Particular Point Space is Path-Connected

Theorem
Let $T = \struct {S, \tau_p}$ be a particular point space.

Then $T$ is path-connected.

Proof
Let $q \in S$.

Let $\mathbb I$ be the closed unit interval in $\R$.

Let $f: \mathbb I \to S$ be the mapping defined as:
 * $\forall x \in \mathbb I: \map f x = \begin{cases}

p & : x \in \hointr 0 1 \\ q & : x = 1 \end{cases}$

Let $U \in \tau_p$.

Then by definition of particular point space, $p \in U$

Either $q \in U$ or $q \notin U$.

If $q \in U$ then $f^{-1} \sqbrk U = \closedint 0 1$ which is open in $\mathbb I$ because $\closedint 0 1 = \mathbb I$.

If $q \notin U$ then $f^{-1} \sqbrk U = \hointr 0 1$ which is half open in $\R$ but open in $\mathbb I$.

So $f: \mathbb I \to S$ is a continuous mapping and so a path from $p$ to $q$.

As $q$ is any point in $S$, it follows from Path-Connected iff Path-Connected to Point that $T$ is path-connected.