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Today in lectures we were doing a brief review of some metric spaces stuff and I'm not quite sure about something we did:

If we define two metrics as $d_1(x,y)= \max_{i=1,\ldots,n} |x_i-y_i|$ and the metric $d_2=\displaystyle\sum_{i=1}^n |x_i-y_i|$ show that a set is open in $(\mathbb{R}^n,d_1)$ iff it is open in $(\mathbb{R}^n,d_2)$

So this is the same in $\mathbb{R}^n$ as in $\mathbb{R}^2$ so for simplicity I'm just looking at it in $\mathbb{R}^2$ just now.

So as $\max_{i=1,2} |x_i-y_i| \leq \displaystyle\sum_{i=1}^2 |x_i-y_i| \leq \sqrt{2\times\max_{i=1,2} |x_i-y_i|^2}=\sqrt{2} \max_{i=1,2} |x_i-y_i|$

Which gives $d_1(x,y)\leq d_2(x,y)\leq\sqrt{2}d_1(x,y)$

So I understand why this shows that any open set in $(\mathbb{R}^2,d_2(x,y))$ is open in $(\mathbb{R}^2,d_1(x,y))$ as we can make a smaller open ball round points but I can't see why this shows the other direction?

Thanks for any help

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Please: Don't write \mbox{max}. Just write \max. This not only prevents italicization, it also results in proper spacing before and after "$\max$" and in some contexts affects positioning of subscripts, thus: $\displaystyle\max_{x\in A}$ versus $\displaystyle\mbox{max}_{x\in A}$. I fixed this in the posted question. –  Michael Hardy Sep 21 '12 at 15:52
    
Are you familiar with the concept of equivalent norms? –  Inquest Sep 21 '12 at 15:54
    
@MichaelHardy Yes I normally would but it wasn't rendering properly on my computer and I didn't know if it was my browser being funny or something Thanks for the edit though –  hmmmm Sep 21 '12 at 15:54
    
@Inquest Not really, this is just the start of a toplology course that I am doing, so only in that the norms will be equivalent if the define the same topology. I am familiar with equivalent metrics though i.e. there is a homemorphism between them. So i guess I don't see why from this inequality we have a homemorphism between the two spaces –  hmmmm Sep 21 '12 at 15:59

2 Answers 2

up vote 1 down vote accepted

Suppose $d_1,d_2$ are two metrics on a set, and there is a constant $c>0$ such that for all points $x,y$ we have $d_1(x,y)\le c d_2(x,y)$. Consider a $d_1$-open ball $B=\{z : d_1(x,z)\le r\}$. If $z$ is in the $d_2$-open ball of radius $r/c$ about $x$, then $z$ is in $B$. Thus $B$ is also a $d_2$-open set. Thus every $d_1$-open ball is a $d_2$ open set, and therefore every $d_1$-open set is a $d_2$-open set.

The same proposition says that if there is also a constant $b>0$ such that for all $x,y$ we have $d_2(x,y)\le b d_1(x,y)$, then every $d_2$-open set is a $d_1$-open set.

The only thing you need to know about $c$ and $b$ is that both are positive.

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Thanks very much, I was just getting a little confused here. –  hmmmm Sep 21 '12 at 16:08
    
"Thus every $d_2$-open ball is a $d_1$ open set" This only shows every $d_2$-ball is contained in a $d_1$-ball. –  Andres Caicedo Sep 21 '12 at 16:10
    
@AndresCaicedo : Correct. I've fixed it. –  Michael Hardy Sep 21 '12 at 16:21

Assume that $md_1\leq d_2 \leq Md_1$ for some $m,M>0$.Define $B_i(x, \epsilon)=\{y\in\mathbb{R}^n:d_i(x,y)<\epsilon\}\;\;,\;i=1,2$.

Let say that $G$ is open in $(\mathbb{R}^n,d_2)$ and let $x \in G$. Exist $\epsilon>0$ s.t. $B_{2}(x,\epsilon))\subseteq G$. Then $B_1(x,\frac{\epsilon}{M}))\subseteq G$ so $G$ is open in $(\mathbb{R}^n,d_1)$.

Conversely if $G$ is open in $(\mathbb{R}^n,d_1)$ and $x \in G$ exist $\epsilon>0$ s.t. $B_{1}(x,\epsilon))\subseteq G$. Since $B_2(x,m\epsilon)\subseteq G \Longrightarrow G$ is open in $(\mathbb{R}^n,d_2)$.

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