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Can you please let me know if my proof is reasonable?

Prove: If $A$ and $B$ are conneted in $\mathbb{R}^n$ and $A\cap B\neq \emptyset$, then $A\cup B$ is connected

Proof: Suppose that $A\cap B$ is not connected. Then there exist sets $U$ and $V$ such that $U\cup V = A\cap B$, $U=C\cap(A\cap B)$, $V=D\cap(A\cap B)$ for some open sets $C$ and $D$, and $U\cap V=\emptyset$.

But $C\cap(A\cap B)\subseteq C\cap A$ and $D\cap(A\cap B)\subseteq D\cap B$, which means that $U$ and $V$ are relatively open in $A$ and $B$, respectively, which is a contradiction.

Now, $A\cup B = A \cup (A\cap B)\cup B$, where all three sets are connected. Thus, $A\cup B$ is connected.

Forever Mozart
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sequence
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2 Answers2

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Your proof appears to have a number of serious errors. The first of which is that $A\cap B$ needn't be connected - for instance, consider the sets $$A=\{(x,y)\in \mathbb R^2:x\geq 0\}\setminus\{(0,0)\}$$ $$B=\{(x,y)\in \mathbb R^2:x\leq 0\}\setminus\{(0,0)\}$$ which have intersection $A\cap B=\{0\}\times(\mathbb R\setminus \{0\})$ which a line, missing the origin, which is decidedly not connected, being partitioned into upper half-plane and lower half-plane.

You start in a reasonable way by assuming there are open $U$ and $V$ with $U\cup V = A\cap B$ and then take open $C$ and $D$ which coincide with $U$ and $V$ on $A\cap B$. You then assert that $C\cap (A\cap B)\subseteq C\cap A$ implies that $U$ is relatively open in $A$, which is incorrect (see the counterexample I gave). Moreover, those sets being open wouldn't be a contradiction unless you could show that $U\cup V=A$ or $U\cup V =B$ (which you couldn't, since it's not necessarily true).

There's also an error on the last line: you write that since $A\cup B$ is a union of three connected sets, it is connected, but this isn't true - for instance, the union of two disjoint connected sets bounded away from each other is not connected. Moreover, the case where they intersect is what you're trying to prove, so this is circular.

I'd suggest that you try rewriting the proof starting with a more direct premise: Suppose that $U\cup V=A\cup B$ is a partition of $A\cup B$ into two disjoint open sets $U$ and $V$. Then, work from there towards a contradiction. Considering the sets $A\cap U$ and $A\cap V$ is a good direction to go in, if you're stuck.

Milo Brandt
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  • How to read the notation ${0} \times (\mathbb{R} \backslash {0})$? – sequence Feb 03 '16 at 02:15
  • Milo: Unfortunately, I don't see how your counterexample is related to $C\cap(A\cap B) \subseteq C\cap A$ not implying $U$ being relatively open in $A$. But I do see that the intersection of two connected sets doesn't have to be connected. – sequence Feb 03 '16 at 02:23
  • Also, why would it be necessary to show that $U\cup V = A$ or $U\cup V = B$, and not that $U\cup V = A\cap B$? – sequence Feb 03 '16 at 02:27
  • What do you mean by "bounded away from each other"? – sequence Feb 03 '16 at 02:29
  • Also, can you please clarify what is exactly circular in my proof? – sequence Feb 03 '16 at 02:30
  • ${0}\times (\mathbb R\setminus {0}) = {(x,y) : x=0 \text{ and } y\neq 0}$. It is a vertical line minus the origin. It has one partition into disjoint open sets, which would be $U={(x,y) : x=0\text{ and }y>0}$ and $V={(x,y) : x=0\text{ and }y<0}$. These sets are decidedly not open in $A$ or $B$. You would need that $C\cap A = U$ or something analogous in order to conclude that, but you have $U=C\cap (A\cap B)\subseteq C\cap A$, which isn’t enough. – Milo Brandt Feb 03 '16 at 15:19
  • I wrote that you needed $U\cup V=A$ or $U\cup V=B$ in your proof because it seemed like you wanted to make a contradiction by saying that if $A\cap B$ wasn’t connected, then one of $A$ or $B$ wasn’t either. Instead you only tried to construct open subsets of $A$ and $B$ which doesn’t contradict anything. “Bounded away from each other” means that there are open sets $A’$ and $B’$ containing $A$ and $B$ and which are disjoint. That is, $[0,1]$ and $(1,2]$ are not bounded away from each other (and are disjoint, but have a connected union) whereas $[0,1]$ and $[2,3]$ are. – Milo Brandt Feb 03 '16 at 15:20
  • @sequence And the circular bit is the last line - the claim you are to prove is that if two sets are connected and not disjoint, their union is connected. However, the last line uses this (actually it uses a stronger property) where it says that the union of three connected sets is connected. – Milo Brandt Feb 03 '16 at 15:20
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Consider $f:A\cup B\rightarrow \{0,1\}$ continuous. Since $A$ is connected, $f(A)=0$ or $f(A)=1$. Suppose $f(A)=0$, then $f(A\cap B)=0$. Since $B$ is connected the restriction of $f$ to $B$ is constant, so $f(B)=f(A\cap B)=0.$. Same argument if $f(A)=1$, so $f$ is constant. Thus $A\cup B$ is connected.

Tsemo Aristide
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