Is the interval topology of $(\mathbb{N}^\mathbb{N}, \leq^*)$ connected? 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?
 A: I claim that this topology has the stronger property of being hyperconnected, 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 $x\in (-\infty,z)\subset V$.
