0

Let $(X_1,d_1)$ and $(X_2,d_2)$ be two metric spaces. Define a new metric space $X=X_1 \times X_2$, such that for $x=(x_1,x_2)$, $y=(y_1,y_2)$, we have

$$d(x,y)=\sqrt{d_1(x_1,x_2)^2+d_2(y_1,y_2)^2}$$

I cannot decide whether $d(x,y) \leq d(x,z)+d(z,y)$ or not. I think that we cannot know it.

and $d(x,y)=0$ if and only if $x=y$.but if we solve,then $x_1=x_2$ and $y_1=y_2$. But this doesn't show that $x=(x_1,x_2)=(y_1,y_2)=y$ because of $x_1$ not equal to $y_1$. Because given information is not enough.So this may not be metric.

Am I right?

Ian Coley
  • 6,000
1214
  • 11
  • 3
    Are you sure you've written your new metric properly? We cannot take $d_1(x_1,x_2)$ since we do not know how to find the distance between two points not lying in the same metric space. – Ian Coley Mar 01 '14 at 16:42
  • we don't know it but we know d1 and d2 are metric.we can use this – 1214 Mar 01 '14 at 16:44
  • Ian was saying that you probably need that $$d(x,y) = \sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}.$$ – Cameron Williams Mar 01 '14 at 16:46
  • I think you don't take my meaning. Suppose I give you a metric on $\mathbb C$ and $\mathbb R$ and I tell you to find the distance between $1+i\in\mathbb C$ and $\sqrt 2\in\mathbb R$. You can't tell me what that is. Similarly, you've tried to put a metric between metric spaces here. I think you should have $$ d(x,y)=\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2} $$ – Ian Coley Mar 01 '14 at 16:46
  • I would have expected to read $d_1(x_1,y_1)$ and $d_2(x_2,y_2)$, and that would have made it a metric. The current formula is not all well-defined - it doesn't make sense. – Arno Mar 01 '14 at 16:47
  • I see,but in my question (x₁,x₂)∈ X₁×X₂, x₁∈X₁ and x₂∈X₂ – 1214 Mar 01 '14 at 16:53
  • @IanColey But you can find the distance between $\sqrt 2$ and $1+i$ because $\mathbb R\subset\mathbb C$. – homegrown Mar 01 '14 at 17:05
  • No, because I am not telling you that the metric on $\mathbb R$ is inherited from the metric on $\mathbb C$. Perhaps I put the discrete metric on $\mathbb R$ but the usual distance metric on $\mathbb C$. – Ian Coley Mar 01 '14 at 17:27

1 Answers1

0

It's easy to prove using the Minkowski inequality $(M)$:

$d(x,y)+d(y,z)=\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}+\sqrt{d_1(y_1,z_1)^2+d_2(y_2,z_2)^2}\stackrel{(M)}{\geq}\sqrt{(d_1(x_1,y_1)+d_1(y_1,z_1))^2+(d_2(x_2,y_2)+d_2(y_2,z_2))^2}\geq\sqrt{d_1(x_1,z_1)^2+d_2(x_2,z_2)^2}=d(x,z)$.

Additionally, $d(x,y)=\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}=\sqrt{d_1(y_1,x_1)^2+d_2(y_2,x_2)^2}=d(y,x)$

And $d(x,y)=0\Longleftrightarrow\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}=0\Longleftrightarrow d_1(x_1,y_1)=0\land d_2(x_2,y_2)=0\Longleftrightarrow{x_1=y_1}\land x_2=y_2\Longleftrightarrow x=y.$

user2345215
  • 16,422