Point is Path-Connected to Itself
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Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $a \in S$.
Then $a$ is path-connected to itself.
Proof
Consider the constant mapping on the closed unit interval $\mathbb I = \left[{0 \,.\,.\, 1}\right]$:
- $\forall x \in \mathbb I: f_a \left({x}\right) = a$
Thus, in particular:
- $f_a \left({0}\right) = a$
- $f_a \left({1}\right) = a$
As a Constant Mapping is Continuous, it follows that $f_a$ is a path in $X$.
Thus $a$ is path-connected to itself.
$\blacksquare$