# Question regarding an equivalent definition of connected sets.

Consider the statement :

In a metric space $$X$$, a subset $$E$$ is disconnected iff $$E\subset A\cup B$$ for some non-empty disjoint open sets $$A,B$$ in $$X$$ such that $$E\cap A\neq \phi$$ and $$E\cap B\neq \phi$$.

Now if I replace $$X$$ to be a topological space rather than a metric space.Then I doubt that this would hold.The if part would hold obviously but the only if part may not hold.Because if $$E$$ is disconnected,then we can find $$A,B$$ open in $$X$$,non-empty such that $$A\cap E$$ and $$B\cap E$$ are also non-empty and $$E=(A\cap E)\cup(B\cap E)\subset A\cup B$$,but we cannot claim that there are $$A,B$$ which are disjoint also.

I just want to clear the doubt because I have yet not studied topology and not capable of coming with a counterexample.Is my claim correct?

I topological spaces,I think I have to be satisfied with,

In a topological space $$X$$, a subset $$E$$ is disconnected if $$E\subset A\cup B$$ for some non-empty disjoint open sets $$A,B$$ in $$X$$ such that $$E\cap A\neq \phi$$ and $$E\cap B\neq \phi$$.

You are correct about the statement not holding for general topological spaces.

Consider the following space $$X$$ : it has $$3$$ points $$a,b,c$$ , and opens $$\emptyset, X, \{a,b\}, \{c,b\}, \{b\}$$.

Then $$\{a,c\}$$ is disconnected : indeed the subspace topology is simply the discrete topology, so you can see $$\{a\}\cup \{b\}$$ is a witness to this fact.

However, any two opens that cover this subspace must intersect (proof : inspect the possible covers !)

In metric spaces, a possible proof of the statement clearly uses the metric : let $$A',B'$$ be opens of $$E$$ such that $$E= A'\sqcup B'$$. Consider for each $$x\in A'$$ a real number $$\epsilon_x >0$$ such that $$B(x,\epsilon_x)\cap E \subset A'$$, and same for $$y\in B'$$ with $$\delta_y$$.

Then consider $$A= \bigcup_{x\in A'}B(x,\frac{\epsilon_x}{2})$$, $$B=\bigcup_{y\in B'}B(x,\frac{\delta_y}{2})$$

If $$z\in A\cap B$$, then it is at distance $$<\epsilon_x/2$$ from some $$x\in A'$$ and $$<\delta_y/2$$ from some $$y\in B'$$. Assume wlog that $$\epsilon_x <\delta_y$$. Then by the triangle inequality, $$x\in B(y,\delta_y)$$, which contradicts $$A'\cap B' = \emptyset$$

You see here that the "ability to divde by $$2$$" is somewhat crucial. It's very likely though that this proof carries over to the context of uniform spaces if you know what those are (they're a bit more general than metric spaces, but you can still "divide by $$2$$")

For general spaces $$A$$ and $$B$$ need not be disjoint but must be disjoint on $$E$$, so

$$E \cap A \cap B = \emptyset$$.

Then we just have a disconnection by relatively open non-empty subsets. We then have for general spaces:

$$E \subset X$$ is disconected iff there exist open sets $$A$$ and $$B$$ such that $$E \subseteq A \cup B, E \cap A \cap B = \emptyset, A \cap E \neq \emptyset, B \cap E \neq \emptyset$$.