5
$\begingroup$

Here's Prob. 2, Sec. 20 in the book Topology by James R. Munkres, 2nd edition:

Show that $\mathbb{R}\times \mathbb{R}$ in the dictionary order topology is metrizable.

The dictionary order on the set $\mathbb{R} \times \mathbb{R}$ is defined as follows:

For any two points $x_1\times y_1$, $x_2 \times y_2$ $\in \mathbb{R} \times \mathbb{R}$, we define $$x_1 \times y_1 \prec x_2 \times y_2$$ if and only if either $x_1 < x_2$, or if $x_1 = x_2$ and $y_1 < y_2$.

Now the dictionary order topology on $\mathbb{R} \times \mathbb{R}$ is the one having as a basis all sets of the fomm $$\left( a \times b, a \times c \right) \colon= \left\{ \ x \times y \in \mathbb{R} \times \mathbb{R} \ \colon \ a \times b \prec x \times y \prec a \times c \ \right\},$$ where $a, b, c \in \mathbb{R}$ such that $b < c$.

Now if we define the funtion $d \colon \left(\mathbb{R} \times \mathbb{R} \right) \times \left( \mathbb{R} \times \mathbb{R} \right) \to \mathbb{R}$ as $$d\left( x_1 \times y_1, x_2 \times y_2 \right) \colon = \begin{cases} 1 \ \mbox{ if } \ x_1 \neq x_2; \\ \min \left(\ \vert y_1 - y_2 \vert, \ 1 \ \right) \ \mbox{ otherwise }, \end{cases} $$ then is this function $d$ a metric? How to verify the triangle inequality?

Does this function give the dictionary order topology on $\mathbb{R} \times \mathbb{R}$?

If $x_1 = x_2 = x_3$, then we have $$ \begin{align} & \ d\left(x_1 \times y_1, x_3 \times y_3 \right) \\ &= \min\left( \vert y_1 - y_3 \vert, \ 1 \right) \\ &\leq \min\left( \vert y_1 - y_2 \vert, \ 1 \right) + \min\left( \vert y_2 - y_3 \vert, \ 1 \right) \\ & \ \ \mbox{ [using the fact that this minimum is the same as the standard bounded metric on $\mathbb{R}$]} \\ &= d\left(x_1 \times y_1, x_2 \times y_2 \right) + d\left(x_2 \times y_2, x_3 \times y_3 \right). \end{align} $$ If $x_1 \neq x_2$ and $x_2 \neq x_3$, then we have $$ \begin{align} d\left(x_1 \times y_1, x_3 \times y_3 \right) &\leq 1 < 1 + 1 = d\left(x_1 \times y_1, x_2 \times y_2 \right) + d\left(x_2 \times y_2, x_3 \times y_3 \right). \end{align} $$ If $x_1 = x_2$ and $x_2 \neq x_3$, then $x_1 \neq x_3$ either, and so $$ \begin{align} d\left(x_1 \times y_1, x_3 \times y_3 \right) &= 1 \\ &\leq d\left(x_1 \times y_1, x_2 \times y_2 \right) + 1 \\ &= d\left(x_1 \times y_1, x_2 \times y_2 \right) + d\left(x_2 \times y_2, x_3 \times y_3 \right). \end{align}$$ And, similarly for the case when $x_1 \neq x_2$ and $x_2 = x_3$.

Is this demonstration of the triangle inequality complete and correct?

PS:

Assuming that the above $d$ is a metric, here is my attempt at showing that the dictionary order topology on $\mathbb{R} \times \mathbb{R}$ is indeed the one induced by the metric $d$ above.

