Topological Space with Generic Point is Path-Connected
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $T$ have a generic point $g \in S$.
Then $T$ is path-connected.
Proof
By Path-Connectedness is Equivalence Relation, it suffices to prove that every point is path-connected with $g$.
Let $x \in S$.
Define a path $\gamma: \closedint 0 1 \to S$ by:
- $\map \gamma t = \begin{cases}
x & : t \le \dfrac 1 2 \\ g & : t > \dfrac 1 2 \end{cases}$
We show that $\gamma$ is indeed continuous.
Let $U \subseteq S$ be open and non-empty.
Because $g$ is a generic point, $g \in U$.
If $x \in U$, then its preimage $\gamma^{-1} \sqbrk U = \closedint 0 1$ is open.
If $x \notin U$, then its preimage $\gamma^{-1} \sqbrk U = \hointl {\dfrac 1 2} 1$ is also open.
Thus $x$ and $g$ are path-connected.
$\blacksquare$