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.

  • 2
    $\begingroup$ Well something's wrong because $A\cap B$ need not be connected. $\endgroup$ – Matt Samuel Feb 3 '16 at 1:49
  • $\begingroup$ @MattSamuel and furthermore the last line of the proof basically assumes the theorem. $\endgroup$ – Forever Mozart Feb 3 '16 at 2:00

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.

  • $\begingroup$ How to read the notation $\{0\} \times (\mathbb{R} \backslash \{0\})$? $\endgroup$ – sequence Feb 3 '16 at 2:15
  • $\begingroup$ 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. $\endgroup$ – sequence Feb 3 '16 at 2:23
  • $\begingroup$ Also, why would it be necessary to show that $U\cup V = A$ or $U\cup V = B$, and not that $U\cup V = A\cap B$? $\endgroup$ – sequence Feb 3 '16 at 2:27
  • $\begingroup$ What do you mean by "bounded away from each other"? $\endgroup$ – sequence Feb 3 '16 at 2:29
  • $\begingroup$ Also, can you please clarify what is exactly circular in my proof? $\endgroup$ – sequence Feb 3 '16 at 2:30

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.

  • $\begingroup$ perfectly acceptable answer $\endgroup$ – Forever Mozart Feb 3 '16 at 1:52
  • 1
    $\begingroup$ This completely ignores the actual question and only answers the title. $\endgroup$ – Matt Samuel Feb 3 '16 at 1:53
  • $\begingroup$ Sorry, I didn't understand your proof at all. Can you please clarify? If $f(A) = 0$ then why is $f(A\cap B)=0$? Also, what is the restriction of $f$ to $B$? $\endgroup$ – sequence Feb 3 '16 at 2:34
  • $\begingroup$ Since $A\cap B\subset A$ if $f$ is constant on $A$ so is the restriction of $f$ on $A\cap B$. $\endgroup$ – Tsemo Aristide Feb 3 '16 at 2:36

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.