Searching on the internet I found the following result: Every Hausdorff space that is path-connected is also arcwise-connected (for every two points of the space there's a path between them that is also an embedding).
I used this result to prove that in the long line the last element isn't path-connected to the rest of the space (I don't know if the long line including the last element is called another way, but I'll just call it long line. Just for clarification, the line has a least element and a last element, it is a line extending only in one direction)
The problem is that the result above is too general and from what I've seen proving it is complicated and requires delicate steps. Is there a way of proving on an easier way, using the properties of the long line, that if this specific space were to be path-connected, then it would have to be arcwise-connected?
I thought of considering a path ending on the last element of the long line, $\omega$, and starting on a distinct element of the space, and then modifying the path in a way that achieves injectivity. If i'm not wrong, because the space is locally homeomorphic to the real numbers with the usual topology, every point strictly between the initial and the ending points of the path is an image of the path. If one could achieve injectivity on those points, then $\omega$ would be forced to only be achieved by the path at the end of it.
If one could in some way eliminate the parts of the path that don't go in the direction of $\omega$ while preserving the continuity, monotonicity of the path may be achieved.
Am I right? if I am, can you help me providing a slightly more formalized version of this ideas? and if I'm wrong, is there a similar way of proceeding?
Note: Just in case, the idea I have for proving that the long line isn't path-connected is to assume there's a path between a point $x$ and $\omega$ (say the path is given by the function $p$, then $p(0) = x$ and $p(1) = \omega$). Then, there has to be a path between those points that is also an embedding, thus only reaching the last element of the long line at the end of the path. Then, by considering a sequence on [0,1] that converges to 1 and by the continuity of the path there will be a sequence of elements distinct from $\omega$ on the space that converges to $\omega$, and knowing that this isn't possible finishes the proof. I don't know if there's a simpler way of proving $\omega$ isn't path-connected to the rest of the space but I like this proof :)
ok so I already figured out it was way easier, i'll just leave the answer here just in case someone needs it (I'm new to the platform, so I don't quite understand all the rules but it makes sense for me to just leave the answer here).
the set $\{ \omega \} $ is closed because it's complement is by definition an open set. Then it's preimage under the path, $p^{-1} (\{ \omega \} ) $ is a closed set on [0,1]. That means the infimum of $p^{-1} (\{ \omega \} ) $ is a minimum of the set, meaning there's a minimum element $t$ such that $p(t) = \omega $. Then, it is enough to restrict p to the set [0,t] to have a path that only reaches $\omega$ at the end, thus being able to proceed with the proof by contradiction (again, by taking a sequence on [0,1] convergent to 1 and thus having a sequence on the long line minus $\omega$ convergent to $\omega$, which is not possible.)