The function defining path-connectedness To consider two points to be path connected in some topological space, there has to be a continuous function from the interval $[0, 1]$ to the path in the topological space.
Using the definition of continuity where the pre-image of any open set in the range has to be an open set in the domain begs some questions:


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*What topology is used for the path in the range?
My guess would be the topology inherited from the overall topology.

*What topology is used for the domain $[0, 1]$?

 A: The path is normally defined to be continuous from $[0,1]$ to the space that contains the two points. If you look at the function from $[0,1]$ to its range, then this is equivalent to being continuous where the topology for the range is the subspace topology.
On $[0,1]$ the topology is that induced as a subspace of $\mathbb{R}$. This is, open sets are arbitrary unions of sets of the form $(a,b)\cap [0,1]$, where $a,b\in\mathbb{R}\cup\{-\infty,+\infty\}$.
A: The topology on $[0,1]$ can be seen in at least three equivalent ways:


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*The subspace topology as a subset of $\Bbb R$ in the Euclidean topology (induced by the metric $(x,y) \to |x-y|$, or by the order topology: all unions of open intervals $(a,b)$). So an Euclidean open set intersected with $[0,1]$.

*As a metric space in its own right, the restricted metric $(x,y) \to |x-y|$, so open sets are unions of metric balls, etc.

*As an ordered space in its own right: with as a base all sets of the form $[0,a), 0<a\le 1$, $(a,1], 0\le a < 1$ and $(a,b)$, $a,b \in [0,1]$.
This all yields the same topology which makes $[0,1]$ connected and compact (connectedness being the most important here, as then path-connectedness becomes a stronger version of connectedness..)
For path-connectedness we just need a continuous function from $[0,1]$ (in this topology) to $X$, the space we are considering for path-connectedness (which presumably already has a topolgoy of its own). We can restrict the co-domain to $f[[0,1]]$ but for continuity that doesn't matter, when $f[[0,1]]$ gets the subspace topology from $X$ (as is standard).
