0
$\begingroup$

Here is Prob. 3, Sec. 21, in the book Topology by James R. Munkres, 2nd edition:

Let $X_n$ be a metric space with metric $d_n$, for $n \in \mathbb{Z}_+$.

(a) Show that $$ \rho ( \mathbf{x}, \mathbf{y} ) = \max \left\{ \ d_1 \left( x_1, y_1 \right), \ldots, d_n \left( x_n, y_n \right) \ \right\} $$ is a metric for the product space $X_1 \times \cdots \times X_n$.

(b) Let $\bar{d}_i = \min \left\{ \ d_i, 1 \ \right\}$. Show that $$ D ( \mathbf{x}, \mathbf{y} ) = \sup \left\{ \ \bar{d}_i \left( x_i, y_i \right) / i \ \right\} $$ is a metric for the product space $\prod X_i$.

Here is the link to my Math SE post on Part (a).

Prob. 3 (a), Sec. 21, in Munkres' TOPOLOGY, 2nd ed: A Finite Cartesian Product Of Metrizable Spaces Is Metrizable

So here I will only be attempting Part (b).

My Attempt:

Part (b)

Let $B \colon= \prod_{i=1}^\infty U_i $ be an arbitrary basis element for the product topology on $\prod_{i=1}^\infty X_i$, where, for each $i \in \{ \ 1, 2, 3, \ldots, \ \}$, the set $U_i$ is a subset of $X_i$ such that $U_i$ is open in $\left( X_i, d_i \right)$, and $U_i \not= X_i$ for at most finitely many values of $i$; let $i \in \left\{ \ i_1, \ldots, i_n \ \right\}$ be the values of $i$ for which $U_i \neq X_i$. Let $\mathbf{p} \colon= \left( p_1, p_2, p_3, \ldots \right)$ be an arbitrary element of $B$.

Then, for each $i \in \left\{ \ i_1, \ldots i_n \ \right\}$, $p_i \in U_i$ and $U_i$ is open in the metric space $\left( X_i, d_i \right)$, which implies that there is a real number $\delta_i > 0$ such that the set $$B_{d_i} \left( p_i, \delta_i \right) \colon= \left\{ \ x \in X_i \colon \ d_i \left( x, p_i \right) < \delta_i \ \right\}$$ is contained in $U_i$. Let us assume that $\delta_i \in (0, 1)$ for each $i \in \left\{ \ i_1, \ldots, i_n \ \right\}$.

Let $$\delta \colon= \min \left\{ \ \frac{\delta_{i_1}}{i_1}, \ldots, \frac{\delta_{i_n}}{i_n} \ \right\}.$$ Then $\delta > 0$ of course.

Moreover, if $\mathbf{x} \colon= \left( x_1, x_2, x_3, \ldots \right)$ is any point in $\prod_{i=1}^\infty X_i$ such that $$\mathbf{x} \in B_{D} ( \mathbf{p}, \delta ) \colon= \left\{ \ \mathbf{y} \in \prod_{i=1}^\infty X_i \ \colon \ D ( \mathbf{y}, \mathbf{p} ) < \delta \ \right\}, $$ then for each $i \in \{ \ i_1, \ldots, i_n \ \}$, we have $$ \frac{\bar{d}_i \left( x_i, p_i \right)}{i} \leq D ( \mathbf{x}, \mathbf{p} ) < \delta \leq \frac{\delta_i}{i}, $$ and so $$ \bar{d}_i \left( x_i, p_i \right) < \delta_i < 1, $$ and hence $$ d_i \left( x_i, p_i \right) = \bar{d}_i \left( x_i, p_i \right) < \delta_i,$$ which implies that $x_i \in B_{d_i} \left( p_i, \delta_i \right)$ and hence $x_i \in U_i$. So $\mathbf{x} \in B$. Moreover, $\mathbf{p} \in B_{D} ( \mathbf{p}, \delta)$.

Thus, we have shown that, for any basis set $B$ for the product topology on $\prod_{i=1}^\infty X_i$ and for any element $\mathbf{p} \in B$, there is a basis element $B^\prime \colon= B_{D} ( \mathbf{p}, \delta)$ for the $D$-metric topology on $\prod_{i=1}^\infty X_i$ such that $$ \mathbf{p} \in B^\prime \subset B. $$ Therefore the $D$-metric topology is finer than the product topology on $\prod_{i=1}^\infty X_i$.

Is this part of the proof correct? Is the presentation easy to follow?

