Connectedness and Closure

Let $ X $ be a topological space and let $ C $ be a connected subset of $ X $. If a set $ A $ contains $ C $ and is contained in the closure of $ C $, \[ C \subset A \subset \operatorname{Cl}(C) \] then $ A $ is a connected subset of $ X $.

Informally, starting from a connected set and adjoining only points that remain in contact with it, without creating gaps or separations, cannot result in a disconnected set.

Here $ C $ is already known to be connected, so it admits no internal separation. Moreover, the set $ A $ contains $ C $, so no points are removed.

Since $ A $ is contained in the closure of $ C $, it can only include points that are not isolated from $ C $, namely points whose every open neighborhood intersects $ C $.

Consequently, the connectedness of $ C $ is inherited by $ A $.

A concrete example
Proof

A concrete example

Consider the topological space $ X = \mathbb{R} $ endowed with the standard topology, and let $ C $ be an interval.

$$ C = (0,1) $$

The set $ C $ is connected in $ \mathbb{R} $, since intervals are connected subsets of the real line.

The closure of $ C $ is

\[ \operatorname{Cl}(C) = [0,1] \]

Now choose a set $ A $ such that $ C \subset A \subset \operatorname{Cl}(C) $. For example:

\[ A = (0,1] \]

In this case, $ C $ is contained in $ A $

$$ C \subset A $$

$$ (0,1) \subset (0,1] $$

Moreover, $ A $ is a subset of the closure of $ C $

$$ A \subset \operatorname{Cl}(C) $$

$$ (0,1] \subset [0,1] $$

Therefore, the set $ A = (0,1] $ is also connected in $ \mathbb{R} $.

In other words, starting from the connected set $ (0,1) $, we have adjoined a single point, namely $ 1 $, which lies in direct contact with the original set. No separation has been introduced.

For this reason, the set $ A $ remains connected in $ \mathbb{R} $.

Proof

Let $ X $ be a topological space and let $ C \subset X $ be a connected subset.

Let $ A $ be a set such that

\[ C \subset A \subset \operatorname{Cl}(C) \]

To show that $ A $ is connected in $ X $, we argue by contradiction and assume that $ A $ is not connected.

If $ A $ is not connected, then there exists a separation of $ A $. That is, there exist open sets $ U $ and $ V $ in $ X $ such that:

$ U $ and $ V $ are open subsets of $ X $
$ A \subset U \cup V $, so that $ U \cup V $ covers $ A $
$ A \cap U \neq \varnothing $ and $ A \cap V \neq \varnothing $
$ (A \cap U) \cap (A \cap V) = \varnothing $, equivalently $ A \cap U \cap V = \varnothing $, meaning that $ U $ and $ V $ do not overlap on $ A $

Now consider the set $ C $.

Since $ C \subset A $, we can write:

\[ C = (C \cap U) \cup (C \cap V) \]

Moreover:

\[ (C \cap U) \cap (C \cap V) = C \cap U \cap V \subset A \cap U \cap V = \varnothing \]

Thus $ C $ is expressed as the union of two disjoint subsets.

The sets $ C \cap U $ and $ C \cap V $ are open in $ C $ with respect to the subspace topology, since they are intersections of $ C $ with open subsets of $ X $.

Therefore, $ C $ would admit a separation given by $ C \cap U $ and $ C \cap V $, unless one of these sets were empty.

However, $ C $ is connected, so it admits no separation.

Consequently, one of the two sets must be empty:

\[ C \cap U = \varnothing \quad \text{or} \quad C \cap V = \varnothing \]

Assume, without loss of generality, that

\[ C \cap V = \varnothing \]

This implies that $ C $ is entirely contained in the open set $ U $.

\[ C \subset U \]

Since $ A \cap V \neq \varnothing $, choose a point

\[ x \in A \cap V \]

From the condition $ A \subset \operatorname{Cl}(C) $, it follows that

\[ x \in \operatorname{Cl}(C) \]

But $ x \in V $, and $ V $ is open in $ X $.

Thus $ V $ is an open neighborhood of $ x $.

Since $ V \cap C = \varnothing $, this open neighborhood of $ x $ does not intersect $ C $.

By definition, a point $ x $ belongs to the closure $ \operatorname{Cl}(C) $ if and only if every open neighborhood of $ x $ intersects $ C $.

Therefore, $ x $ cannot belong to the closure of $ C $.

\[ x \in V \ \text{open}, \ V \cap C = \varnothing \quad \Rightarrow \quad x \notin \operatorname{Cl}(C) \]

This contradicts the earlier conclusion that $ x \in \operatorname{Cl}(C) $.

Hence, the assumption that $ A $ is not connected is false. Therefore, its negation holds:

\[ A \ \text{is connected in} \ X \]

This completes the proof.

And so on.