# Prob. 5, Sec. 24, in Munkres' TOPOLOGY, 2nd ed: Are these sets linear continuua?

Here is Prob. 5, Sec. 24, in the book Toplogy by James R. Munkres, 2nd edition:

Consider the following sets in the dictionary order. Which are linear continuua?

(a) $\mathbb{Z}_+ \times [0, 1)$

(b) $[0, 1) \times \mathbb{Z}_+$

(c) $[0, 1) \times [0, 1]$

(d) $[0, 1] \times [0, 1)$

And, here is the definition of linear continuum.

A simply ordered set $L$ having more than one element is called a linear continuum if the following hold:

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

(2) If $x < y$, there exists $z$ such that $x < z < y$.

My Attempt:

Part (a):

We know that if $X$ is a well-ordered set, then $X \times [0, 1)$ in the dictionary order is a linear continuum. Here is a Math SE post of mine on this very fact; and, here is another Math SE post of mine.

Since the set $\mathbb{Z}_+$ is well-ordered, therefore the set $\mathbb{Z}_+ \times [0, 1)$ is a linear continuum in the dictionary order.

Part (b):

Suppose $x_1 \times n_1$ and $x_2 \times n_2$ be any two elements of $[0, 1) \times \mathbb{Z}_+$ such that $$x_1 \times n_1 \prec x_2 \times n_2.$$ Then either $x_1 < x_2$, $x_1 = x_2$ and $n_1 < n_2$.

If $x_1 < x_2$, then we have $$x_1 < \frac{ x_1 + x_2 }{2} < x_2,$$ and hence $$x_1 \times n_1 \prec \frac{ x_1 + x_2 }{2} \times n_1 \prec x_2 \times n_2.$$

On the other hand, if $x_1 = x_2$ and $n_2 = n_1 + 1$, then we cannot find any element $x \times n$ of $[0, 1) \times \mathbb{Z}_+$ satisfying $$x_1 \times n_1 \prec x \times n \prec x_2 \times n_2.$$

So $[0, 1) \times \mathbb{Z}_+$ does not satisfy the second requirement for a linear continuum.

Let $S$ be a non-empty subset of $[0, 1) \times \mathbb{Z}_+$ such that $S$ is bounded from above in $[0, 1) \times \mathbb{Z}_+$; let $a \times m$ be an upper bound of $S$ in $[0, 1) \times \mathbb{Z}_+$. Then, for every element $x \times n \in S$, we have $$x \times n \preceq a \times m,$$ which implies that $x \leq a$. Thus the set $$\pi_1 (S) = \left\{ \ x \in [0, 1) \ \colon \ x \times n \in S \ \mbox{ for some } n \in \mathbb{Z}_+ \ \right\}$$ is a non-empty, bounded above subset of $[0, 1)$, with $a$ as an upper bound. Let $a_0$ denote the least upper bound (in $\mathbb{R}$) of the set $\pi_1(S)$. Then $a_0 \in [0, 1)$ also. [Am I right?]

If $a_0 \not\in \pi_1(S)$, then $a_0 \times 1$ is the least upper bound of our set $S$. On the other hand, if $a_0 \in \pi_1(S)$, then the set $$\left( \left\{ a_0 \right\} \times \mathbb{Z}_+ \right) \cap S$$ is non-empty; moreover $\left\{ a_0 \right\} \times \mathbb{Z}_+$ has the order type of $\mathbb{Z}_+$. In this case, we cannot guarantee the existence of a least upper bound for $S$.

For example, let $S$ be the subset of $[0, 1) \times \mathbb{Z}_+$ given by $$S = \left[ 0, \frac{1}{2} \right] \times \mathbb{Z}_+ = \left\{ \ x \times n \ \colon \ 0 \leq x \leq \frac{1}{2}, n \in \mathbb{Z}_+ \ \right\}.$$ This set $S$ is bounded above by any element of the form $a \times m$, where $a \in \left( \frac{1}{2}, 1 \right)$ and $m \in \mathbb{Z}_+$. However, this set has no least upper bound in $[0, 1) \times \mathbb{Z}_+$, for if $a \times m$ is an upper bound of this set, then so is the element $$\frac{a+ 1/2}{2} \times m.$$

In short, $[0, 1) \times \mathbb{Z}_+$ fails to satisfy either of the requirements for a linear continuum.

