Let $(P,\leq)$ be a poset. The interval topology $\tau_i(P)$ on $P$ is generated by $$\{P\setminus\downarrow x : x\in P\} \cup \{P\setminus\uparrow x : x\in P\},$$ where $\downarrow x = \{y\in P: y\leq x\}$ and $\uparrow x = \{y\in P: y\geq x\}$.

Define ${\cal L}$ as in this question: the set of functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $f(n)\leq f(n+1)\leq f(n)+1$, where two functions are considered equal if they differ at at most finitely many points, and $f\prec g$ is defined to mean $f(n)\leq g(n)$ for all but at most finitely many exceptions $n$.

Is the space $({\cal L}, \tau_i({\cal L}))$ Hausdorff? Is it connected?


$\mathcal L$ is not connected or Hausdorff, but if you delete its countably many isolated points the result is a closed, connected, convex sublattice (which is still not Hausdorff).

To see this, call a function $f\in \mathbb N^{\mathbb N}$ passive at $n$ if $f(n+1)=f(n)$, and aggressive at $n$ if $f(n+1)=f(n)+1$. If $\mathcal L_0$ is the sublattice of $\langle \mathbb N^{\mathbb N}, \max, \min\rangle$ consisting of functions either passive or aggressive at each $n$, then $\mathcal L$ is the quotient of $\mathcal L_0$ modulo the equivalence relation of eventual equality.

If a function is passive at almost all $n$, then it is equivalent to a unique constant function. The set of these is an $\omega$-chain at the bottom of $\mathcal L$: all other elements of $\mathcal L$ lie above this $\omega$-chain.

If a function is aggressive at almost all $n$, then it is equivalent to a unique function of the form $f_k(n) = n+k$ where $k\in \mathbb Z$. Here, when $k\geq 0$, I really mean $f_k(n) = n+k\in \mathbb N^{\mathbb N}$, but for negative subscripts what I mean is $f_{-k}(n) = \max\{0, n-k\}\in \mathbb N^{\mathbb N}$. In any case, the equivalence classes of functions $f_k(n)$ are ordered the same way the subscripts are, and this yields a $\mathbb Z$-chain at the top of $\mathcal L$. All elements of $\mathcal L$ not in this $\mathbb Z$-chain lie below the entire chain.

All of this shows that $\mathcal L$ looks like the ordinal sum $\mathbb N + C + \mathbb Z$, where $C$ is the ``core'' of passive-aggressive functions, i.e. those that are passive infinitely often and aggressive infinitely often.

The chains at the top and bottom of $\mathcal L$ consist of points isolated in the interval topology, so of course $\mathcal L$ is not connected. I want to argue that $C$, the passive-aggressive core of $\mathcal L$, is connected but not Hausdorff. ($C$ is the intersection of all closed intervals of the form $[c,f_k(n)]$, where $c$ denotes an equivalence class represented by a constant function, so $C$ is closed, convex, and the subspace topology on $C$ is the interval topology on $C$.)

The assertions about $C$ follow from:

Claim. $C$ is not equal to a finite union of proper closed intervals.

Proof of Claim. Here a closed interval in $C$ has the form $[f):=[f,\infty)$, $(g]:=(-\infty,g]$ or $[f,g]$. (I will use $f$ and $g$ to denote functions and also their equivalence classes in $\mathcal L$.) For the claim we can restrict attention to intervals of the first two types, since if $[f,g]$ is proper, then one of the larger intervals $[f)$ or $(g]$ is proper, and we can use that in place of $[f,g]$ in our union.

So assume that $C = [f_1)\cup \cdots \cup [f_r)\cup (g_1]\cup\cdots\cup (g_s]$ where each interval is proper. Since the functions $f_i, g_j$ represent elements of $C$ they are passive-aggressive. We build a passive-aggressive function $h$ that is not in the union. Start by defining $h$ at $0, 1, 2, \ldots$ so that it is aggressive at every $n$ until we reach a value $n_0$ where $h(n_0)$ strictly majorizes every $g_j(n_0)$. This is possible, since the $g_j$'s are passive infinitely often. Now continue defining $h$ so that it is passive at $n_0+1, \ldots, n_1$ until $h(n_1)$ is strictly majorized by every $f_i(n_1)$. This is possible, since the $f_i$'s are aggressive infinitely often. Continue this, alternating back and forth, to construct an $h$ that strictly majorizes every $g_j$ infinitely often and which is strictly majorized by every $f_i$ infinitely often. This $h$ is passive-aggressive, but does not represent an element of $[f_1)\cup \cdots \cup [f_r)\cup (g_1]\cup \cdots\cup (g_s]$. \\

Corollary. $C$ has no pair of disjoint nonempty open sets.

Proof of Cor: If $U$ and $V$ are such, then we may assume they are nonempty, basic, open sets. The complements $C\setminus U$ and $C\setminus V$ are proper, basic, closed sets whose union is $C$. But a basic closed set is a finite union of closed intervals, so $(C\setminus U) \cup (C\setminus V)$ now expresses $C$ as a finite union of proper closed intervals, contrary to the Claim. \\

But if $C$ has no pair of disjoint nonempty open sets, it is connected and not Hausdorff.

  • $\begingroup$ That's very well written, thanks a lot Keith! $\endgroup$ – Dominic van der Zypen Aug 28 '15 at 6:51

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