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I would like feedback on the following proof I just wrote up:

Let $C$ be a connected subset of the metric space $X, d$ and let $C \subset B \subset \bar C$. Prove that $B$ is connected.

Let $B, C$ be sets in a metric space $X, d$ such that $C \subset B \subset \bar C$ and $C$ is connected. To show $B$ is connected we must show that there exist two sets $B_1, B_2 : B_1 \cup B_2 =\bar B_1 \cap B_2 \neq \emptyset \land \bar B_2 \cap B_1 \neq \emptyset$.

Because $C$ is connected $C$ can be written as the union of two non-empty non-separated sets $C_1, C_2 : \bar C_1 \cap C_2 \neq \emptyset \land \bar C_2 \cap C_1 \neq \emptyset$. But then $B$ contains $\bigcup_{i = 1}^2 C_i$ (call this union $D$).

Suppose there exists a $B_1 \subset B$ such that $\bar D \cap B_1 = \emptyset$. Then by DeMorgan's law $(\bar C_1 \cap B_1) \cup (\bar C_2 \cap B_1) = \emptyset$. This is impossible because $\bar C$ contains $B$, so the intersection is necessarily resultant to $B$. Therefore any other subsets of $B$ are necessarily connected to $D$.

Now assume that there exists a $B_1$ such that $D \cap \bar B_1 = \emptyset$. Then $D \cap B = \emptyset$. But $D \subset B$ and as shown before this is impossible. From the contradiction it must be true that $B$ is connected and is also the union of two non-empty non-separated arbitrary sets. RAA.

Don Larynx
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1 Answers1

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I think you have the right idea! In your proof you say, "assume that there exists a $B_1$ such that $D\cap B_1=\varnothing.$ Then $D \cap B = \varnothing$. Is this the same $B_1$ as before? How do you draw this conclusion?

Perhaps, if you want to draw a contradiction, assume that $B_1, B_2 \subset B$ are open, disjoint, and non-empty such that $B_1 \cup B_2 = B$. Now, since $C$ is connected, either $B_1 \cap C = \varnothing$ or $B_2 \cap C = \varnothing$ (why!?). Without losing generality, suppose that $B_1 \cap C = \varnothing$. Then $C \subset B_1^c$. However, $B_1^c$ is a closed set containing $C$ and therefore $\overline{C} \subset B_1^c$ meaning that $B_1 \not\subset \overline{C}$. However, this creates a contradiction (why!?).

Tom
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