📂Topology

What Is Connectedness in Topology

Definition ¹

In a topological space $X$, if there exist open sets $A \ne \emptyset$, $B \ne \emptyset$ satisfying $A \cap B = \emptyset$ and $A \cup B = X$, then $X$ is called a disconnected space. If $X$ is not a disconnected space, it is called a connected space.

Theorems

• [1]: Connectedness is a topological property.
• [2]: Every trivial space is a connected space.
• [3]: Every discrete space is a disconnected space.
• [4]: Every singleton set is connected.

Explanation

This is a fairly intuitive definition for expressing that something is not connected, and its negation, connectedness, can also be easily accepted. Graph theory defines connectedness in a similar way.

As an example, consider the Euclidean space $( \mathbb{R} , d )$. No matter which open interval we consider, it fails to satisfy the condition for being disconnected, so it is a connected space. On the other hand, consider its subspace $( \mathbb{Q}, d )$. Since $( \mathbb{Q} , d ) = ( \mathbb{Q} , \mathscr{P} ( \mathbb{Q} ) )$ is a discrete space, it can easily be shown to be a disconnected space.

Proof

[1]

Suppose there exists a homeomorphism $f : X \to Y$ and that $X$ is a connected space. The proof is complete once we show that $Y$ is a connected space.

Assume that $Y$ is a disconnected space. Then there exist open sets $A, B \subset Y$ satisfying $$ A \cap B = \emptyset \\ A \cup B = Y $$

If $f$ is a continuous function, then for every open set $V \subset Y$, $f^{-1} (V)$ is an open set in $X$.

Since $Y$ is a continuous function, $f^{-1} (A)$ and $f^{-1} (B)$ are open sets in $X$. However, $$ f^{-1} (A) \cap f^{-1} (B) = f^{-1} (A \cap B) = f^{-1} ( \emptyset ) = \emptyset \\ f^{-1} (A) \cup f^{-1} (B) = f^{-1} (A \cup B) = f^{-1} ( Y ) = X $$ holds. In the end, $X$ is a disconnected space, which contradicts the premise.

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[2]

Since the topology $\mathscr{T} = \left\{ \emptyset , X \right\}$ of a trivial space $X$ does not contain two nonempty open sets, $X$ is a connected space.

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[3]

If $X$ has only one element, it is a trivial space before it is a discrete space, so we should assume that $X$ has two or more elements. In a discrete space $X$, for every nonempty open set $U$, $V = X \setminus U$ is an open set in $X$, so it is a disconnected space.

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[4]

For $A , B \subset \left\{ x \right\}$ to satisfy $A \cap B = \emptyset$, either $A$ or $B$ must necessarily be the empty set, so it cannot be a disconnected space.

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