# Prob. 10 after Sec. 16 in Munkres' TOPOLOGY, 2nd edition: Which of these topologies is finer than which?

Let $I = [0,1]$, the unit closed interval on the real line with its usual order. Compare the product topology on $I \times I$, the dictionary order topology on $I \times I$, and the topology $I \times I$ inherits as a subspace of $\mathbb{R} \times \mathbb{R}$ in the dictionary order topology.

My work:

Here $(a,b)$ denotes an open interval, etc.

For the product topology, we have the following sets as a basis:

$[0,a) \times [0,b), \text{ where } 0 < a, b \leq 1$;

$[0,a) \times (b_1, b_2), \text{ where } 0 < a \leq 1, \text{ and } 0 \leq b_1 < b_2 \leq 1$;

$[0,a) \times (b_1, 1], \text{ where } 0 < a \leq 1, \text{ and } 0 \leq b_1 < 1$;

$(a_1, a_2) \times [0,b), \text{ where } 0 \leq a_1 < a_2 \leq 1, \text{ and } 0 < b \leq 1$;

$(a_1, a_2) \times (b_1, b_2), \text{ where } 0 \leq a_1 < a_2 \leq 1, \text{ and } 0 \leq b_1 < b_2 \leq 1$;

$(a_1, a_2) \times (b_1, 1], \text{ where } 0 \leq a_1 < a_2 \leq 1, \text{ and } 0 \leq b_1 < 1$;

$(a_1, 1] \times [0,b_2), \text{ where } 0 \leq a_1 < 1, \text{ and } 0 < b_2 \leq 1$;

$(a_1, 1] \times (b_1, b_2) \text{ where } 0 \leq a_1 < 1, \text{ and } 0 \leq b_1 < b_2 \leq 1$;

$(a_1, 1] \times (b_1, 1] \text{ where } 0 \leq a_1 < 1, \text{ and } 0 \leq b_1 < 1$.

Am I right?

A basis for the dictionary order topology on $I \times I$ is as follows:

We denote an element of $I \times I$ by $x \times y$.

$[0\times 0, x \times y), \text{ where } 0 < x, y \leq 1$;

$(x_1 \times y_1, x_2 \times y_2 ), \text{ where } x_1, x_2, y_1, y_2 \in I \text{ such that either } x_1 < x_2, \text{ or } x_1 = x_2 \text{ and } y_1 < y_2$;

$(x \times y, 1 \times 1], \text{ where } 0 \leq x, y < 1$.

Am I right?

Finally, since a basis for the dictionary order topology on $\mathbb{R} \times \mathbb{R}$ is $$\left\{ \ \{x \} \times (y_1, y_2) \ \colon \ x, y_1, y_2 \in \mathbb{R}, y_1 < y_2 \ \right\},$$ we see that a basis for the subspace topology that $I \times I$ inherits is as follows: $$\left\{ \ \{x \} \times (y_1, y_2) \ \colon \ x, y_1, y_2 \in I, \text{ and } y_1 < y_2 \ \right\}.$$

Am I right?

• @Brian M. Scott could you please take time answering this question as well? Mar 4, 2015 at 10:43

You’ve correctly described bases for the product topology on $I\times I$ and the dictionary order topology on $\Bbb R\times\Bbb R$. Your base for the dictionary order on $I\times I$ is almost correct, provided that it’s understood that the intervals are with respect to the dictionary order, but it’s not very helpful in answering the question, because it doesn’t describe those intervals in more familiar terms.

For instance, if $x_1<x_2$, then

$$(x_1\times y_1,x_2\times y_2)=\big(\{x_1\}\times(y_1,1]\big)\cup\big((x_1,x_2)\times I\big)\cup\big(\{x_2\}\times[0,y_2)\big)\;,$$

where $(y_1,1]=\varnothing$ if $y_1=1$, and $[0,y_2)=\varnothing$ if $y_2=0$.

If $x_1=x_2$ and $y_1<y_2$, matters are simpler: then

$$(x_1\times y_1,x_2\times y_2)=\{x_1\}\times(y_1,y_2)\;.$$

Similarly,

$$[0\times 0,x\times y)=\big([0,x)\times I\big)\cup\big(\{x\}\times[0,y)\big)$$

if $0<x,y\le 1$, and

$$(x\times y,1\times 1]=\big(\{x\}\times(y,1]\big)\cup\big((x,1]\times I\big)$$

if $0\le x,y<1$.

