Can path connectedness be defined without using the unit interval or more generally the real numbers?
I.e., do we need Dedekind cuts or Cauchy convergence equivalence classes of the rational numbers (metric space completion) in order to define any object topologically equivalent to the unit interval?
Compared to the definition of connectedness, which only uses open and closed sets, having to use the unit interval to define path connectedness seems somewhat like using a sledgehammer.
I suspect that the answer might be no, since for every Hausdorff path connected space, the paths are homeomorphic to the unit interval (at least according to the relevant Wikipedia article). In particular, every locally path-connected Hausdorff space has a bunch of 1-manifolds as subsets.
Still it is unclear to me, since it seems like it should be able to specify all of the unit interval's topological properties without having to recourse to its analytic definition.
Your thoughts or help would both be greatly appreciated.
EDIT: this question probably has something to do with homotopy theory: https://en.wikipedia.org/wiki/Homotopy, with which I am rudimentarily familiar at best.
There are really two separate questions here: can you define the unit interval space without talking about real numbers, and can you define path-connectedness without talking about the unit interval space? The answer to both is yes; let me address the second question first.
Let $P$ be a topological space and let $a,b\in P$ be two points. Say that a space $X$ is $(P,a,b)$-connected if for any $x,y\in X$, there is a continuous map $f:P\to X$ such that $f(a)=x$ and $f(b)=y$. Of course, for $(P,a,b)=([0,1],0,1)$, this is just the usual definition of path-connectedness.
However, there is a more "universal" characterization of path-connectedness that doesn't require you to know about the space $[0,1]$. Namely, a space $X$ is path-connected iff it is $(P,a,b)$-connected for all compact Hausdorff spaces $P$ with two distinct points $a,b\in P$.
To prove this, suppose $X$ is path-connected, $P$ is a compact Hausdorff space, $x,y\in X$, and $a,b\in P$ are distinct. Since $X$ is path-connected, there is a path $g:[0,1]\to X$ such that $g(0)=x$ and $g(1)=y$. By Urysohn's lemma, there is a continuous map $h:P\to [0,1]$ such that $h(a)=0$ and $h(b)=1$. The composition $gh:P\to X$ is then continuous and satisfies $g(a)=x$ and $g(b)=y$.
The idea here is that you could use any space $P$ with two chosen points $a$ and $b$ to define a notion of "paths" in a space. However, if you restrict to compact Hausdorff spaces $P$, then the ordinary interval $[0,1]$ is the "strongest possible kind of path" you can have in a space: if you have a $[0,1]$-path between two points, then you have a $P$-path for every other compact Hausdorff space $P$ as well.
(Of course, all we used about compact Hausdorff is that we know there is a map $P\to [0,1]$ separating $a$ and $b$. However, I phrased everything in terms of the compact Hausdorff condition since this is a natural condition you can define without already knowing about the space $[0,1]$.)
OK, now let me say a little about the first question. There are in fact many different ways to uniquely characterize the space $[0,1]$ up to homeomorphism without reference to the reals or anything that is essentially equivalent to constructing the reals. In fact, you can deduce one from the answer I gave to the second question above.
Namely, say that a compact Hausdorff space $(P,a,b)$ equipped with two distinct points is a universal path if it has the special property of $[0,1]$ noted above: whenever there is a $(P,a,b)$-path between two points $x$ and $y$ of an arbitrary space, there is also a $(Q,c,d)$-path from $x$ to $y$ for any compact Hausdorff space $Q$ with two distinct points. Say that a universal path $(P,a,b)$ is minimal if for any other universal path $(Q,c,d)$, there is an embedding $P\to Q$ sending $a$ to $c$ and $b$ to $d$.
I now claim that $([0,1],0,1)$ is the unique minimal universal path (up to homeomorphism). We know it is universal. To show that it is minimal, let $(Q,c,d)$ be any universal path. Since the identity map $Q\to Q$ is a $(Q,c,d)$-path from $c$ to $d$ in $Q$, universality implies there is a $([0,1],0,1)$-path from $c$ to $d$ in $Q$. But if there is a path between two points of a Hausdorff space, there is also a path which is an embedding (see Does path-connected imply simple path-connected?). Thus there is an embedding $[0,1]\to Q$ sending $0$ to $c$ and $1$ to $d$.
Now suppose $(P,a,b)$ is any minimal universal path. The previous paragraph shows that there is a $([0,1],0,1)$-path from $a$ to $b$ in $P$. Now since $([0,1],0,1)$ is a universal path, minimality of $P$ says that $P$ embeds in $[0,1]$ sending $a$ to $0$ and $b$ to $1$. But since $P$ contains a path from $a$ to $b$, the image of this embedding contains a path from $0$ to $1$, and thus contains all of $[0,1]$. Thus the embedding is actually a homeomorphism $P\to [0,1]$.
As I mentioned, this is just one of many ways of characterizing $[0,1]$. For another characterization that also relates closely to the intuitive notion of "paths", see this answer by Tom Leinster on MO.