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I came up with two weak incomplete solutions to the problem.

If $E$ and $F$ are connected subsets of a metric space $M$ with $E \cap F \neq \emptyset$. Show $E \cup F$ is connected.

failed proof1:

If $E \cup F$ is not connected, let $W$ and $V$ be disjoint nonempty open sets that separate $E \cup F = W \cup V$. So if $x \in E \cap F \subset E\cup F = W \cup V$. First neither $W, V$ can equal either $E$ or $F$, if it does we have the first contradiction. Notice the situation $W \subset E$ would mean that $E \cup F = V$, so the sets $W$ and $V$ would not be empty. So we are left with (without any loss of generality), $E \subset W$, then $E \not\subset V.$

if $x \in W, x \notin V$. The sets $V$ and $W$ are open relative to $E \cup F$. So we must have,

$$W = (E \cup F) \cap O_m^1,$$ $$V = (E \cup F) \cap O_m^2$$ for open sets $O_m^1, O_m^2 \in M$.

So $\emptyset = E \cap V = E \cap [(E \cup F) \cap O_m^2 ] = E \cap F \cap O_m^2 \neq \emptyset$, contradiction.

faled proof2:

Construct a continuous function (this is automatically surjective) $f: M \to M$. Assuming $E\cup F$ is disconnected, the image $f(E \cup F) = f(E) \cup f(F)$. I got nowhere with this.

Lemon
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2 Answers2

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I think you are closer with your first attempt. In my experience, proofs regarding connectedness are often done with contradiction. I agree with where you are going with the proof up to where you state "$E \cup F = W \cup V$".After this point, I feel that you introduce some unnecessary cases. One nice result to apply here instead is that:

If $C$ is a connected subset and $C \subset A \cup B$ where $A \cap B = \emptyset$ then either $C \subset A$ or $C\subset B$.

Applying that to what you have, it is obvious that $E \subset W \cup V$ so (without loss of generality) we know $E \subset W$ and $E \cap V = \emptyset$. Similarly $F \subset W \cup V$. $F$ is also connected, so $F \subset W$ or $F \subset V$. We can now introduce cases.

Case 1: $F \subset W$ and $F \cap V = \emptyset$

Then $E\cup F = W$ so $E\cup F = W \cup V$ implies that $V = \emptyset$ which is a contradiction as you already noted.

Case 2: $F \subset V$ and $F \cap W = \emptyset$

Then $$(E \cap F) \subset (W \cap V) = \emptyset$$ so $E \cap F = \emptyset$ which is also a contradiction. We now conclude that $E \cup F$ is connected.

graydad
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  • so what did I do wrong? Which cases are not relevant that I introduced? Isn't your case (2) essentially the same as (1)? – Lemon Oct 05 '14 at 20:03
  • I did not say they are not relevant, just that you don't need to make as many cases to complete the proof. In my opinion, the cases of "if $x \in E\cap F$", "if $x \in W,x\notin V$" and $W \subset E$ are unnecessary distinctions to make. I also feel this way where you note that $V$ and $W$ are open relative to $E \cup F$ and introducing the open sets $O_1^m,O_2^m$. Having so many scenarios makes the proof difficult to read is all. Not saying it isn't right, just that it can be streamlined. How are my two cases essentially the same? – graydad Oct 05 '14 at 20:11
  • How do you justify $E \cap F \subset W \cap V$ – Lemon Oct 06 '14 at 00:53
  • In that second case we have $E \subset W$ and $F \subset V$. Hence $$W \cap V = \left[E \cup(W \backslash E) \right]\bigcap \left[F \cup(V \backslash F) \right] \supseteq E \cap F$$ – graydad Oct 06 '14 at 01:04
  • Can't you say $E \cap F \subset V \cap E = \emptyset$ instead? – Lemon Oct 06 '14 at 02:55
  • Yup! That works too. – graydad Oct 06 '14 at 03:38
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If $E\cup F=V\cup W$ for $V,W$ nonempty, open and disjoint, then $V\cap E$ and $W\cap E$ splits $E$ into two open disjoint parts, so either $V\supseteq E$ or $V\subseteq F\setminus E$. Similarly with $V,F$ resp. $W,E$ and $W,F$. All these cases lead to contradictions: if, for example, $V\supseteq E$, then $W\subseteq F\setminus E$, so it is not true that $W\supseteq F$, so it must be $W\subseteq E\setminus F$, hence $W=\emptyset$.

Peter Franek
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