However, you also need the intervals $[0\times 0,0\times y)$ with $0<y\le 1$ and $(1\times y,1\times 1]$ with $0\le y<1$; in more familiar terms they are $\{0\}\times[0,y)$ and $\{1\}\times(y,1]$.

Let $\tau$ be the product topology on $I\times I$, $\tau_d$ the dictionary order topology on $I\times I$, and $\tau_s$ the topology that $I\times I$ inherits from $\Bbb R\times\Bbb R$ with its dictionary order topology.

• Show that $\left(\frac12,1\right]\times\left[0,\frac12\right)\in\tau\setminus\tau_d$, so $\tau\nsubseteq\tau_d$.

• Show that $\tau_d\nsubseteq\tau$.

• Show that $\tau_d\subseteq\tau_s$; note that this implies that $\tau_s\nsubseteq\tau$.

• Show that $\{0\}\times(0,1]\in\tau_s\setminus\tau_d$, so $\tau_d\subsetneqq\tau_s$.

• Show that $\tau\subseteq\tau_s$.

@Brian M. Scott the set $\left( \tfrac{1}{2}, 1 \right] \times \left[0, \tfrac{1}{2} \right)$, being the Cartesian product of two open sets in $I = [0,1]$ in the order topology, is open in the product topology on $I \times I$; in fact it is a basis element for this topology.

But how come this set is not in the dictionary order topology on $I \times I$?

The dictionary order topology on $I \times I$ is not contained in the product topology on $I \times I$ because the set $\{1\} \times \left( \tfrac{1}{3}, \tfrac{1}{2} \right)$, for example, is in the former topology but not in the latter. Am I right?

The dictionary order topology on $I \times I$ is contained in the subspace topology that $I \times I$ inherits from the dictionary order topology on $\mathbb{R} \times \mathbb{R}$ because we can express each of the basis elements in the former topology as an element of the latter (i.e. as the intersection with $I \times I$ of an open set in the dictionary order topology on $\mathbb{R} \times \mathbb{R}$).

The details are as follows:

Whenever $0 < x \leq 1$, we have $$\left[ 0 \times 0, 0 \times x \right) = \{0\} \times [0, x) = \left( I \times I \right) \ \cap \ \left[ \ \{0\} \times (-x, x) \ \right].$$

Whenever $0 \leq y < 1$, we have $$\left(1 \times y , 1\times 1 \right] = \{ 1 \} \times (y, 1] = \left( I \times I \right) \ \cap \ \left[ \ \{1\} \times (y , 2) \ \right].$$

Whenever $0 < x \leq 1$ and $0 < y \leq 1$, we have \begin{eqnarray*} [ \ 0 \times 0, \ x \times y \ ) &=& \left( [0,x) \times I \right) \ \cup \ \left( \{ x \} \times [0,y) \right) \\ &=& \left( I \times I \right) \ \ \cap \ \ \left[ \left( \ ( -x, x ) \times \mathbb{R} \ \right) \ \cup \ \left( \ \{x \} \times (-y, y) \ \right) \right] \\ &=& \left[ \ \left( I \times I \right) \ \cap \ \left( \ ( -x, x ) \times \mathbb{R} \ \right) \ \right] \ \ \cup \ \ \left[ \ \left( I \times I \right) \ \cap \ \left( \ \{x \} \times (-y, y) \ \right) \right]. \end{eqnarray*} Here by $( -x, x ) \times \mathbb{R}$ we mean the open interval $\left( \ (-x) \times y_1, \ x \times y_2 \ \right)$ in the dictionary order topology of $\mathbb{R} \times \mathbb{R}$; that is, $$( -x, x ) \times \mathbb{R} \ \colon= \ \{ \ u \times v \in \mathbb{R} \times \mathbb{R} \ \colon \ (-x) \times y_1 <_{\mathbb{R} \times \mathbb{R}} u \times v <_{\mathbb{R} \times \mathbb{R}} x \times y_2 \ \},$$

where $y_1, y_2 \in \mathbb{R}$ are arbitrary.