Now let $B_{D}(\mathbf{p}, \epsilon )$ be an arbitrary basis element for the $D$-metric topology on $\prod_{i=1}^\infty X_i$, where $\mathbf{p} \colon= \left( p_1, p_2, p_3, \ldots \right)$ is an arbitrary point of $\prod_{i=1}^\infty X_i$ and $\epsilon$ is a positive real number.

Now for each point $\mathbf{x} \in B_{D} ( \mathbf{p}, \epsilon)$, we need to find a basis element $B_{\mathbf{x}}$ for the product topology on $\prod_{i=1}^\infty X_i$ such that $$\mathbb{x} \in B_{\mathbf{x}} \subset B_{D}( \mathbf{p}, \epsilon). \tag{1} $$

Let $\mathbf{x} \colon= \left( x_1, x_2, x_3, \ldots \right)$ be an arbitrary point in $B_D ( \mathbf{p}, \epsilon)$. Then $\mathbf{x} \in \prod_{i-1}^\infty X_i$ and $D( \mathbf{x}, \mathbf{p} ) < \epsilon$; so for any real number $\delta$ such that $0 < \delta < \epsilon - D( \mathbf{x}, \mathbf{p})$, we have the open ball $B_D ( \mathbf{x}, \delta)$ such that $$ B_D ( \mathbf{x}, \delta) \subset B_D ( \mathbf{p}, \epsilon ). \tag{2} $$

Let $n$ be a natural number such that $n > 2/\delta$, and then let $$ B_{\mathbf{x}} \colon= B_{d_1} \left( x_1, \frac{\delta}{2} \right) \times \cdots \times B_{d_n} \left( x_n, \frac{\delta}{2} \right) \times X_{n+1} \times X_{n+2} \times \cdots. \tag{A} $$ As every open ball in a metric space is an open set, so this set $B_{\mathbf{x}}$ is a basis element for the product topology on $\prod_{i=1}^\infty X_i$; and $\mathbf{x} \in B_{\mathbf{x}}$ of course.

Moreover, if $\mathbf{y} \colon= \left( y_1, y_2, y_3, \ldots \right) \in B_{\mathbf{x}}$, then for each $i \in \{ \ 1, \ldots, n \ \}$, we have $$ \frac{ \bar{d}_i \left( y_i, x_i \right) }{ i } \leq \bar{d}_i \left( y_i, x_i \right) \leq d_i \left( y_i, x_i \right) < \frac{\delta}{2}; \tag{3} $$ and for $i \in \{ \ n+1, n+2, n+3, \ldots \ \}$, we have $$ \frac{ \bar{d}_i \left( y_i, x_i \right) }{ i } \leq \frac{1}{i} \leq \frac{1}{n} < \frac{\delta}{2}. \tag{4} $$ Since we have chosen $n$ to be a natural number such that $n > 2/\delta$, so we have the inequality $$ \frac{1}{n} < \frac{\delta}{2},$$ which we have used in (4).

From (3) and (4) we can conclude that $$ D ( \mathbf{y}, \mathbf{x} ) = \sup \left\{ \ \frac{ \bar{d}_i \left( y_i, x_i \right) }{i} \ \colon \ i \in \mathbf{N} \ \right\} \leq \frac{\delta}{2} < \delta, $$ which shows that $\mathbf{y} \in B_D ( \mathbf{x}, \delta)$ and hence $ \mathbf{y} \in B_D( \mathbf{p}, \epsilon)$, by virtue of (2) above.

Since $\mathbf{y}$ was an arbitrary point of the set $B_{\mathbf{x}}$ as defined in (A) above, therefore we can conclude $B_{\mathbf{x}} \subset B_D ( \mathbf{p}, \epsilon)$, and that (1) holds with this set $B_{\mathbf{x}}$.

So the product topology on $\prod_{i-1}^\infty X_i$ is finer than the $D$-metric topology.

Is this part of the proof the right one in terms of its logic and presentation?

Hence these two topologies on $\prod_{i=1}^\infty X_i$ are the same.

Is my proof sound enough? Is every step in both parts of this proof correct? Is the presentation accessible enough too, especially for a student who might be studying topology for the first time?

$\endgroup$
  • $\begingroup$ If the main purpose of the question is asking for comments on your proof and its correctness (as opposed to asking for any solution of the given problem), you should probably add (proof-verification) tag. $\endgroup$ – Martin Sleziak Oct 1 '17 at 13:29
  • $\begingroup$ @MartinSleziak thank you for your advice. I've just added the tag. $\endgroup$ – Saaqib Mahmood Oct 1 '17 at 14:33

Your Answer

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

Browse other questions tagged or ask your own question.