# Proving a metric induces the product topology

Let $(M,d)$ and $(N,d')$ be metric spaces. Prove that the product topology is induced by the metric $d_1((x,y),(x',y')=d(x,x')+d(y,y')$ and $d_2((x,y),(x',y'))=\operatorname{max}\{d(x,y),d'(x',y')\}$.

I have to say, that I truly have no idea how to prove that a metric induces a certain topology, my guess here is to just consider the topology induced by the metric as well as the product topology and show that they are included in each other.

I don't have a complete proof, but I wrote a couple of things following that idea:

$(1)$ Let $T$ be the product topology and $T_i$ be the topology induced by the metric $d_i$.

A basis for the product topology of $M\times N$ contains sets of the form $U\times V$ where $U$ is open in $M$ and $V$ is open in $N$; and a basis for the topology induced by $T_1$ would be the sets of open balls according to the metric $d_1$, this is, the set of all open balls where each ball si given by $$B_{1}((x,y),r) = \{(x',y')\in M\times N:d_1((x,y),(x',y'))<r\}$$ In terms of the metric $d,d'$ the balls could be expressed as $$B_{1}((x,y),r) = \{(x',y')\in M\times N:d(x,y)+d'(x',y')<r\}$$

Which means I need to show that for $(x,y)\in U\times V$ open in $M\times N$ there is a ball such that $$(x,y)\in B_1((x,y),r)\subset U\times V$$

Now, what are exactly those open sets $U$ and $V$?. $M$ and $N$ are metric spaces which means that $U$ and $V$ are open balls according to their respective metric, then I could write $$U=B_{U}(x,\delta)=\{y\in M:d(x,y)<\delta\}$$ $$V=B_{V}(x',\epsilon)=\{y'\in N:d'(x',y')<\epsilon\}$$

I would like to pick $r$ in order that $\pi_{1}^{-1}(B_1((x,y),r))\subset U$ and $\pi_{2}^{-1}(B_1((x',y'),r))\subset V$. I have thought two different ways of working this part, but I don't know which of them is right

$(1a)$ One of the ideas to solve this problem was to set $r<\displaystyle\frac{1}{2}\operatorname{min}\{\delta,\epsilon\}$. Doing this I believe that from $d(x,y)+d'(x',y')<r$ I can conclude that $d(x,y)<\delta$ and $d'(x',y')<\epsilon$, therefore the ball is included in the set.

$(2a)$ On second thoughts, maybe $(1a)$ was excessive and considering that the set $U\times V$ would have a diameter of $\delta+\epsilon$ according to the metric $d_1$, maybe it would be enough just to sed $r<\delta+\epsilon$ so the open ball would lie inside $U\times V$?

Now I will assume the open ball is given, how can I pic $\delta$ and $\epsilon$ to have $U\times V$ inside the ball?.Taking $\delta=\epsilon < \frac{1}{2}r$ seems to be enough.

Does this prove that $T=T_1$?

$(2)$ Now the ball $B_2$ would be given by $$B_2((x,y),r)=\{(x',y'):\operatorname{max}\{d(x,x'),d'(y,y')\}<r\}$$

Here I don't know how to proceed, my problem with this is that I need to find a relation between $r$ and $\delta,\epsilon$ in order to define the inclusions as I did before, but I believe I need to know more about $d(x,x')$ and $d'(y,y')$ if I were to define $r$.

• TL:DR but I left a partial answer below :) May 5, 2014 at 6:49

my guess here just consider the topology induced by the metric as well as the product topology and show that they are included in each other.

Yes, exactly. As you noted, if $T$ is the product topology then an open set in $T$ is of the form $U \times V$ where $U$ is open in $(M,d)$ adn $V$ is open in $(N,d')$. To show $T=T_1$ you can show that

(i) for $(x,y) \in U \times V$ there exists $\delta$ such that $B_{d_1}((x,y),\delta ) \subseteq U \times V$

and

(ii) If $B_{d_1}((x,y), r)$ is an open ball (in $T_1$) and $(x',y') \in B_{d_1}((x,y), r)$ then there exists $U \times V$ in the product topology such that $(x',y') \in U \times V \subseteq B_{d_1}((x,y), r)$

The same two proofs with $d_1$ replaced by $d_2$ will yield $T = T_2$.

Now regarding $T = T_1$:

$(1a)$ One of the ideas to solve this problem was to set $r<\displaystyle\frac{1}{2}\operatorname{min}\{\delta,\epsilon\}$. Doing this I believe that from $d(x,y)+d'(x',y')<r$ I can conclude that $d(x,y)<\delta$ and $d'(x',y')<\epsilon$, therefore the ball is included in the set.

