Why must a continuous function on the Long Line eventually be constant?

Question

I've seen a couple of texts, including this answer on another question mention that any continuous real-valued function on the Long Line must eventually be constant, but I've not seen the reasoning as to why.

Why must this be true?

Answer 1

The key relevant property of the long line here is that any countable subset has a supremum. Indeed, any countable subset is bounded (since the first coordinates will be bounded in $\omega_1$) and thus has a supremum (the long line is Dedekind complete).

Now let $X$ denote the long line and suppose $f:X\to\mathbb{R}$ is a continuous function. Fix $\epsilon>0$ and suppose that for all $x\in X$, there exist $y,z>x$ such that $|f(y)-f(z)|>\epsilon$. We can then use this iteratively to choose $y_1<z_1<y_2<z_2<y_3<z_3<\dots$ such that $|f(y_n)-f(z_n)|>\epsilon$ for all $n$. But this increasing sequence converges to a limit $x=\sup_n y_n=\sup_n z_n$, so $f(y_n)$ and $f(z_n)$ must both be converging to $x$. This is a contradiction.

Thus for any $\epsilon>0$, there exists $x_\epsilon\in X$ such that $|f(y)-f(z)|\leq\epsilon$ for all $y,z>x_\epsilon$. If $x$ is the supremum of the elements $x_1,x_{1/2},x_{1/3},x_{1/4},\dots$, then for any $y,z>x$ we have $|f(y)-f(z)|\leq 1/n$ for all $n$ so $f(y)=f(z)$. Thus $f$ is eventually constant.

Answer 2

This post has two purposes; to paraphrase the explanation in the linked explanation and also to generalise the result. We work in $\mathsf{ZFC}$.

I use the convention that cardinals are ordinals which are not equipotent with any lesser ordinal.

Theorem:

Suppose $X$ is a linearly ordered space. If there is a regular, uncountable cardinal $\kappa$ and a continuous, weakly monotone $\phi:\kappa\to X$ with cofinal image, then all real-valued continuous functions on $X$ are eventually constant.

This can be further adapted. If the (linearly ordered) space $Y$ contains a dense subspace $X$ with the above property then $Y$ has it too; by an easy adjustment of the proof we may replace $\Bbb R$ by any metric space and further by any Hausdorff uniform space with a countable base of entourages.

This subsumes the usual claim about the long line (where “eventually” can be understood to apply at both the positive and negative ends of the line) $\mathscr{L}$ since the mapping: $$\phi:\omega_1\ni\lambda\mapsto\lambda\times\{0\}\in\mathscr{L}$$Easily satisfies all hypotheses. The generalisation shows that this works even if $\omega_1$ is replaced with a suitable larger ordinal; the line can be “lengthened”. $\omega_1$ is uncountable of course, and it is regular since a subset of it is bounded iff. it is countable, since the countable union of countable sets is again countable. Note that the “length” of the line is a bit misleading; although the line obtained with $\omega_1+1$ rather than $\omega_1$ is perhaps longer than the long line, it does not share this property that all continuous functions thereon are eventually constant.

Note that the ordered space $\kappa$ has that every successor ordinal is an isolated point, so continuity of $\phi$ at these points is trivial. It is only at the limit ordinals $\alpha<\kappa$ where we should be careful; these have a local base of the form $(\lambda,\alpha]$ for $\lambda<\alpha$.

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Lemma:

If $\kappa$ is an ordinal with uncountable cofinality then the set of all limit ordinals less than $\kappa$ is stationary in $\kappa$.

In fact, this set has non-empty intersection with any infinite closed set (a weaker condition than being club). Let $C$ be such a set. E.g. by using the well-ordering of sets of ordinals, we can find a strictly increasing ($\omega$-)sequence $\gamma_1<\gamma_2<\cdots$ which is drawn from $C$; by the hypothesis on cofinality, this converges to some $\gamma<\kappa$ and as the sequence is strictly increasing, $\gamma$ must be a limit ordinal. Since $C$ is closed, it’s easy to see $\gamma\in C$ is also true.

Theorem proof:

Let $f:X\to\Bbb R$ be any continuous function. Let $\kappa,\phi$ be as in the hypotheses.

Define $S_0$ to be the set of all limit ordinals less than $\kappa$. By hypothesis on $\kappa$, $S_0$ is stationary. We inductively will create a sequence of stationary subsets $S_0\supseteq S_1\supseteq S_2\supseteq\cdots$ of $\kappa$, a sequence of pressing-down functions $g_n:S_{n-1}\to\kappa,n\in\Bbb N$ and a sequence of ordinals $(\lambda_n)_{n\in\Bbb N}\subset\kappa$. Given $S_0,\cdots,S_n$ and $g_1,\cdots,g_{n-1},\lambda_1,\cdots,\lambda_{n-1}$, we proceed as follows; by continuity of $f$ and of $f\circ\phi$ and by the fact each element of $S_{n-1}$ is a limit ordinal, we may create a choice function $g_n:S_{n-1}\to\kappa$ that chooses for each $\alpha$ some $g_n(\alpha)<\alpha$ with: $$f([\phi(g_n(\alpha)),\phi(\alpha)])\subseteq\left(f(\phi(\alpha))-\frac{1}{2n},f(\phi(\alpha))+\frac{1}{2n}\right)$$

Since $\kappa$ is a regular uncountable cardinal, by the pressing-down lemma/Fodor’s lemma there is an ordinal $\lambda_n\in\kappa$ making $g_n^{-1}\{\lambda_n\}=:S_n$ a stationary set in $\kappa$. This concludes the induction.

Now let $\lambda:=\sup_{n\in\Bbb N}\lambda_n$; $\kappa$ must have uncountable cofinality, so $\lambda<\kappa$ follows. Suppose $x\in X$ has $\phi(\lambda)\le x$. Since $\phi$ has cofinal image, there is some $\beta\ge\lambda$ with $x\le\phi(\beta)$. For any $n\in\Bbb N$, since $S_n$ is stationary and $[\beta,\kappa)$ is club, there must be some $\alpha\in S_n$ with $\beta\le\alpha$. Then $x$ and $\phi(\lambda)$ are both elements of the interval $[\phi(g_n(\alpha)),\phi(\alpha)]$ so that: $|f(x)-f(\phi(\lambda))|\le\frac{1}{n}$ follows from the triangle inequality.

As this holds for any $n$, it must be that $f(x)=f(\phi(\lambda))$. Thus $f$ is constant on the ray $[\phi(\lambda),\longrightarrow)$; $f$ is “eventually constant”.