Let $$B \colon= \{ a \} \times (b, c) = \{ \ a \times t \in \mathbb{R} \times \mathbb{R} \ \colon \ b < t < c \ \} = ( a \times b, a \times c) $$ be a basis element for the dictionary order topology on $\mathbb{R} \times \mathbb{R}$, and let $x \times y \in B$. Then of course $x = a$ and $b < y < c$. Let us put $$ \epsilon \colon= \min \{ \ y-b, c-y, 1 \}. $$ Then if $s \times t \in B_d ( x \times y, \epsilon)$, then $s \times t \in \mathbb{R} \times \mathbb{R}$, and $$ d( s \times t, x \times y ) < \epsilon. \tag{1}$$ and, as $\epsilon \leq 1$, so $$ d( s \times t, x \times y ) < 1, $$ which implies that $s = x$, that is, $s = a$, and also from (0) we can conclude that $$ d( s \times t, x \times y ) = \min \{ \ \lvert t-y \rvert, 1 \ \}. $$ Then (1) implies that $$ d( s \times t, x \times y ) = \min \{ \ \lvert t-y \rvert, 1 \ \} < \epsilon = \min \{\ y - b, c - y , 1 \ \}. $$ So $$ d( s \times t, x \times y ) = \lvert t-y \rvert < \min \{ \ y-a, b-y \} , $$ The last relation implies that $b < t < c$. So $s \times t \in B$.

Thus for any basis set $B$ for the dictionary order topology and for any element $x \times y \in B$, we have a basis element $B_d ( x \times y, \epsilon)$ for the $d$-metric topology such that $$ x \times y \in B_d ( x \times y, \epsilon ) \subset B. $$ So the $d$-metric topology is finer than the dictionary order topology on $\mathbb{R} \times \mathbb{R}$.

Now let us consider an open ball $B_d( a \times b, \epsilon )$, where $a \times b \in \mathbb{R} \times \mathbb{R}$ and $\epsilon > 0$ are arbitrary. Let $x \times y \in B_d ( a \times b, \epsilon )$. Then if we choose a real number $\delta$ such that $$ 0 < \delta < \min \{ \ \epsilon - d( a \times b, x \times y), \ 1 \ \}, $$ then we note that $\delta < 1$ and also that $$ B_d ( x \times y, \delta ) \subset B_d ( a \times b, \epsilon ). \tag{2} $$

Now let us put $$ B \colon= \{ \ x \ \} \times ( y-\delta, y + \delta) = \big( \ x \times (y-\delta), \ x \times (y+ \delta) \ \big). $$ Then this $B$ is a basis element for the dictionary order topology on $\mathbb{R} \times \mathbb{R}$ such that $x \times y \in B$.

Moreover, if $s \times t \in B$, then $s = x$ and $y-\delta < t < y+\delta$ and hence $\lvert t-y \rvert < \delta$. But $\delta < 1$. So $$ d( s \times t, x \times y ) = \min \{ \lvert t-y \rvert, 1 \} = \lvert t-y \rvert < \delta, $$ which implies that $ s \times t \in B_d( x \times y, \delta )$ and hence also that $s \times t \in B_d( a \times b, \epsilon)$ by virtue of (2) above. Therefore $B \subset B_d( a \times b, \epsilon)$.

Thus we have shown that for any basis element $B_d( a \times b, \epsilon)$ for the $d$-metric topology and for any element $x \times y \in B_d( a \times b, \epsilon)$, there is a basis element $B$ for the dictionary order topology on $\mathbb{R} \times \mathbb{R}$ such that $$ x \times y \in B \subset B_d( a \times b, \epsilon). $$ Thus the dictionary order topology is finer than the $d$-metric topology.

The preceding few paragraphs show that the $d$- metric topology is the same as the dictionary order topology on $\mathbb{R} \times \mathbb{R}$.

Is this proof correct? Is each and every step of it correct in its logic and presentation? If not, then where lies the problem?

