Exercise 16.10

Proof. First, consider the point $0\times 1\in I\times I$ and $B_p=[0,1∕2)\times (1∕2,1]$, which is a basis element of $T_p$ that clearly contains $0\times 1$. Note that $B_p$ is a basis element because $[0,1∕2)$ and $(1∕2,1]$ are both basis elements in the order topology on $I$ since $0$ and $1$ are the smallest and largest elements of $I$, respectively. Now consider any interval $B_d=(a\times b,a^{\prime }\times b^{\prime })$ in the dictionary order on $I\times I$ that contains $0\times 1$, which is of course a basis element of $T_d$. Then we have that $a\times b$ and $a^{\prime }\times b^{\prime }$ are in $I\times I$ with $a\times b\prec 0\times 1\prec a^{\prime }\times b^{\prime }$. Hence $0<a^{\prime }$ or $0=a^{\prime }$ and $1<b^{\prime }$. As $1$ is the largest element of $I$, the latter case is not possible so that it must be that $0<a^{\prime }$. Let $x=(0+a^{\prime })∕2=a^{\prime }∕2$ so that clearly $0<x<a^{\prime }$. Then we have that $a\times b^{\prime }\prec x\times 0\prec a^{\prime }\times b^{\prime }$ so that the point $x\times 0$ is in $B_d$. However, clearly $0\notin (1∕2,1]$ so that $x\times 0\notin B_p$. This shows that $B_d$ cannot be a subset of $B_p$.

Here we note that, in the dictionary order on $I\times I$, the smallest element is $0\times 0$ while the largest is $1\times 1$. With this in mind, the above argument for an open interval also applies to the half-open intervals $[0\times 0,a\times b)$ and $(a\times b,1\times 1]$, which are of course also basis elements of $T_d$. This then shows that $T_d$ is not finer than $T_p$ by the negation of Lemma 13.3.

Now consider the point $0\times 1∕2$ and the interval $B_d=(0\times 0,0\times 1)$ in the dictionary ordering, which is, therefore, a basis element of $T_d$, and clearly also contains $0\times 1∕2$. Consider also any basis element $B_p=A\times B$ of $T_p$ that contains $0\times 1∕2$. Since $0\in A$ and $A$ must be a basis element of the order topology on $I$, it has to be that $A=[0,a)$ for some $0<a\le 1$. Then let $x=(0+a)∕2=a∕2$ so that $0<x<a$, and thus $x\in A$. Then, since $1∕2\in B$ (since $0\times 1∕2\in A\times B$), we have that $x\times 1∕2\in A\times B=B_p$ as well. However, we also clearly have that $0\times 1\prec x\times 1∕2$ since $0<x$ so that $x\times 1∕2\notin (0\times 0,0\times 1)=B_d$. This shows that $B_p$ cannot be a subset of $B_d$. As $B_p$ was an arbitrary basis element of $T_p$, this shows by the negation of Lemma 13.3 that $T_p$ is not finer than $T_d$.

This suffices to show that $T_d$ and $T_p$ are incomparable. □

Proof. Consider any $x\times y\in I\times I$ and suppose that $B_p=A\times B$ is a basis element of $T_p$ that contains $x\times y$. First suppose that $A=(a,b)$ and $B=(c,d)$ so that of course $a,b,c,d\in I$, $a<x<b$, and $c<y<d$. It is then trivial to show that the interval $B_s=(x\times c,x\times d)$ in the dictionary order also contains $x\times y$, is a basis element of $T_s$ (since $B_s\subset I\times I$ so that $B_s\cap (I\times I)=B_s$), and is a subset of $B_p$. A similar argument can be made if $A$ is an interval of the form $[0,a)$ or $(a,1]$. If $B=(c,1]$ and $A$ is still $(a,b)$, then let $X$ be the interval $(x\times c,x\times 2)$ in the dictionary order so that we have $B_s=X\cap (I\times I)=\left\{x\right\}\times (c,1]$ is a basis element of $T_s$ that contains $x\times y$ and is a subset of $B_p$. A similar argument applies if $B=[0,d)$ and/or when the interval $A$ is half-open. This shows that $T_s$ is finer than $T_p$ by Lemma 13.3.