Yes that's also correct. This appears to be direction (ii) above: Let $U \times V$ be open in $T$ and $(x,y) \in U \times V$ and consider $B((x,y),r)$ where $r < {\min (\delta , \varepsilon) \over 2}$. Then for any $(x',y') \in B((x,y),r)$ we have $$d_1 ((x,y), (x',y')) = d(x,x') + d(y,y') < r$$

In particular, $d(x,x') < \delta$ and $d(y,y') < \varepsilon$ hence $x' \in U$ and $y' \in V$ and hence $(x',y') \in U \times V$, showing that $B((x,y),r) \subseteq U \times V$ hence $U \times V$ is open in $T_1$.

(2a)...

From the above you can see that $r = \delta + \varepsilon$ does not work.

Now I will assume the open ball is given, how can I pick $\delta$ and $\epsilon$ to have $U\times V$ inside the ball? Taking $\delta=\epsilon < \frac{1}{2}r$ seems to be enough.

Does this prove that $T=T_1$?

No but almost. Let $B_{d_1}((x,y),r)$ be an open ball in $T_1$ and $(x',y') \in B((x,y),r)$. You want to find an open set in $T$ that contains $(x',y')$ and is contained in $B((x,y),r)$. Let $R = {\min (r-d_1((x,y),(x',y'), d_1((x,y),(x',y'))\over 2}$ and consider the set $B_d(x', R) \times B_{d'}(y',R)$ (open in $T$). Then $d_1((x,y),(x',y')) = d(x,x') + d(y,y') < \min( r-d_1((x,y),(x',y'), d_1((x,y),(x',y'))) < r$.

(2)...

(2) is easier than (1): Note that if $\max (\delta, \varepsilon) < r$ then both $\delta < r$ and $\varepsilon < r$.

I hope this helps and if not leave me a comment.

• Thanks, but I'm not sure about the last part where you show that $B_{d}(x',r/2)\times B_{d'}(y',r/2)\subset B_{d_1}((x,y),r)$, what if $(x',y')$ is really close to the boundary of $B_{d_1}((x,y),r)$?, this is, looks like there is no restriction to have taken $(x',y')$ such that $\frac{1}{2}<d_1((x,y),(x',y')<r$ and then in particular $B_d(x',\frac{1}{2}r)$ wouldn't be a subset of $U$. I was considering it in a way that $U\times V$ would be also centered in $B_{d_1}$, something like $B_{d}(x,\frac{1}{2})\times B_{d'}(y,\frac{1}{2}r)$.
– Cure
May 5, 2014 at 11:58
• @Dante You're absolutely right, I was a bit in a hurry when I wrote this. The $r/2$ is a typo, let me correct it (the claim is still true of course). May 5, 2014 at 12:05

To prove that $d$ induces the product topo on $A \times B$, you want to show that every open set in $A\times B$ is a union of open balls $B_d(x_0, r) = \{ x \in A\times B$ such that $d(x,x_0) \lt r\}$. Well every open set in $A\times B$ is a union of some basis open sets $U_i \times V_i$, you said. And each $U_i, V_i$ is a union of open balls in the respective two given topologies. So we have $U_i = \bigcup_{j \in J} B_1(a_j, r_j), \ \ V_i = \bigcup_{k \in K} B_2(b_k, s_k)$. But taking the cartesian product of arbitrary unions is a union of cartesian products, of, in this case, products of the form $B_1 (a_j, r_j) \times B_2 (b_k, s_k)$. So given a point in the latter, it suffices to show that we can fit a $d$-ball in there, and hence in the larger union. Well, let $x_0 = (a,b)$ be the point. Choose $\epsilon \gt 0$ such that $B_d(x_0, \epsilon) = \{ (x,y) : d((x,y), (a,b)) \lt \epsilon\}$ is contained in that product. Let's look at the defining property of the ball more explicitly: $$d((x,y), (a,b)) \lt \epsilon \\ \iff \\ d_1(x,a) + d_2(y,b) \lt \epsilon$$ We want the above to imply that $(x,y) \in B_1 \times B_2$ (above), so simply choose the minimum radius of the two balls $B_1, B_2$, i.e. $\min\{r_j, s_k\} = \epsilon$, then by non-negativity of metrics you have that $(x,y)$ lies in the product of the two balls.

Since every point $z$ of every open set in $A\times B$, has such a $d$-ball around it, contained completely within the open set, then the open set is clearly an arbitrary union of those balls, and thus the topology is induced by that metric.