$\endgroup$
  • $\begingroup$ I have never seen the topology book by Munkres, but it must be very popular, because many questions on this site refer to it. Tell me, does Munkres really use the strange notation $x\times y$ instead of the normal $(x,y)$ for an element of $\mathbb R\times\mathbb R?$ $\endgroup$ – bof Sep 26 '17 at 9:17
  • $\begingroup$ @bof yes, he uses $x \times y$ as well as $(x,y)$, as the latter can also mean an open interval. $\endgroup$ – Saaqib Mahmood Sep 26 '17 at 13:09
  • $\begingroup$ @SaaqibMahmuud Does this book has a solution manual ? You should find one, if available, since many of your questions go unanswered. $\endgroup$ – A---B Sep 28 '17 at 19:32
  • $\begingroup$ @A---B yes, this book does have a solution manual, but sometimes the solutions there are not clear enough to me. That's why I try to come up with my own attempts and show my work to the Math SE community. Very often my questions do get answered, I must also acknowledge. $\endgroup$ – Saaqib Mahmood Sep 29 '17 at 5:32
  • $\begingroup$ im here...,,,,,,sir@ saaqib mahmood sir $\endgroup$ – user469754 Mar 27 '18 at 14:02
5
$\begingroup$

Here's an outline:

It might be enlightening to convince yourself that the dictionary order topology on $\mathbb{R}\times\mathbb{R}$ is the homeomorphic to the disjoint union of continuum many copies of $\mathbb{R}$.

One may check, and my recollection is that Munkres does this, that if $d$ is a metric, then $d' = \min(d,1)$ is a metric inducing the same topology. Thus, as far as the topology of metric spaces is concerned, it is entirely sufficient to consider metrics which are $\leq 1$.

Now, suppose that $(X_\alpha)_{\alpha \in I}$ are metric spaces and that the corresponding metrics $d_\alpha$ all have $d_\alpha \leq 1$. Form the disjoint union $X = \bigsqcup_{\alpha \in I} X_\alpha$ and give it its natural topology (open sets in $X$ are disjoint unions $\bigsqcup_{\alpha \in I} U_\alpha$ where $U_\alpha$ is open in $X_\alpha$). You may find it less distracting to check in this setting that $d$ defined by $d(x,y) = \begin{cases} d_\alpha(x,y) & \text{ if } x,y \in X_\alpha \\ 1 & \text{ otherwise} \end{cases}$ is a metric on $X$ and induces the aforementioned topology.

To put it in a slogan: "the disjoint union of metrizeable spaces is metrizeable".

$\endgroup$
  • $\begingroup$ what do you mean by disjoint union? Are these metric spaces $X_\alpha$ disjoint? Or, in the case of a point of intersection of two distinct spaces, we ought to consider it to be in either but not both? I'm not familiar with the latter formulation, which Munkres does not discuss either. $\endgroup$ – Saaqib Mahmood Feb 15 '16 at 8:21
  • 1
    $\begingroup$ If $(X_\alpha)_{\alpha \in I}$ is a collection of sets (not necessarily disjoint), indexed by a set $I$, then one can make sense of their disjoint union. There are various ways to make this precise, but the point is just to do some operation on the collection to "disjointify" it before taking the union. A common thing is to define $\bigsqcup_{\alpha \in I} X_\alpha := \bigcup_{\alpha \in I} X_\alpha \times \{\alpha\}$. The elements of this set look like pairs $(x,\gamma)$, so I can always use the 2nd coordinate to figure out which set the element is supposed to belong to. $\endgroup$ – Mike F Feb 15 '16 at 8:28
  • $\begingroup$ Now, if you have a collection of topological spaces $(X_\alpha)_{\alpha \in I}$, there is a natural way to put a topology on the disjoint union $\bigsqcup_{\alpha \in I} X_\alpha$. I mentioned briefly how to do this in my answer, but you'll get a more fleshed out version by googling "disjoint union topology". In a sense, this construction is like a generalization of the "discrete topology" which you are probably familiar with. Every individual space $X_\alpha$ ends up being open in the disjoint union (like points in a discrete space) but now there is also topology on each piece. $\endgroup$ – Mike F Feb 15 '16 at 8:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.