Excluded Point Space is Locally Path-Connected

Theorem
Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.

Then $T$ is locally path-connected.

Proof
Consider the set $\mathcal B$ defined as:


 * $\mathcal B = \left\{{\left\{{x}\right\}: x \in S \setminus \left\{{p}\right\}}\right\} \cup \left\{{S}\right\}$

Then $\mathcal B$ is a basis for $T$.

Let $H \in \mathcal B$.

Then $\exists x \in S: H = \left\{{x}\right\}$.

From Point is Path-Connected to Itself we have that $H$ is path-connected.

Now consider the open set $S \in \mathcal B$.

Let $a, b \in S$.

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

Let $f: I \to S$ be the mapping defined as:
 * $\forall x \in I: f \left({x}\right) = \begin{cases}

a & : x \in \left[{0. . 1}\right) \\ b & : x = 1 \end{cases}$

Then $f^{-1} \left({S}\right) = \left[{0 \,.\,.\, 1}\right]$ which is open in $I$ because $\left[{0 \,.\,.\, 1}\right] = I$.

As $a$ and $b$ are any two points of $S$ it follows that $S$ is path-connected.

So we have shown that all elements of a basis of $T$ are path-connected.

Hence by definition $T$ is locally path-connected.