Am I right?

Part (c):

Let $S$ be a non-empty, bounded above subset of $[0, 1) \times [0, 1]$. Then there exists an element $a \times b \in [0, 1) \times [0, 1]$ such that $$x \times y \preceq a \times b$$ for every element $x \times y \in S$. Then $a$ is an upper bound in $[0, 1)$ for the set $$\pi_1 (S) = \left\{ \ x \in [0, 1) \ \colon \ x \times y \in S \ \mbox{ for some } y \in [0, 1] \ \right\},$$ which is a subset of $[0, 1)$ of course. Let $a_0$ denote the least upper bound (in $\mathbb{R}$) of the set $\pi_1(S)$. Then $a_0 \in [0, 1)$ also.

If $a_0 \not\in \pi_1(S)$, then $a_0 \times 0$ is the least upper bound of set $S$.

On the other hand, if $a_0 \in S$, then the set $$\left( \ \left\{ a_0 \right\} \times [0, 1] \ \right) \cap S$$ is a non-empty subset of $\left\{ a_0 \right\} \times [0, 1]$, and $\left\{ a_0 \right\} \times [0, 1]$ has the order type of $[0, 1]$; so the set $$\pi_2 \left( \left( \ \left\{ a_0 \right\} \times [0, 1] \ \right) \cap S \right) = \left\{ \ y \in [0, 1] \ \colon a_0 \times y \in S \ \right\},$$ being a non-empty, bounded above subset of $[0, 1]$, has a least upper bound $b_0 \in [0, 1]$. Then the element $a_0 \times b_0$ is the least upper bound of $S$ in $[0, 1) \times [0, 1]$.

Thus we have shown that every non-empty, bounded above subset $S$ of $[0, 1) \times [0, 1]$ has a least upper bound in $[0, 1) \times [0, 1]$.

Am I right?

Now let $x_1 \times y_1$ and $x_2 \times y_2$ be any two elements of $[0, 1) \times [0, 1]$ such that $$x_1 \times y_1 \prec x_2 \times y_2.$$ Then either $x_1 < x_2$, or $x_1 = x_2$ and $y_1 < y_2$.

If $x_1 < x_2$, then we have $$x_1 < \frac{ x_1 + x_2 }{2} < x_2,$$ and therefore $$x_1 \times y_1 \prec \frac{ x_1 + x_2 }{2} \times y_1 \prec x_2 \times y_2.$$

On the other hand, if $x_1 = x_2$ and $y_1 < y_2$, then we have $$y_1 < \frac{ y_1 + y_2 }{2} < y_2,$$ and hence $$x_1 \times y_1 \prec x_1 \times \frac{ y_1 + y_2 }{2} < x_2 \times y_2.$$

In either case, we have shown that for any two elements $x_1 \times y_1$ and $x_2 \times y_2$ in $[0, 1) \times [0, 1]$ such that $x_1 \times y_1 \prec x_2 \times y_2$, there exists an element $x \times y \in [0, 1) \times [0, 1]$ such that $$x_1 \times y_1 \prec x \times y \prec x_2 \times y_2.$$

Hence $[0, 1) \times [0, 1]$ is a linear continuum.

Am I right?

Part (d):

Let $S$ be a non-empty, bounded above subset of $[0, 1 ] \times [0, 1)$. Then there exists an element $a \times b \in [0, 1 ] \times [0, 1 )$ such that $$x \times y \preceq a \times b$$ for every element $x \times y \in S$. Then $a$ is an upper bound in $[0, 1]$ for the set $$\pi_1 (S) = \left\{ \ x \in [0, 1] \ \colon \ x \times y \in S \ \mbox{ for some } y \in [0, 1 ) \ \right\},$$ which is a subset of $[0, 1]$ of course. Let $a_0$ denote the least upper bound (in $\mathbb{R}$) of the set $\pi_1(S)$. Then $a_0 \in [0, 1]$ also.

If $a_0 \not\in \pi_1(S)$, then $a_0 \times 0$ is the least upper bound of set $S$.

