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The version of disconnected space I was given is as follows:

A topological space $(X, \tau)$ is disconnected if there exists disjoin, nonempty open sets $A, B \subseteq X$ such that $X = A\sqcup B$ and $\overline A \cap B = A \cap \overline B = \varnothing$

I need to show that above is equivalent that

There exists disjoint open sets $A,B \subseteq X$ such that $X = A \sqcup B$


I think the reverse implication is easy so I am going to prove that first:

  1. Suppose that $A, B \subseteq X$ are open and disjoint.

Then $\overline A \cap B = (\partial A \cup A) \cap B = \partial A \cap B$, and $A \cap \overline B = A \cap \partial B$

Then $(\overline A \cap B) \cup (A \cap \overline B) = (\partial A \cap B) \cup (A \cap \partial B) = (\partial A \cup A) \cap (\partial A \cap \partial B) \cap (B \cup A) \cap (\partial B \cup B) = A \cap (\partial A \cap \partial B) \cap X \cap B$

But $A \cap B = \varnothing$, so $(\overline A \cap B) \cup (A \cap \overline B) = \varnothing$, $\implies \overline A \cap B = \varnothing$ and $A \cap \overline B = \varnothing$.

Can someone check if this is okay?

  1. Suppose that there exists disjoint, nonempty sets $A, B \subseteq X$ such that $X = A\sqcup B$ and $\overline A \cap B = A \cap \overline B = \varnothing$. We wish to produce open disjoint sets such that $A \sqcup B = X$

(Initial attempt: Since $A, B$ are disjoint, then their interiors are disjoint. Then $\text{int} A, \text{int} B$ are disjoint open sets...but they do not necessarily cover $X$)

Does anyone see how to do this?

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Your first argument is not correct: it’s not necessarily true that $A\cup\operatorname{bdry}A=A$. In any case there’s a much easier approach. Suppose that $x\in B$; $B$ is an open nbhd of $x$ disjoint from $A$, so $x\notin\operatorname{cl}A$. Thus, $B\cap\operatorname{cl}A=\varnothing$, and a similar argument shows that $A\cap\operatorname{cl}B=\varnothing$.

For the second problem, if $X=A\sqcup B$, and $A\cap\operatorname{cl}B=\varnothing$, then $\operatorname{cl}B=B$. (Why?) But then $A=X\setminus\operatorname{cl}B$ is open. Clearly we can make a similar argument for $B$.

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  • $\begingroup$ You are a genius! $\endgroup$ – Shamisen Expert Aug 6 '16 at 19:50
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  1. What you want to show equivalence to is : There exist disjoint NON-empty open $A,B$ with $A\cup B=X.$

    1. If $X=A\cup B$ where $A,B$ are non-empty, open, and $A\cap \bar B=\phi = \bar A \cap B,$ then $A\cap B\subset \bar A\cap B=\phi.$ So $X=A\cup B$ where $A,B$ are disjoint non-empty and open.

    2. If $X=A\cup B$ where $A,B$ are non-empty, open, and disjoint, note that $X$ \ $A$ and $X$ \ $B$ are closed. But $X$ \ $A=B$ and $X$ \ $B=A.$ So $A$ and $B$ are closed: $A=\bar A$ and $B=\bar B .$ So $A\cap \bar B =A\cap B=\phi $ and $\bar A \cap B=A\cap B=\phi.$

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