connected components equivalence relation

Question

Prove that the relation $x \sim y$ iff $y$ is an element of the connected component of $x$ is an equivalence relation.

This question is confusing me, do I simply go about showing the relation is reflexive, symmetric, and transitive? I don't really see how to do this for this question. Any suggestions or hints are appreciated!

Answer 1

HINT: Perhaps the easiest way to prove that this is an equivalence relation is to show that the connected components of a space $X$ partition $X$: they are pairwise disjoint, and their union is $X$. It will then follow immediately from the relationship between equivalence relations and partitions that $x\sim y$ is an equivalence relation: it’s the equivalence relation induced on $X$ by the partition of $X$ into connected components.

If you need a review of the relationship between equivalence relations and partitions, I discussed it at some length in this answer.

Answer 2

Yes. Whenever you are asked to verify that a certain mathematical object (in this case a relation) has a specific property (in this case, an equivalence relation), you really just need to verify that the definition of this property hold for this object.

In this case, you simply need to show that the relation is reflexive, symmetric and transitive.

Of course, sometimes it is diffcult to verify the definitions, or it might be long and tedious. And there are sometimes ways to go around this issue. For example, you might know that if the object (or a related object) have a certain property, you could conclude what you want. For example, sometimes showing that something is connected is complicated, and it's easier to show that it is path connected which would imply this.

As Brian suggests in his answer, if you show that the collection of connected component is a partition then you can conclude that the relation you are after is an equivalence relation. This is because equivalence relations and partitions are strongly connected mathematically.

Answer 3

So define $x \sim y$ iff $y$ is in the connected component of $x$, where the connected component of $x$ is the largest connected subset of $X$ that contains $x$.

To see this is well-defined, we can define for fixed $x$ the collection $\mathcal{C}_x = \left\{C \subset X \mid C \mbox { connected and } x \in X \right\}$, and note that any 2 members of $\mathcal{C}_x$ intersect (in $x$) and so their union $C(x) = \cup \mathcal{C_x}$ is also connected, by a standard theorem on connected subsets, and clearly $C \subset C(x)$ for all $C \in \mathcal{C}_x$, as $C$ is one of the sets we take the union of to construct $C(x)$, so $C(x)$ is the component of $x$ (by construction it is the maximal connected subset that contains $x$).

So $x \sim y$ iff $y \in C(x)$.

Clearly for all $x$, $x \sim x$, as $x$ is in $C(x)$ by definition.

Suppose $y \sim x$ so $y \in C(x)$. Then $C(x)$ is a connected subset of $X$ that contains $y$, and $C(y)$ is the maximal such subset, so $(x \in) C(x) \subset C(y)$, which means that $x \sim y$. This shows symmetry. Note that by reversing the roles of $x$ and $y$ we also get $C(y) \subset C(x)$ and so $x \sim y$ iff $C(x) = C(y)$ as sets.

The last reformulation clearly shows transitivity as well: $x \sim y$ and $y \sim z$, then $C(x) = C(y) = C(z)$, so $x \sim z$.

This shows that thus is indeed an equivalence relation.

Answer 4

Let's call our space $S$. For any $x\in S,$ we can define $C_x$, the connected component of $x$, to be the $\subseteq$-greatest connected subset $A$ of $S$ such that $x\in A.$ Put another way, $C_x$ is the union of all connected subsets of $S$ containing $x$ as an element--we can show that such a union is connected in various ways.

Then $x\sim y$ if and only if $y\in C_x$. Clearly, $x\in C_x$, so $x\sim x$.

If $x\sim y$, then $C_x$ is a connected set containing $y$, so $C_x\subseteq C_y$ (since $C_y$ is the $\subseteq$-greatest subset of $S$ containing $y$), and so $x\in C_y$, meaning $y\sim x$.

If $x\sim y$ and $y\sim z$, then $y\in C_x$ and $z\in C_y$. As with the symmetry proof, we then have $C_x\subseteq C_y$ and $C_y\subseteq C_z$, so it follows that $x\in C_z$. But then $C_z$ is a connected set containing $x$, so $C_z\subseteq C_x$, and so $z\in C_x$, meaning $x\sim z$.