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 $r$.
Best Answer
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 $T = T_1$:
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 $T_1$.
From the above you can see that $r = \delta + \varepsilon$ does not work.
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) 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.