Exercise 16.9

Question

Show that the dictionary order topology on the set $\mathbb{R} 	imes \mathbb{R}$ is the same as the product topology $\mathbb{R}_d 	imes \mathbb{R}$, where $\mathbb{R}_d$ denotes $\mathbb{R}$ in the discrete topology. Compare this topology with the standard topology on $\mathbb{R}^2$.

Amit Awasthi · Accepted Answer

\begin{proof}
  We see that the basis elements of $(\mathbb{R} 	imes
  \mathbb{R})_{\mathrm{lex}}$ consist of intervals of the form $(a 	imes b, c
  	imes d)$ for $a < c$, and for $a = c$ and $b < d$, as in Example 14.2.
  These basis elements are open in $\mathbb{R}_d 	imes \mathbb{R}$ since
  \begin{equation*}
    (a 	imes b,c	imes d) = (a,c) 	imes \mathbb{R} \cup \{a\} 	imes (c,\infty)
    \cup \{b\} 	imes (-\infty,d) \in \mathcal{T}_{\mathbb{R}_d 	imes
    \mathbb{R}}.
  \end{equation*}
  For the reverse situation, consider the basis elements for $\mathbb{R}_d
  	imes \mathbb{R}$; these consist of all $\{a\} 	imes (b,c)$ since $\{a \mid
    a \in \mathbb{R}\}$ forms a basis for $\mathbb{R}_d$ by Example 13.3. But then, $\{a\} 	imes (b,c)$ are open in $\mathbb{R} 	imes \mathbb{R}$ with the order topology since it is of the form $(a 	imes b, c 	imes d)$ for $a = c$.
  \par We now compare this to the standard topology on $\mathbb{R}^2$. Since $(a,b) 	imes (c,d) \in \mathbb{R}_d 	imes \mathbb{R}$, we see that $\mathbb{R}^2 \subseteq \mathbb{R}_d 	imes \mathbb{R}$. Moreover, since $\{a\} 	imes (b,c) \in (\mathbb{R}_d 	imes \mathbb{R}) \setminus \mathbb{R}^2$, we see that $\mathbb{R}^2 \subsetneq \mathbb{R}_d 	imes \mathbb{R}$.
\end{proof}

Dan Whitman · Answer

In what follows let $\mathcal{T}_d$ denote the dictionary order topology on ${\mathbb{R}} 	imes {\mathbb{R}}$, and let $\mathcal{T}_p$ denote the product topology on ${\mathbb{R}}_d 	imes {\mathbb{R}}$.
Also let $\prec$ denote the dictionary ordering of ${\mathbb{R}} 	imes {\mathbb{R}}$.
First we show that $\mathcal{T}_d = \mathcal{T}_p$.
\begin{proof}
First, we note that clearly the dictionary order on ${\mathbb{R}} 	imes {\mathbb{R}}$ has no largest or smallest elements so that, by definition, $\mathcal{T}_d$ has as basis elements intervals $((x,y), (x',y'))$, that is the set of all points $z \in {\mathbb{R}} 	imes {\mathbb{R}}$ where $(x,y) \prec z \prec (x',y')$.
Clearly the set $\left\{\left\{xight\} \mid x \in {\mathbb{R}}ight\}$ is a basis for ${\mathbb{R}}_d$.
Hence, by Theorem~15.1, the set $\mathcal{B}_p = \left\{\left\{xight\} 	imes (a,b) \mid 	ext{$x \in {\mathbb{R}}$ and $a < b$}ight\}$ is a basis for the product topology $\mathcal{T}_p$.

So consider any $(x,y) \in {\mathbb{R}} 	imes {\mathbb{R}}$ and any basis element $B_d = ((a,b), (a',b'))$ of $\mathcal{T}_d$ that contains $(x,y)$.
Hence $(a,b) \prec (x,y) \prec (a',b')$.

Case: $a = x$:
Then since $(a,b) \prec (x,y)$, it has to be that $b < y$.
\begin{adjustwidth}{1cm}{}
Case: $a = x = a'$.
Then it also has to be that $y < b'$ since $(x,y) \prec (a',b')$.
Then the set $B_p = \left\{xight\} 	imes (b,b')$ is a basis element of $\mathcal{T}_p$ that contains $(x,y)$ and is a subset of $B_d$.