The argument above that shows that $T_p$ is not finer than $T_d$ using the negation of Lemma 13.3 applies equally well to show that $T_p$ is not finer than $T_s$. This of course suffices to show the desired result that $T_s$ is strictly finer than $T_p$. □

Proof. First, consider any point $x\times y$ in $I\times I$ and let $B_d$ be a basis element of $T_d$ that contains $x\times y$ so that it is some kind of interval with endpoints $a\times b$ and $a^{\prime }\times b^{\prime }$ in $I\times I$. We note here that, since $ℝ\times ℝ$ has no smallest or largest elements, basis elements of the dictionary order topology there can only be open intervals. Now, if $B_d$ is an open interval in $I\times I$ then clearly then the same interval $B=(a\times b,a^{\prime }\times b^{\prime })$ is a basis element in the dictionary order topology of $ℝ\times ℝ$, though the two intervals can, in general, be different sets. For example the interval $(0\times 0,1\times 1)$ in $ℝ\times ℝ$ contains the point $0\times 100$ whereas the same interval in $I\times I$ does not since $0\times 100\notin I\times I$. It is, however, trivial to show that $B\cap (I\times I)=B_d$ so that $B_d$ is basis element of $T_s$.

If we have that $B_d$ is the half-open interval $[0\times 0,a^{\prime }\times b^{\prime })$ then let $B=(0\times -1,a^{\prime }\times b^{\prime })$, which is clearly a basis element of the dictionary order topology on $ℝ\times ℝ$. It is then easy to see that $B\cap (I\times I)=B_d$ again so that it is a basis element of $T_s$. If $B_d$ is the half-open interval $(a\times b,1\times 1]$, then the open interval $(a\times b,1\times 2)$ is a basis element of the dictionary order topology on $ℝ\times ℝ$ and has the same result. Hence in all cases, $B_d$ is also a basis element of $T_s$, and it trivially is a subset of itself, and it contains $x\times y$. This shows that $T_s$ is finer than $T_d$ by Lemma 13.3.

To show that it is strictly finer, consider the point $0\times 1$ and the open interval $B=(0\times 0,0\times 2)$, which is clearly a basis element of the dictionary order topology in $ℝ\times ℝ$. It is then easy to prove that $B_s=B\cap (I\times I)=\left\{0\right\}\times (0,1]=(0\times 0,0\times 1]$ so that $B_s$ is a basis element of $T_s$. Now consider any basis element $B_d$ of $T_d$ that contains $0\times 1$ so that $B_d$ is some type of dictionary-order interval with endpoints $a\times b$ and $a^{\prime }\times b^{\prime }$, both in $I\times I$. The only way the interval can be closed above is if $a^{\prime }\times b^{\prime }=1\times 1$, in which case clearly $1\times 1\in B_d$ but $1\times 1\notin B_s$. So assume that it is open above so that $0\times 1\prec a^{\prime }\times b^{\prime }$, and hence either $0<a^{\prime }$ or $0=a^{\prime }$ and $1<b^{\prime }$. The latter case cannot be since $1$ is the largest element of $I$ and $b^{\prime }\in I$. Therefore it has to be that $0<a^{\prime }$. So let $x=(0+a^{\prime })∕2=a^{\prime }∕2$ so that $0<x<a^{\prime }$ and thus $0\times 1\prec x\times 0\prec a^{\prime }\times b^{\prime }$. From this, it follows that $x\times 0$ is in $B_d$. However, clearly $x\times 0\notin B_s$ since $0\times 1\prec x\times 0$.

Hence, in any case, we have shown that, while they both contain $0\times 1$, $B_d$ cannot be a subset of $B_s$. Since $B_d$ was an arbitrary basis element, this shows that $T_d$ is not finer than $T_s$ by the negation of Lemma 13.3. This shows the desired result that $T_s$ is strictly finer than $T_d$. □