# If A complement is the union of two separated sets, prove that the union of those separated sets with A is connected.

Let $$A$$ be a connected subset of a connected metric space $$(X,d)$$.

Assume $$A^{c}$$ is the union of two separated sets $$B$$ and $$C$$.

Prove that $$A \cup B$$ and $$A \cup C$$ are connected.

Attempt

Proving $$A \cup B$$ is connected is sufficient.

Assume towards a contradiction that $$A \cup B$$ is not connected. (So it is disconnected).

There exists open sets $$G_{1}$$ and $$G_{2}$$ such that $$A \cup B \subseteq G_{1} \cup G_{2}$$, $$(A\cup B) \cap G_{1} \neq \phi$$, $$(A\cup B) \cap G_{2} \neq \phi$$ and $$(A\cup B)\cap G_{1} \cap G_{2} = \phi$$.

Since $$A \subseteq A \cup B$$ and $$A$$ is connected, then either $$A$$ lies in $$G_{1}$$ or $$G_{2}$$.

WLOG, assume $$A$$ lies in $$G_{1}$$.

I do not know how to proceed from here. I need to obtain a contradiction to finish my proof.

Edit

I have corrected my proof.

• You cannot prove that $A \cup B$ is connected for any subset $B$ of $A^{c}$. You don't use the fact that $B$ and $C$ are separated anywhere in your proof so this cannot be correct. @Snop D. – Kabo Murphy Dec 4 '18 at 9:25

You are almost there. If $$A$$ lies entirely in $$G_1$$, and $$A\cap G_2\neq \emptyset$$, then you can easily conclude that $$G_1\cap G_2$$ cannot be an empty set, therefore, a contradiction follows.

You can conclude this because you know that $$A\cap G_2\subseteq G_1\cap G_2$$, and $$\emptyset$$ has no non-empty subsets.

• The hypothesis that $B$ and $C$ are separated and their union is $A^{c}$ is necessary for this result. Where is this used in this proof? – Kabo Murphy Dec 4 '18 at 9:10
• @KaviRamaMurthy OP already reached a point where he proved that $A$ lies in either $G_1$ or $G_2$. I just continued from there. – 5xum Dec 4 '18 at 9:14
• @5xum That seems to work! Perhaps I need the $A^{c}$ part to prove the condition in my Edit. – Snop D. Dec 4 '18 at 9:19
• I had made a mistake in my proof. I have corrected it now. – Snop D. Dec 4 '18 at 22:19

Observation:if $$X$$ is disconnected, then there exists non-empty subsets $$U, V$$ which, in addition to satisfying that $$X=U \cup V$$, satisfy the following equivalent conditions:

1. $$U, V$$ are disjoint and $$U, V$$ are open in $$X$$
2. $$U, V$$ are disjoint and $$U, V$$ are closed in $$X$$
3. $$U$$ $$\cup$$ $$cl(V)_X = \emptyset$$ and $$V$$ $$\cup$$ $$cl(U)_X = \emptyset$$

Lemma:suppose $$X$$ is disconnected, so that there exists non-empty subsets $$U, V$$ such that $$X=U \cup V$$, $$U, V$$ are disjoint, and $$U, V$$ are open in $$X$$. If $$A$$ is a connected subset of $$X$$, then $$A\subseteq U$$ or $$A\subseteq V$$.

Suppose, toward contradiction, that $$A\cup B$$ is disconnected, so that there exists non-empty subsets $$G, H$$ such that $$A\cup B=G \cup H$$, $$G, H$$ are disjoint, and $$G, H$$ are open in $$A \cup B$$.

Since $$A$$ is connected, w.l.o.g. suppose that $$A\subseteq G$$, so that $$H \subseteq B$$.

Claim 1: $$H$$ is closed in $$X$$.

Proof 1:

Since $$B$$ is closed in $$X\setminus A$$ and $$cl(B)_{B\cup C}\cap C=\emptyset$$, $$cl(H)_{A\cup B\cup C}=cl(H)_{A}\cup cl(H)_{B\cup C}=cl(H)_{A}\cup cl(H)_{B}=cl(H)_{A}\cup H$$.

If $$x\in cl(H)_{A}\setminus H$$, then $$x\notin H, x\in A$$, so that $$x\in G$$. This implies $$x \in cl(H)_A\cap G \subseteq cl(H)_{A\cup B}\cap G$$, so that $$cl(H)_{A\cup B}\cap G \neq \emptyset$$, which is a contradiction. Therefore, $$cl(H)_A=H$$.

Putting everything together, we see that $$cl(H)_{A\cup B\cup C}=cl(H)_{A}\cup H=H \cup H=H$$. Therefore, $$H$$ is closed in $$A\cup B\cup C=X$$.

Claim 2: $$H$$ is open in $$X$$.

Proof 2:

If $$x\in H$$ and $$x\notin int(H)_{X}$$, then every neighbourhood of $$x$$ contains some point in $$X\setminus H=(A\cup B \cup C)\setminus H$$, so that every neighbourhood of $$x$$ contains some point in $$C\setminus H$$. This implies that $$x\in cl(C)_{B\cup C}$$, so that, since $$x\in H\cap cl(C)_{B\cup C}\subseteq B\cap cl(C)_{B\cup C}$$, $$B\cap cl(C)_{B\cup C}=B\cap cl(C)_{X\setminus A}\neq \emptyset$$, which is a contradiction. Therefore, $$H=int(H)_{X}$$, so that $$H$$ is open in $$X$$.

Conclusion: since $$H$$ is a nontrivial clopen subset of $$X$$, $$X$$ is disconnected, which is a contradiction. Therefore, the supposition that $$A\cup B$$ is disconnected must be false.

Isn't the complement of $$A \cup B$$ just C? And I think it's easy to see that C is open.