This question might have been answered before but I need different proof using the following fact:

Suppose $C$ is a connected set in a metric space (M,d). Let $A$ and $B$ be two separated sets in $(M,d)$, i.e., $A \cap \bar{B} = \emptyset = A \cap \bar{B} $ such that $C \subseteq A \cup B$. Show that $C \subseteq A$ or $C \subseteq B$.

We want to show if $C$ is a connected set then closure of $C$, i.e. $\bar{C}$ is connected.

Claim : $\bar{C}$ is connected

To show the claim we assume that $\bar{C}$ is disconnected. If we do so, we need to reach a contradiction which is possibly would be $C$ is disconnected.


Suppose $C$ is a connected set and $\bar{C}$ is not connected.

Since $\bar{C}$ is a disconnected set, there exist two non-empty sets $A$ and $B$ which satisfy the following:

1- $A \ne \emptyset$ and $B \ne \emptyset$

2- $A \cap \bar{B} = \emptyset = A \cap \bar{B} $

3- $\bar{C}= A \cup B$

Using the fact $C \subseteq \bar{C}$ $\rightarrow C \subseteq A \cup B$.

Now using the fact that $A$ and $B$ are two separated sets and $ C \subseteq A \cup B$, we can use the very first fact so

$C \subseteq A$ or $C \subseteq B$

WLOG, asume that $C \subseteq A$ so $C \cap B = \emptyset $

From this point how can we prove $C$ is disconnected to reach to contradiction to the first assumption.

My guess is we need two non-empty sets $U$ and $V$ which have the following criteria:

1- $U \ne \emptyset$ and $V \ne \emptyset$

2- $U \cap \bar{V} = \emptyset = U \cap \bar{V} $

3- $C= U \cup V$

But what are they?

  • $\begingroup$ The title seems to have a little mistake. You mean connected, not closed. $\endgroup$
    – Dog_69
    Oct 23 '18 at 21:00
  • $\begingroup$ Since $B$ is open you can coclude that $\bar{C}\subset A$ $\endgroup$
    – ALG
    Oct 23 '18 at 21:06
  • $\begingroup$ We don't prove $C$ is disconnected at all, we use connectedness of $C$ to prove one of $A$ or $B$ to be empty which is a direct contradiction. $\endgroup$ Oct 23 '18 at 21:35

In this answer I show that $A$ and $B$ are both open (and closed) in the subspace $A \cup B$. Now if $C \subseteq A \cup B$, it's immediately clear that $A \cap C$ and $B \cap C$ cannot both be non-empty, or else these two sets would form a disconnection of $C$ by definition. So one intersection is empty and $C$ is contained in the other.

This reproves the fact.

Following your proof outline, you suppose that $\overline{C} = A \cup B$ with $A$ and $B$ completely separated in $\overline{C}$ (!), not $X$ (connectedness is an intrinsic property). Condition 2 has closures taken in $\overline{C}$.

Then $A$ and $B$ are disjoint and closed in $\overline{C}$ and so disjoint and closed in $X$ as well, so separated. Apply the fact to conclude that (say) $C \subseteq A$ and thus $\overline{C} \subseteq \overline{A} = A$ and this implies $B= \emptyset$, a contradiction. Done.

The structure of the proof is thus : assume $\overline{C}$ disconnected using some disconnection, then use that $C$ is connected to see that the disconnection couldn't be a real disconnection. The connectedness of $C$ use is "hidden" in the lemma, so the final proof is almost nothing.

We can slightly generalise the result by stating it as (which is the proposition I have been taught, way back in "metrische topologie"):

Let $C$ be connected and let $D$ be such that $C \subseteq D \subseteq \overline{C}$. Then $D$ is also connected.

which has almost the same proof.

  • $\begingroup$ @ Henno: I do not want to proceed with the implication of $A$ and $B$ are both open (an closed) in $\bar{C}$, I want direct proof based on my assumptions, could you help me to do that? $\endgroup$
    – Saeed
    Oct 23 '18 at 23:22
  • $\begingroup$ @Saeed I rephrased the proof. Check it out. $\endgroup$ Oct 24 '18 at 6:33

Assume that $C$ is connected, but that $\overline{C}$ is disconnected.

Hence $C\,\not =\,\overline{C}.\,\,$ Let us write $$\overline{C} = A \,\,\, {\veebar}_s \,\, B$$

to mean that $\overline{C}$ is the indicated union of separated (hence disjoint) sets $A$ and $B$.

Hence we assume that $\overline{C} = A \,\,\, {\veebar}_s \,\, B$ and $A \neq \emptyset$ and $B \neq \emptyset$, seeking a contradiction.

The lemma is

Suppose $C$ is a connected set in a metric space. Let $C \subset A \,\,\, {\veebar}_s \,\, B,\,\,$ meaning that $A \cap \overline{B} = \emptyset = A \cap \overline{B} $ and that $C \subset A \cup B$. Then either $C \subset A$ or $C \subset B$ must hold.

Because $C \,\subset\, A \,\,\, {\veebar}_s \,\, B,$ we apply the given lemma to say that $C \subset A$ or $C \subset B.$ Without loss say $C \subset A$, and pass to the derived sets: $$C' \subset A'.$$

Now observe, there must be at least one limit point $p$ for $C$ to be found inside $B$, since $B \neq \emptyset,$ $C \subset A$ and $C\,\cup\,C' = A \,\,\, {\veebar}_s \,\, B.\,$ This $p$ is also a limit point for $A$.

But we know $\overline{A} \cap B$ is empty. We reach a contradiction.


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