Whenever $0 < x \leq 1$, we have $$[ \ 0\times 0 , \ x \times 0 \ ) = [0, x) \times I = \left( I \times I \right) \ \cap \left[ \ (-x, x) \times \mathbb{R} \ \right].$$

Whenever $0 \leq x < 1$ and $0 \leq y < 1$, we have \begin{eqnarray*} ( \ x \times y, 1 \times 1 \ ] &=& \left[ \ \{x \} \times (y, 1] \ \right] \ \ \cup \ \ \left[ (x, 1] \times I \right] \\ &=& \left[ \ \left(I \times I \right) \ \cap \ \left( \{ x \} \times (y,2) \right) \ \right] \ \ \cup \ \ \left[\ \left( I \times I \right) \ \cap \ \left( (x, 2) \times \mathbb{R} \right) \ \right]. \end{eqnarray*}

Whenever $0 \leq x < 1$, we have $$(x \times 1, 1 \times 1 ] = (x,1] \times I = \left( I \times I \right) \ \ \cap \ \ \left( \ (x, 2) \times \mathbb{R} \ \right).$$

Whenever $0 \leq x \leq 1$ and $0 \leq y_1 < y_2 \leq 1$, we have $$(x \times y_1, x \times y_2) = \{x\} \times (y_1, y_2) = \left( I \times I \right) \ \cap \ \left(\{x\} \times (y_1, y_2) \right).$$

Whenever $0 \leq x_1 < x_2 \leq 1$, $0 \leq y_1 < 1$, and $0< y_2 \leq 1$, we have \begin{eqnarray*} (x_1 \times y_1, x_2 \times y_2) &=& \left( \{x_1\} \times (y_1, 1] \right) \cup \left( (x_1, x_2) \times I \right) \cup \left(\{x_2\} \times [0, y_2) \right) \\ &=& \left( I \times I \right) \ \ \cap \ \ \left[ \ \left( \{x_1\} \times (y_1, 2) \right) \ \cup \ \left( (x_1, x_2) \times \mathbb{R} \right) \ \cup \ \left( \{x_2\} \times (-y_2, y_2) \right) \ \right]. \end{eqnarray*}

Whenever $0 \leq x_1 < x_2 \leq 1$ and $0 \leq y_1 < 1$, we have \begin{eqnarray*} (x_1 \times y_1 , x_2 \times 0 ) &=& \left( \{x_1\} \times (y_1, 1] \right) \cup \left( (x_1, x_2) \times I \right) \\ &=& \left( I \times I \right) \ \ \cap \ \ \left[ \ \left( \{x_1\} \times (y_1, 2) \right) \ \cup \ \left( (x_1, x_2) \times \mathbb{R} \right) \ \right]. \end{eqnarray*}

Whenever $0 \leq x_1 < x_2 \leq 1$ and $0 < y_2 \leq 1$, we have \begin{eqnarray*} (x_1 \times 1, x_2 \times y_2) &=& \left( (x_1, x_2) \times I \right) \cup \left(\{x_2\} \times [0, y_2) \right) \\ &=& \left( I \times I \right) \ \ \cap \ \ \left[ \ \left( (x_1, x_2) \times \mathbb{R} \right) \ \cup \ \left( \{x_2\} \times (-y_2, y_2) \right) \ \right]. \end{eqnarray*}

And finally, whenever $0 \leq x_1 < x_2 \leq 1$, we have \begin{eqnarray*} (x_1 \times 1, x_2 \times 0) &=& \left( (x_1, x_2) \times I \right) \\ &=& \left( I \times I \right) \ \cap \ \left( \ (x_1, x_2) \times \mathbb{R} \ \right). \end{eqnarray*}

Have I proceeded correctly so far?

The set $\{0\} \times (0,1]$ is not open in the dictionary order topology on $I \times I$ because there is no basis element for this topology that contains the element $0 \times 1$ and is a subset of the set $\{0\} \times (0,1]$. Am I right?

However, the set $\{0\} \times (0,1]$ can be imagined as the intersection with $I \times I$ of the set $\{0\} \times (0,2)$, and the latter is open in $\mathbb{R} \times \mathbb{R}$ with the dictionary order topology. Am I right?

It's getting too late in the night; it's almost a quarter to one; so I'd continue from here later on in the day.

• @Brian M. Scott could you please have a look at my answer and then comment on any problems with it? Mar 9, 2015 at 12:00