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I am a bit confused by this whole question I have in front of me. It defines a distance, $d$, on a product topology $X= \Pi_i X_i $, where $\Pi_iU_i$ forms a basis of open sets and $U_i=X_i$ except for finitely many $i$'s and $d_i$ preserves the topology on $X_i$. I need to show that the distance defined by $d$ coincides with the product topology.

I guess I'm very confused, I know I've asked a similar question, but can someone explain this so I can understand what I need to do? Please include as much detail as possible.

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Are you given a specific function $d$? If so, we need to know what it is. If not, we need to know exactly what information about $d$ you were given. –  Brian M. Scott Nov 5 '12 at 2:40
    
Well I'm avoiding giving details for a reason, but yeah one property of the distance is that $d_i(x,y) \le 1$. The distance is also an inifinite sum of the $d_i$'s –  Atreyu Nov 5 '12 at 2:43
    
If you don’t give those details, we can give only the most general guidance. –  Brian M. Scott Nov 5 '12 at 2:45
    
I kind of suspected as much, though $d_i$ might also have been $\min\{\rho_i,1\}$; if you get stuck on the details, don’t hesitate to ask. –  Brian M. Scott Nov 5 '12 at 3:00
    
Actually to be honest I don't really understand how I am supposed to go about doing this. Do I first take an open ball w.r.t $d_i$ and show that I can find an open ball w.r.t. $d$ contained in this ball, and then vice-versa (hopefully that makes sense)? Thanks very much for your help by the way. –  Atreyu Nov 5 '12 at 3:09

1 Answer 1

up vote 3 down vote accepted

You have a base $\mathscr{B}$ for the product topology. You know that $\mathscr{B}_m=\{B(x,r):x\in X\text{ and }r>0\}$ is a base for the metric topology. One way to show that the topologies are identical is to show that every member of $\mathscr{B}$ is a union of members of $\mathscr{B}_m$ and vice versa. The usual way to show that $B\in\mathscr{B}$ is a union of members of $\mathscr{B}_m$ is to show that for each $x\in B$ there is an $r>0$ such that $B(x,r)\subseteq B$. Similarly, the usual way to show that each $B(x,r)\in\mathscr{B}_m$ is a union of members of $\mathscr{B}$ is to show that for each $y\in B(x,r)$ there is a $B_y\in\mathscr{B}$ such that $y\in B_y\subseteq B(x,r)$. There’s not much more that can be said without some specifics concerning the spaces and metrics involved.

Added: Let $B\in\mathscr{B}$. In other words, there are open sets $U_i\in X_i$ for $i\in\Bbb N$ such that $B=\prod_iU_i$, and $F=\{i\in\Bbb N:U_i\ne X_i\}$ is finite. We want to show that $B$ is open in the metric topology induced by $d$, and the most straightforward way to do this is to show that for each $x\in B$ there is an $s_x>0$ such that $B(x,s_x)\subseteq B$: then we’ll have $B=\bigcup_{x\in B}B(x,s_x)$, which is certainly open in the metric topology.

So let $x\in B$. Then $x_i\in U_i$ for each $i\in\Bbb N$. In most cases this doesn’t say much, since in most cases $U_i=X_i$, but for $i\in F$ it does actually restrict $x_i$. Now $U_i$ is an open set in $X_i$, so there is an $r_{i,x}>0$ such that $B_{d_i}(x_i,r_{i,x})\subseteq U_i$. It’s a bit messy having all these different radii lying about, so let $r_x=\min\{r_{i,x}:i\in F\}$; then $B_{d_i}(x_i,r_x)\subseteq U_i$ for each $i\in F$.

The idea now is to find a radius $s_x$ such that if $d(x,y)<s_x$, then $d_i(x_i,y_i)<r_x$ for each $i\in F$. Why? Because then

$$\begin{align*} y\in B(x,s_x)&\Rightarrow d(x,y)<s_x\\ &\Rightarrow\forall i\in F\big(d(x_i,y_i)<r_x\big)\\ &\Rightarrow\forall i\in F\big(y_i\in B_{d_i}(x_i,r_x)\subseteq U_i\big)\\ &\Rightarrow\forall i\in\Bbb N\big(y_i\in U_i)\\ &\Rightarrow y\in B\;, \end{align*}$$

and we’ll have shown that $B(x,s_x)\subseteq B$, as desired.

Recall that $$d(x,y)=\sum_{i\in\Bbb N}\frac{d_i(x_i,y_i)}{2^i}\;,$$

and consider a particular $k\in\Bbb N$: how big can $d_k(x_k,y_k)$ be compared with $d(x,y)$? Even if $d_i(x_i,y_i)=0$ for all $i\in\Bbb N\setminus\{k\}$, at worst we’ll still have $$\frac{d_k(x_k,y_k)}{2^k}\le d(x,y)\;,$$ or $d_k(x_k,y_k)\le 2^kd(x,y)$. Thus, if we had $d(x,y)<\dfrac{r_x}{2^k}$, we’d be able to conclude that $d_k(x_k,y_k)\le 2^kd(x,y)<r_x$. To make this happen for each $k\in F$, let $m=\max F$ and set $s_x=\dfrac{r_x}{2^m}$. Now let $y\in B(x,s_x)$. Then for each $k\in F$ we have

$$d_k(x_k,y_k)\le 2^kd(x,y)<2^ks_x=2^k\cdot\frac{r_x}{2^m}=\frac{2^k}{2^m}r_x\le r_x$$ (since $m\ge k$), and therefore $y_k\in B_{d_k}(x_k,r_x)\subseteq U_k$.

Taking a deep breath and looking back, we see that we’ve just shown that if $y\in B(x,s_x)$, then $y\in B$ and hence that $B(x,s_x)\subseteq B$, which is what we set out to do several paragraphs ago.


That’s actually the harder direction, I think. You should have a bit less work proving that if $y\in B(x,r)$ for some $x\in X$ and $r>0$, then there is a $B_y\in\mathscr{B}$ such that $y\in B_y\subseteq B(x,r)$. The key idea is that in order to make $d(z,y)$ small, you need only ensure that $d_i(z_i,y_i)$ is small enough on a large enough finite set of coordinates: by taking $m$ large enough, you can make the contribution to $d(z,y)$ from the coordinates with $i\ge m$ as small as you like.

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@Atreyu: What I’ve just posted may be a bit more complicated than it absolutely had to be: it’s been a long day, and I’m a bit tired. But at worst it will give you something to start with. –  Brian M. Scott Nov 5 '12 at 4:03
    
No worries thanks for your help –  Atreyu Nov 5 '12 at 4:18

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