I am trying to show that the divisor topology is path-connected. This is the set $X=\{2,3,4,\ldots\}$ where the open sets are the set of unions of the sets $U_n = \{d \in X : d |n\}$.
So for any $N,M \in X$ I need to find a continuous function $f:[0,1] \to X$ such that $f(0)=N$ and $f(1)=M$. Hmmm, so I have thought maybe of splitting $[0,1]$ up and having a kind of step function between $N$ and $M$. I just need to make sure it is continuous at the "jumps". I am a bit stuck here.
Let $m,n\in X$, with $m\ne n$.
Define $f:[0,1]\to X$ by $$ f(x)= \begin{cases} m&\;\;\text{if}\;x\in[0,\frac{1}{3})\\[4pt] mn&\;\;\text{if}\;x\in[\frac{1}{3},\frac{2}{3}]\\[4pt] n&\;\;\text{if}\;x\in(\frac{2}{3},1] \end{cases} $$ Clearly, $f(0)=m$, and $f(1)=n$.
To verify that $f$ is continuous . . .
Without loss of generality, assume $m < n$, and let $A=\{m,mn,n\}$.
As a subspace of $X$, the open subsets of $A$ are $$\varnothing,\;\{m\},\;\{n\},\;\{m,n\},\;A$$ unless $m{\mid\,}n,\;$in which case, the open subsets of $A$ are $$\varnothing,\;\{m\},\;\{m,n\},\;A$$
In either case, it's easily verified that if $U$ is open in $A$, then $f^{-1}(U)$ is open in $[0,1]$.