I have a finite topological space $X= \{ 1,...n \}$ with the following topology that I created

$T=\{ \emptyset, X \} \bigcup \{A \subseteq X | 1\in A\} $

It is connected because $\bigcap_{A \in X, A \mbox{ } open} A=\{1\}$ So it's impossible find two open sets with $A \cup B=X$ such that $A \cap B= \emptyset$

Is it a path-connected space?

By definition we have to find (for each $p,q \in X$) a continuous function $f:[0,1] \longrightarrow X$ such that $f(0)=p$ and $f(1)=q$

Is it possible? And which topology has $[0,1]$? The Euclidian one?


How about this path:

$f([0,1/3])=p$, $f((1/3,2/3))=1$, $f([2/3,1])=q$

This should work when both $p$ and $q$ are not $1$ and you can tweak it a little bit for when it's 1 too.

  • 1
    $\begingroup$ Actually, the exact same map works when $p$ or $q$ or both of them are equal to $1$... $\endgroup$ Mar 29 '15 at 23:52

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