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! $\endgroup$ Aug 28 '15 at 6:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.