Relationship between Component Types

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $p \in S$.

Let:

$A$ be the arc component of $p$

$P$ be the path component of $p$

$C$ be the component of $p$

$Q$ be the quasicomponent of $p$.

Then:

$A \subseteq P \subseteq C \subseteq Q$

In general, the inclusions do not hold in the other direction.

Proof

Let $f \in A$.

By Arc in Topological Space is Path we have that $f \in P$.

That is, $A \subseteq P$.

$\Box$

---

Let $f \in P$.

From Path-Connected Space is Connected we have directly that $P \subseteq C$.

$\Box$

---

Let $f \in C$.

From Connected Space is Connected Between Two Points we have directly that $C \subseteq Q$.

$\Box$

Hence the result.

$\blacksquare$

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Also see

• Path Component is not necessarily Arc Component
• Component is not necessarily Path Component
• Quasicomponent is not necessarily Component

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness