1. ## Metric Spaces

Let $\displaystyle X=\bf R^3$ and define the metric d: $\displaystyle \bf R^3 \times \bf R^3 \rightarrow \bf R$ by
$\displaystyle d((x_1,y_1,z_1),(x_2,y_2,z_2)) = \max\left\{|x_1-x_2|, |y_1-y_2|, |z_1-z_2|\right\}$.
Describe the neighborhood $\displaystyle N(0; 1)$, where $\displaystyle 0$ is the origin in $\displaystyle \bf R^3$

My professor says that it's an open cube with vertices at (±1, ±1, ±1), but I can't see why it's not an open sphere with radius 1.

I'm thinking $\displaystyle N(0; 1) = \left\{y \in X: d(0,y) <1\right\}$. Wouldn't there be points whose distance from the origin exceed 1 in an open cube?

2. ## Re: Metric Spaces

"Distance" in this case means the metric function, not "ordinary distance". Let's take a point "as far away as possible" from the origin, say (1,1,1).

Then d((0,0,0),(1,1,1)) = max{|0 -1|,|0-1|,|0-1|} = max{|-1|,|-1|,|-1|} = max{1,1,1} = 1.

Convince yourself that if (x,y,z) is any point with two of the 3 coordinates with absolute value less than 1, and the third coordinate with absolute value equal to 1, that d((0,0,0),(x,y,z)) = 1, as well.

Hopefully this allows you to see that the boundary of N(0,1) is precisely the faces of said cube, so all the interior points comprise N(0,1).

3. ## Re: Metric Spaces

Let $\displaystyle X=\bf R^3$ and define the metric d: $\displaystyle \bf R^3 \times \bf R^3 \rightarrow \bf R$ by
$\displaystyle d((x_1,y_1,z_1),(x_2,y_2,z_2)) = \max\left\{|x_1-x_2|, |y_1-y_2|, |z_1-z_2|\right\}$.
Describe the neighborhood $\displaystyle N(0; 1)$, where $\displaystyle 0$ is the origin in $\displaystyle \bf R^3$
My professor says that it's an open cube with vertices at (±1, ±1, ±1), but I can't see why it's not an open sphere with radius 1.
I'm thinking $\displaystyle N(0; 1) = \left\{y \in X: d(0,y) <1\right\}$. Wouldn't there be points whose distance from the origin exceed 1 in an open cube?
Consider the vertex $V=(-1,1,-1)$ that is not in the "open cube" ${\frak{B}}(0;1)$$= \{X| d(0,X)<1\} In that metric d(0,V)=1. Correct? So V is on the boundary of that cube. SO? 4. ## Re: Metric Spaces Originally Posted by Plato Consider the vertex V=(-1,1,-1) that is not in the "open cube" {\frak{B}}(0;1)$$ = \{X| d(0,X)<1\}$

In that metric $d(0,V)=1$. Correct? So $V$ is on the boundary of that cube. SO?
So any vertex V such that $\displaystyle d(0,V)=1$ is not in $\displaystyle N(0;1)$. Which is why $\displaystyle N(0;1)$is an open cube with all such vertices.

5. ## Re: Metric Spaces

Originally Posted by Deveno
Convince yourself that if (x,y,z) is any point with two of the 3 coordinates with absolute value less than 1, and the third coordinate with absolute value equal to 1, that d((0,0,0),(x,y,z)) = 1, as well.

Hopefully this allows you to see that the boundary of N(0,1) is precisely the faces of said cube, so all the interior points comprise N(0,1).
I can see it now. Thanks for clearing that up.