Definition: Let $L$ be simply ordered set with more than one element. We say that $L$ is a linear continuum iff. the following two properties holds:

  1. $L$ has the least upper bound property.

  2. If $x,y\in L$ with $x<y$, then $\exists z.\ x<z<y$.

In some sense linear continuums are a generalization of $\mathbb{R}$. And it's easy to see that axioms of $\mathbb{R}$ satisfy the definition of a linear continuum.

But if we take $I^2,$ where $I=[0,1]$ with dictionary order and denote it as $I^2_o$ then I claim that it is also a linear continuum.

The property 2. is easy to check.

Let's show that 1. also holds. If $A\subset I^2_o$ s.t. $A\neq \emptyset$ and $A$ is bounded above by $(x_0,y_0)$ then $\pi_1(A)$ is nonempty and bounded by $x_0$ then it has supremum which we call $l_1$ and using the same for $\pi_2(A)$ we get supremum $l_2$. Then one can show that $A$ is bounded above by $(l_1,l_2)$ and this upper bound is the minimal one.

Sorry that I missed most details. I just want to check is my reasoning ok?

Would be very thankful for any comments!

  • $\begingroup$ See the pictures in Munkres (2nd ed. ) in 24.3 why your sup approach is too naive. $\endgroup$ Sep 28, 2019 at 22:19

2 Answers 2


The sup property needs to be argued a bit more carefully.

Theorem: Suppose that $(X, <_X)$ and $(Y, <_Y)$ are linear continua in Munkres' sense. Then $X \times Y$ in the lexicographic order $$(x,y) <_l (x',y') \iff (x <_X x' \lor (x=x' \land y <_Y y'))$$ is also a linear continuum, provided that $\min(Y)$ and $\max(Y)$ exist.

Proof: let $(x,y)<_l (x',y') \in X \times Y$. If $x <_X x'$ we can find $x'' \in X$ such that $x <_X x'' <_X x'$ and then it follows that $(x,y) <_l < (x'',y) <_l (x',y')$ and property 2 holds for these points. Otherwise $x=x'$ and $y <_Y y'$ and $y'' \in Y$ exists with $y <_Y y'' <_Y y'$ and $(x,y) <_l < (x,y'') <_l (x',y')$ and we're done for property 2 (the denseness)

Now assume $A \subseteq X \times Y$ is upper bounded by some $(p,q)$. So in particular $\pi_X[A]$ is upperbounded in $X$ by $p$ and $a=\sup \pi_X[A]$ exists in $X$. Two cases: $a \in \pi_X[A]$, so that we have $A_a:=(\{a\} \times Y) \cap A \neq \emptyset$. The set $\{a\} \times Y$ is order isomorphic to $Y$ and $(a,b):=\sup A_a$ exists (as $(a,\max(Y))$ is an upper bound and $Y$ is a continuum) and $(a,b)=\sup A$ as well.

Otherwise $a \notin \pi_X[A]$ and then $(a,\min(Y))=\sup A$ (it's an upperbound for $A$ and if $(a',b)$ would be a smaller upperbound for $A$ for some $a' < a$ (the only way to be smaller than $(a,\min(Y)$), then $a'$ would be a smaller upperbound for $\pi_X[A]$ too...


I agree that the $l_1$ in $(l_1, l_2)$ is correct. However, I am not convinced with $l_2$. How about $\sup \{(0.9, 0.09), (0.99, 0.099), (0.999, 0.0999), \ldots\}$? Then you would have $(l_1, l_2) = (1, 0.1)$. However, $(1, 0)$ is a smaller upper bound.

Do you see where the problem came from? The values of the second components are completely irrelevant if the first component never attains the least upper bound of the first components!

Here's how we can fix it:

  • Choose $l_1 = \sup \pi_1(A)$ as before.
  • Choose $l_2 = \sup \pi_2(\{(x, y) \in A\ |\ x = l_1\})$, where by convention we choose $0$ if the set is empty.

I claim $(l_1, l_2)$ is the least upper bound of $A$ as follows:

  • It is an upper bound.
    Let $(x, y) \in A$. By construction we have $x \leq l_1$. Now if $x < l_1$, we are already done. In case $x = l_1$, we have $y \leq l_2$ by construction.

  • Every other upper bound $(k_1, k_2)$ fulfills $(l_1, l_2) \leq (k_1, k_2)$.

    Since $(k_1, k_2)$ is an upper bound, we have for every $(x, y) \in A$ that $x \leq k_1$. Hence, by the very definition of $l_1$ as the least upper bound in the first component, we can conclude $l_1 \leq k_1$. If $l_1 < k_1$, we are done due to the given lexicographic ordering.

    So assuming $l_1 = k_1$, we have to show $l_2 \leq k_2$. Since $(k_1, k_2)$ is an upper bound of $A$, we have for every $(l_1, y) \in A$ that $(l_1, y) \leq (k_1, k_2)$. Since we know $l_1 = k_1$, we can deduce $y \leq k_2$.
    Note that this is vacuously true if such tuples do not exist. If such tuples do not exist, we have $l_2 = 0$ anyway, which immediately implies $l_2 = 0 \leq k_2$. If such tuples exist, $k_2$ is an upper bound for the set $\pi_2(\{(x, y) \in A | x = l_1\})$. By the very definition of $l_2$ as the least upper bound of that we have $l_2 \leq k_2$.

  • $\begingroup$ Thanks a lot for an answer! But what's wrong with my reasoning? I have done it on the paper and everything seems correct. $\endgroup$
    – ZFR
    Sep 28, 2019 at 17:32
  • $\begingroup$ @ZFR I gave a counterexample on the first line. $\endgroup$
    – ComFreek
    Sep 28, 2019 at 17:32
  • $\begingroup$ @ZFR He's right you know... $\endgroup$ Sep 28, 2019 at 22:23

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