# 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?

• Does it has the stronger property that any two nonempty open sets has nonempty intersection? – Fan Zheng Dec 22 '15 at 20:34
• Good question @FanZheng - I don't know...! – Dominic van der Zypen Dec 22 '15 at 20:38
• @FanZheng What sequence could belong both to the open set of sequences which are eventually less than the constant 1 sequence and to the open set of sequences which are eventually greater than the constant 2 sequence? – Todd Trimble Dec 22 '15 at 21:22
• @ToddTrimble If I didn't get it wrong, the open set is defined as the set of sequences not eventually less/greater than a given sequence. So this is a very weak topology, which suggested my question above. – Fan Zheng Dec 22 '15 at 21:25
• Read "any of finitely many given sequences" for "a given sequence" above. – Fan Zheng Dec 22 '15 at 21:31

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 $x\in (-\infty,z)\subset V$.