Case: $a = x < a'$.
Then it is easy to show that the set $B_p = \left\{xight\} 	imes (b, y+1)$ is a basis element of $\mathcal{T}_p$ that contains $(x,y)$ and is a subset of $B_d$.
\end{adjustwidth}

Case: $a < x$:
\begin{adjustwidth}{1cm}{}
Case: $x = a'$.
Then it has to be that $y < b'$ since $(x,y) \prec (a',b')$.
Then it is easy to show that the set $B_p = \left\{xight\} 	imes (y-1, b')$ is a basis element of $\mathcal{T}_p$ that contains $(x,y)$ and is a subset of $B_d$.

Case: $a = x < a'$.
Then it is easy to show that the set $B_p = \left\{xight\} 	imes (y-1, y+1)$ is a basis element of $\mathcal{T}_p$ that contains $(x,y)$ and is a subset of $B_d$.
\end{adjustwidth}
In every case and sub-case, it follows from Lemma~13.3 that $\mathcal{T}_d \subset \mathcal{T}_p$.

Now suppose $(x,y) \in {\mathbb{R}} 	imes {\mathbb{R}}$ and $B_p = \left\{xight\} 	imes (a,b)$ is a basis element of $\mathcal{T}_p$ containing $(x,y)$.
Also let $B_d$ be the interval in the dictionary order $((x,a), (x,b))$, which is clearly a basis element of $\mathcal{T}_d$.
It is then trivial to show that $B_p = B_d$ so that $x \in B_d \subset B_p$, which shows that $\mathcal{T}_p \subset \mathcal{T}_d$ by Lemma~13.3.
This suffices to show that $\mathcal{T}_d = \mathcal{T}_p$ as desired.
\end{proof}

We now claim that this topology $\mathcal{T}_d = \mathcal{T}_p$ is strictly finer than the standard topology on ${\mathbb{R}} 	imes {\mathbb{R}}$.
We denote the latter by simply $\mathcal{T}$.
\begin{proof}
Since it was just shown that $\mathcal{T}_d = \mathcal{T}_p$, it suffices to show that either one is strictly finer than the standard topology.
It shall be most convenient to use the product topology $\mathcal{T}_p$.
So first consider any $(x,y) \in {\mathbb{R}}^2$ and  any basis element $B = (a,b) 	imes (c,d)$ of $\mathcal{T}$ containing $(x,y)$.
Hence $a < x < b$ and $c < y < d$.
It is then trivial to show that the set $\left\{xight\} 	imes (c,d)$, which is clearly a basis element of $\mathcal{T}_p$, contains $(x,y)$ and is a subset of $B$.
This shows that $\mathcal{T}_p$ is finer than $\mathcal{T}$ by Lemma~13.3.

To show that it is strictly finer, consider the point $(0,0)$ and the set $B_p = \left\{0ight\} 	imes (-1,1)$, which clearly contains $(0,0)$ and is a basis element of $\mathcal{T}_p$.
Now consider any basis element $B = (a,b) 	imes (c,d)$ of $\mathcal{T}$ that also contains $(0,0)$.
It then follows that $a < 0 < b$ and $c < 0 < d$.
Consider then the point $x = (a+0)/2 = a/2$ so that clearly $a < x < 0 < b$ and hence $x \in (a,b)$.
Thus the point $(x,0) \in B$, but also $(x,0) 
otin B_p$ since $x < 0$ so that $x 
eq 0$.
This shows that $B$ cannot be a subset of $B_p$.
Since $B$ was an arbitrary basis element of $\mathcal{T}$, this shows that $\mathcal{T}$ is \emph{not} finer than $\mathcal{T}_p$ by the negation of Lemma~13.3.

This suffices to show that $\mathcal{T}_p$ is strictly finer than $\mathcal{T}$ as desired.
\end{proof}

Exercise 16.9

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Exercise 16.9

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