# 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.

$\blacksquare$