Connectedness Between Two Points is an Equivalence Relation

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $a \sim b $ denote the relation:

$a \sim b \iff \exists S \subseteq T: S$ is connected between the two points $a$ and $b$

where $a, b \in X$.


Then $\sim$ is an equivalence relation.


Proof

Checking in turn each of the criteria for equivalence:


Reflexivity

Trivially, any separation of $T$ is a partition of $S$.

So any $a \in S$ will be in exactly one of the open sets forming the partition.

So $a \sim a$ and $\sim$ is seen to be reflexive.

$\Box$


Symmetry

If $a \sim b$ then each separation of $T$ includes a single open set $U \in \varnothing$ which contains both $a$ and $b$.

Trivially it follows that $b \sim a$ and so $\sim$ is seen to be symmetric.

$\Box$


Transitivity

Let $a \sim b$ and $b \sim c$.

Then by definition:

Each separation of $T$ includes a single open set $U_1 \in \varnothing$ which contains both $a$ and $b$.
Each separation of $T$ includes a single open set $U_2 \in \varnothing$ which contains both $b$ and $c$.

So $b \in U_1$ and $b \in U_2$ and so $U_1 \cap U_2 \ne \varnothing$.

So by the definition of partition it follows that $U_1 = U_2$ and so $a \sim c$.

So $\sim$ has been shown to be transitive.

$\Box$


$\sim$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\blacksquare$


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