Show that the closure of a connected set is connected

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.

Proof:

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?

• The title seems to have a little mistake. You mean connected, not closed. Oct 23 '18 at 21:00
• Since $B$ is open you can coclude that $\bar{C}\subset A$
– ALG
Oct 23 '18 at 21:06
• 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. 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.

• @ 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? Oct 23 '18 at 23:22
• @Saeed I rephrased the proof. Check it out. 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.