1
$\begingroup$

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 $E\cap B\neq \phi$.

$\endgroup$
2
$\begingroup$

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 $2$")

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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 $E \subseteq A \cup B, E \cap A \cap B = \emptyset, A \cap E \neq \emptyset, B \cap E \neq \emptyset$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.