Given a quasi-ordered set $(Q,\leq)$ the *interval topology* on $Q$ is generated by
$$\{Q\setminus\downarrow x : x\in Q\} \cup \{Q\setminus\uparrow x : x\in Q\},$$
where $\downarrow x = \{y\in Q: y\leq x\}$ and $\uparrow x = \{y\in Q: y\geq x\}$.


Let $\mathbb{N}^\mathbb{N}$ denote the set of all functions $f:\mathbb{N}\to\mathbb{N}$ and define a quasi-ordering relation $\leq^\*$ on $\mathbb{N}^\mathbb{N}$ by $$f\leq^\* g \text{ iff } \exists k\in \mathbb{N} \text{ such that } f(n) \leq g(n) \text{ for all } n\geq k.$$


Let $\tau\_i(\mathbb{N}^\mathbb{N})$ be the interval topology of $(\mathbb{N}^\mathbb{N}, \leq^\*)$.


Is $(\mathbb{N}^\mathbb{N}, \tau\_i(\mathbb{N}^\mathbb{N}))$ connected?



Is the interval topology of $(\mathbb{N}^\mathbb{N}, \leq^*)$ connected?
I claim that this topology has the stronger property of being [hyperconnected](https://en.wikipedia.org/wiki/Hyperconnected_space), i.e., the intersection of any two nonempty open sets is nonempty.


Indeed, any open set contains a set of the form


$$ U=\bigcap\_{i=1}^m \{x:x\not\le^\*y\_i\}\bigcap\bigcap\_{j=1}^n \{x:x\not\ge^\*z\_j\} $$


so the same can be said of the intersection of two open sets. I claim that 



> 
> $U$ is nonempty iff every $z\_j$ has infinitely many nonzero terms.
> 
> 
> 


It is easy to see this claim implies that the topology is hyperconnected. Now we show the claim.


Clearly this condition is necessary, for if $z\_j$ has finitely many nonzero terms, then for every $x\in\mathbb N^{\mathbb N}$ we have $x\ge^\*z\_j$. Conversely, assume all of $z\_j$ has infinitely many nonzero terms, then we can produce an $x\in U$ by the following algorithm:


`x=0 //initialize x to the zero array
k=0 //k indexes x
while true do
 for i = 1 to m
 x(k)=y(i)(k)+1 //Repeating this infinitely makes x\not\le^*y_i
 k=k+1
 end for
 for j = 1 to n
 while z(j)(k)=0 do
 k=k+1
 end while //This loop must end because z(j) has infinitely many nonzero terms
 x(k)=0 //Repeating this infinitely makes x\not\ge^*z_j as z(j)(k)>0
 end for
end while`


**Edit**: As Todd noted in the comment, another natural definition of the interval topology is generated by the open sets $\{x:x^\*>y\}$ and $\{x:x^\*<z\}$, where $x<^\*y$ iff there is $n\_0$ such that for all $n>n\_0$, $x\_n<y\_n$.


This topology is very different from the previous one. In fact it is *not* connected. It can be written as $U\cup V$, where $U$ is the set of all sequences tending to $+\infty$ and $V$ is its complement. We will show that both $U$ and $V$ are open.


Take $x\in U$. Then there is $n\_0$ such that for all $n>n\_0$, $x\_n>0$. Let


$$ y\_n=\begin{cases}x\_n & n\le n\_0\\ x\_n-1 & n>n\_0 \end{cases}. $$


Then $y\in U$ and $x\in (y,+\infty)\subset U$.


Take $x\in V$. Let $z=x+1$ (termwise). Then $z\in V$ and $x\in (-\infty,z)\subset V$.


