you can never ignore topology when talking about it. – Jakobian Jul 29 at 12:41

To answer the question in the title:

$$\mathbb R=(-\infty, 0]\cup (0,\infty)$$, and $$(-\infty, 0]$$ and $$(0,\infty)$$ are disjoint sets.

Does this mean $$\mathbb R$$ is disconnected?

To go into detail into your argument:

You proved that $$S$$ is a union of two disjoint sets. You did not prove that the two disjoint sets are both closed (or that they are both open, or that one is both closed and open).

• Ahhh okay good example. Also something tells me that the relative complements of open isn't necessarily closed right? – Hawk Jul 29 at 12:53
• Actually I got it, $A - S' = A \cap (S')^c$ is the intersection of two open sets, hence they are open are now the proof is done? – Hawk Jul 29 at 12:58