On the other hand, if $a_0 \in S$, then we cannot guarantee the existence of a least upper bound for set $S$ in $[0, 1] \times [0, 1)$. For example, let $$S \colon= [0, 1/2 ] \times [0, 1) = \{ \ x \times y \ \colon \ 0 \leq x \leq 1/2, 0 \leq y < 1 \ \}.$$ Then $S$ is bounded above in $[0, 1] \times [0, 1)$ by any (and only an) element of the form $a \times b$, where $1/2 < a \leq 1$ and $0 \leq b < 1$. However, this set $S$ has no least upper bound in $[0, 1] \times [0, 1)$: If $a \times b$ is an upper bound for $S$, then $1/2 < a$ and so $$\frac{1}{2} < \frac{ 1/2 + a }{2} < a,$$ and so $$\frac{ 1/2 + a }{2} \times b \prec a \times b,$$ but $$\frac{ 1/2 + a}{2} \times b$$ is also an upper bound for set $S$.

Thus $[0, 1] \times [0, 1)$ fails to satisfy the first requirement for a linear continuum.

Am I right?

Now let $x_1 \times y_1$ and $x_2 \times y_2$ be any two elements of $[0, 1 ] \times [0, 1)$ such that $$x_1 \times y_1 \prec x_2 \times y_2.$$ Then either $x_1 < x_2$, or $x_1 = x_2$ and $y_1 < y_2$.

If $x_1 < x_2$, then we have $$0 \leq x_1 < \frac{ x_1 + x_2 }{2} < x_2 \leq 1,$$ and therefore $\frac{ x_1 + x_2 }{2} \times y_1 \in [0, 1] \times [0, 1)$ and $$x_1 \times y_1 \prec \frac{ x_1 + x_2 }{2} \times y_1 \prec x_2 \times y_2.$$

On the other hand, if $x_1 = x_2$ and $y_1 < y_2$, then we have $$0 \leq y_1 < \frac{ y_1 + y_2 }{2} < y_2 < 1 ,$$ and hence $x_1 \times \frac{ y_1 + y_2}{2} \in [0, 1] \times [0, 1)$ and $$x_1 \times y_1 \prec x_1 \times \frac{ y_1 + y_2 }{2} < x_2 \times y_2.$$

In either case, we have shown that, for any two elements $x_1 \times y_1$ and $x_2 \times y_2$ in $[0, 1 ] \times [0, 1 )$ such that $x_1 \times y_1 \prec x_2 \times y_2$, there exists an element $x \times y \in [0, 1] \times [0, 1)$ such that $$x_1 \times y_1 \prec x \times y \prec x_2 \times y_2.$$

Am I right?

Is what I have done in parts (a) through (d) correct? If not, then where have I gone awry?

• a) is just $[0,\infty)$ in disguise (send $x$ to $(\lfloor x \rfloor, x - \lfloor x \rfloor)$ for an order isomorphism. – Henno Brandsma May 21 '18 at 9:05

## 1 Answer

Part a) is correct, given the exercise you already did. In this case it’s easy to see that the set is just $[0,\infty)$ in disguise, as we can use the order isomorphism $x \in [0,\infty) \to ( \lfloor x \rfloor , \{x\})$ (see here for notations), and $[0,\infty)$ is a standard example of a linear continuum, being an order convex subset of $\mathbb{R}$.

As to others, one example for one property is enough is refute being a linear continuum. You don’t need to do so much. So in b) X = $[0,1) \times Z^+$ and there are clearly no points between $(0,1)$ and $(0,2)$. Done.

As to c), $[0,1) \times [0,1]$ is just an order convex subset (as we remove a set $\{1\} \times [0,1]$ of elements all larger than the subspace, it's essentially the half-closed interval $[(0,0),(1,0))$ in the lexicographic order) of the well-known linear continuum $[0,1] \times [0,1]$ in the lexicographic order (Munkres shows this in the text IIRC (page 154 in the second edition, example 1), so that space is also a linear continuum. (This fact is implicit in the proof in the text that a linear continuum is connected, but the proof is easy from the definitions too)

As to d), again one example suffices, in $[0,1] \times [0,1)$ your set $S$ has an upperbound but no least upper bound. Case closed. The exercise stops right there. No need to check other properties or find conditions when a bounded set does have a supremum.

So your answers are correct, but way too long. Find a counterexample or reduce a proof to theorems you already know which you did well for a), and could have done for c) too, as I showed.