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Let $\langle X, \prec \rangle$ be a linear order, and $\langle X, \tau \rangle$ be the induced order topology.

Are the following two statements equivalent?

  1. Every continuous (in the sense of the topology) bijection is either increasing or decreasing (in the sense of the ordering).
  2. The space $X$ is connected (in the sense of the topology).
Kenny Lau
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  • 2 $\implies$ 1 I think it is true. otherwise it violate the injectivity right? am I correct? –  Jan 02 '18 at 08:49
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    Indeed, if $x<y<z$ and $f(z)<f(x)<f(y)$, then $(y,z)$ can be divided into $(y,z) \cap {a \mid f(a) < f(x)}$ and $(y,z) \cap {a \mid f(a) > f(x)}$, and this generates a partition of the whole space into two disjoint open subsets. – Kenny Lau Jan 02 '18 at 08:53
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    Your above comment should be included in your question, and you should be more specific as to whether you want your proof (if you have one) to be verified or whether you can't prove the other implication and want some direction. – user21820 Jan 02 '18 at 08:56
  • How can we think the converse? –  Jan 02 '18 at 08:56
  • "Every continuous...bijection..." from $\langle X,\tau \rangle$ or to $\langle X,\tau \rangle$? –  Jan 02 '18 at 14:03
  • For the reverse, if $X$ is not connected, we can write it as $X = A \cup B$ with $A,B$ order convex and non-empty, $A \prec B$ and either $\max(A) < \min(B)$ (a jump), or $A$ has no maximum, and $B$ has no minimum (a gap). We can define a bijection by a increasing map on $A$ and decreasing on $B$, say. – Henno Brandsma Jan 02 '18 at 19:14
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    @HennoBrandsma. Given a linear order S, how do I construct a decreasing bijection from S to S? – William Elliot Jan 04 '18 at 07:40

2 Answers2

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They are not equivalent, even if you require $X$ to have more than two points. For instance, suppose $X$ is a linear order such that every point of $X$ has a distinct uncountable cofinality from below and cofinality $\omega$ from above. (Such a linear order can be constructed by inductively adding new points to give the requisite cofinalities from below in an induction of length $\omega$; see my first example in this answer.)

Such an $X$ cannot be Dedekind-complete (otherwise it would have points with cofinality $\omega$ from below), so it is disconnected in the order topology. However, I claim the only continuous injection $X\to X$ is the identity map. Indeed, suppose $f:X\to X$ is a continuous injection and $f(x)=y\neq x$ for some $x\in X$. Let $\kappa$ be the cofinality of $x$ from below and let $\lambda$ be the cofinality of $y$ from below. Let $(x_\alpha)_{\alpha<\kappa}$ be a strictly increasing sequence approaching $x$ from below and let $(y_\beta)_{\beta<\lambda}$ be a strictly increasing sequence approaching $y$ from below. Also let $(y^n)_{n<\omega}$ be a strictly decreasing sequence approaching $y$ from above.

By our choice of $X$, $\kappa\neq \lambda$. First, suppose $\kappa>\lambda$. By continuity of $f$, for each $\beta<\lambda$ there exists $\gamma_\beta<\kappa$ such that $x_\alpha>y_\beta$ for all $\alpha>\gamma_\beta$. Also, for each $n<\omega$, there exists $\gamma^n<\kappa$ such that $x_\alpha<y^n$ for all $\alpha>\gamma^n$. Let $\gamma$ be the supremum of all the $\gamma_\beta$s and $\gamma^n$s. Since $\kappa$ is regular and $\kappa>\lambda$, $\gamma<\kappa$. But then for $\alpha>\gamma$, we have $f(x_\alpha)=y$, since $y$ is the only element of $X$ which is greater than every $y_\beta$ and less than every $y^n$. This contradicts injectivity of $f$.

Now suppose $\kappa<\lambda$. Replacing $(x_\alpha)$ with a cofinal subsequence, we may assume either $f(x_\alpha)<y$ for all $\alpha$ or $f(x_\alpha)>y$ for all $\alpha$. In the case $f(x_\alpha)>y$, we can use the same argument as the previous paragraph (using just the $\gamma^n$s) to reach a contradiction. In the case $f(x_\alpha)<y$, note that for every $\alpha$ there exists $\beta_\alpha<\lambda$ such that $f(x_\alpha)<y_{\beta_\alpha}$. Since $\kappa<\lambda$, the supremum $\beta$ of all these $\beta_\alpha$s is less than $\lambda$. But then $f(x_\alpha)<y_\beta$ for all $\alpha$. This means $(f(x_\alpha))$ does not converge to $y$, contradicting continuity of $f$.

If you have difficulty understanding the example above, here's the gist of it. You can have a linear order $X$ which is disconnected but highly asymmetrical, so that every point of $X$ "looks different" topologically. This asymmetry makes $X$ very "rigid", so there are no nontrivial continuous bijections $X\to X$ at all, so rather vacuously they are all increasing or decreasing.

Eric Wofsey
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I don't think $(1)\implies (2)$ in general. The following is a counterexample.

Let $X=\{0,1\}$ and $\tau:=\{\emptyset,\{0\},\{1\}, X\}$ (i.e., $\tau$ is the subspace topology that $X$ inherits as a subspace of $\mathbb{R}$ equipped with usual (order) topology). Then observe that every continuous bijection from $\langle X,\tau\rangle$ to $\langle X,\tau\rangle$ is order preserving (or order reversing). However the space $\langle X,\tau\rangle$ is not connected because $\{1\}$ is a non-trivial clopen set in the topological space.