Components, Partitions, and Separations

Components and Partitions

1. Partitions
Let $X$ be a space. A partition of $X$ is the collection of disjoint subspaces of $X$ such that the union equals $X$. 

Example: $[0,1]$ and $(1,2]$ are partitions of $[0,2]$.

Example: $[0,1]$ and $[1,2]$ are not partitions because they are not disjoint.

2. Components
Components are equivalence classes by defining the equivalence relation $\sim$ so that $x\sim y$ if there exists a connected subspace of $X$ such that $x,y\in X$.

Example: $[0,1]$ and $[2,3]$ are the components of $[0,1]\cup [2,3]$.

Note that there can be only one way of constructing components of a space. Components satisfy the following characteristics.

i. Partition : Components are partitions, meaning they are disjoint and the union becomes $X$.
ii. Connected : If a component is separated, picking each element (say $x, y$) of the separation makes $x\not\sim y$.
iii. Maximality : If there is a larger connected subspace that contains $A$, $A$ cannot be a component.


Example: $[0,1]$ and $(1,2]$ are not the components of $[0,2]$ because $[0,2]$ is connected, a larger subspace of $[0,2]$ that contains $[0,1]$ and $(1,2]$. Indeed the components of $[0,2]$ is only $[0,2]$.


Example: $\{0\}\cup\{1\}$ and $(0,1)$ are not the components of $[0,1]$ because $\{0\}\cup\{1\}$ is not connected.

3. Separation

Recall that a separation of a space $X$, say $A$ and $B$ are clopen partitions. This does NOT make a separation components, because a separation is not necessarily connected.

 

4. Closedness of Components

Components are always closed. 

 

Proof.

If $A$ is a component of $X$ not closed, $\bar{A}$ is a larger connected subspace of $X$ that contains $A$, which contradicts the maximality of $A$. So we can add the 4th characteristic of components.

iv. Closedness : Each component is closed.

5. Components in a locally connected space

By Theorem 25.3 (Munkres), we know that all components are open if and only if $X$ is locally connected. 

6. The if and only if condition

We have found out the four characteristics of components. Now we can question: if each partition is connected and closed, are they the components? (Think about it.)

The answer is no. Finding a counterexample was quite hard, but here is the road to finding one.

Now question this: if each partition is connected and OPEN, are they the components?
Let $A_i$ be a partition. Then the union of all other partitions, $B:=\bigcup_{\alpha\not=i}{A_\alpha}$ open. Then $A_i$ and $B$ is a separation of $X$. Assume for contradiction that $A_i$ is not a component. Then for a $x_0\in A_i$, there exists $y\in B$ such that $x_0\sim y$. But a connected space containing $x_0$ should be contained in $A_i$, so it cannot contain $y$, a contradiction.

See that $A_i$ is closed because $B$, an arbitrary union of open sets, is open. We can conclude if partitions are open, they are also closed. But does the inverse also hold? No, because the union of closed sets are not necessarily closed if there are infinte of them. 

If partitions are connected and open, they are components. So let us find partitions that are closed and connected, but not open. This means that the number of partitions must be infinite.

Finally, the a counterexample we've been looking for: $\mathbb{R}$ can be partitioned into singletons (i.e. $ \{ \{ x\}:x\in \mathbb{R} \}$) which are connected and closed. But they are not maximal, so they cannot be components.

7. Conclusion : Omitting Maximality

Throughout the steps so far, we can now discuss components without the maximality characteristic. 

This is important because proving the maximality of a given partition is usually difficult.

1. Component $\implies$ Partition + Connected + Closed 

True.

2. Partition + Connected + Closed $\implies$ Component

Not necessarily. But if there are finite partitions, true.

3. Component $\implies$ Partition + Connected + Open

Not necessarily. This holds if and only if the space is locally connected.

Also holds if there are finite components.

4. Partition + Connected + Open $\implies$ Component

True.

 

이 내용은 Munkres 책 Lemma 61.1에 이용이 되지만, 별다른 증명 없이 사용되었기 때문에 증명을 적어보았다.