1. ## Metric Space

The topologies on $\displaystyle \mathbb{R}^n$ induced by the euclidean metric d and the square metric rho are the same as the product topology on $\displaystyle \mathbb{R}^n$.

Euclidean metric
$\displaystyle d(x,y)=\sqrt{(x_1-y_1)^2+\cdots +(x_n-y_n)^2}$

Square metric
$\displaystyle \rho(x,y)=\max\{|x_1-y_1|,\cdots ,|x_n-y_n|\}$

So proof says:

Let $\displaystyle x=(x_1,\cdots ,x_n)$ and $\displaystyle y=(y_1,\cdots ,y_n)$ be two points in $\displaystyle \mathbb{R}^n$. It is simple algebra to check that

$\displaystyle \rho(x,y)\leq d(x,y)\leq\sqrt{n}\rho(x,y)$

(1) Where and why did they have this inequality?
(2) I know it is stated that it is simple algebra to check it, but I don't see it. How is checked?

2. ## Re: Metric Space

Well : $\displaystyle \rho (x,y) = \max \{|x_1-y_1| ,..., |x_n - y_n|\} = |x_j - y_j|$ for some $\displaystyle j\in \{1,...,n\}$ (such a j always exists)

Now: $\displaystyle \rho(x,y) = |x_j - y_j| = \sqrt{(x_j - y_j)^2} \leq \sqrt{ (x_1-y_1)^2 +... + (x_j - y_j)^2 + ... (x_n-y_n)^2}$ since the square root function is increasing and the squares are non-negative.

Finally $\displaystyle (x_1-y_1)^2 = |x_1 - y_1|^2 \leq |x_j-y_j|^2 = (x_j-y_j)^2$ and so on ... $\displaystyle (x_n-y_n)^2 \leq (x_j - y_j)^2$, thus:

$\displaystyle (x_1-y_1)^2 + ... + (x_n-y_n)^2 \leq n \cdot (x_j - y_j)^2$ and taking square roots on both sides we get the other inequality.

(1) They probably want to prove that the metrics are equivalent... by showing that the identity between them is a homeomorphism (hence they give rise to the same topology). And going through these inequalities is a simple mean of doing so.

3. ## Re: Metric Space

Originally Posted by PaulRS
Well : $\displaystyle \rho (x,y) = \max \{|x_1-y_1| ,..., |x_n - y_n|\} = |x_j - y_j|$ for some $\displaystyle j\in \{1,...,n\}$ (such a j always exists)

Now: $\displaystyle \rho(x,y) = |x_j - y_j| = \sqrt{(x_j - y_j)^2} \leq \sqrt{ (x_1-y_1)^2 +... + (x_j - y_j)^2 + ... (x_n-y_n)^2}$ since the square root function is increasing and the squares are non-negative.

Finally $\displaystyle (x_1-y_1)^2 = |x_1 - y_1|^2 \leq |x_j-y_j|^2 = (x_j-y_j)^2$ and so on ... $\displaystyle (x_n-y_n)^2 \leq (x_j - y_j)^2$, thus:

$\displaystyle (x_1-y_1)^2 + ... + (x_n-y_n)^2 \leq n \cdot (x_j - y_j)^2$ and taking square roots on both sides we get the other inequality.

(1) They probably want to prove that the metrics are equivalent... by showing that the identity between them is a homeomorphism (hence they give rise to the same topology). And going through these inequalities is a simple mean of doing so.
Continue the book says:

The first inequality shows that

$\displaystyle B_d(x,\epsilon)\subset B_{rho}(x,\epsilon)$

How do they get this from the inequality? What is consider the first? Are the reading left to right?

4. ## Re: Metric Space

They meant the left-most part, yes.

Pick an element of $\displaystyle y\in B_d (x,\epsilon)$ , then by definition $\displaystyle d(x,y)<\epsilon$.

But $\displaystyle \rho(x,y)\leq d(x,y) < \epsilon$ hence $\displaystyle y\in B_{\rho}(x,\epsilon)$ .

Thus in all : $\displaystyle B_d(x,\epsilon) \subset B_{\rho}(x,\epsilon)$.

5. ## Re: Metric Space

Originally Posted by PaulRS
They meant the left-most part, yes.

Pick an element of $\displaystyle y\in B_d (x,\epsilon)$ , then by definition $\displaystyle d(x,y) \leq \epsilon$.

But $\displaystyle \rho(x,y)\leq d(x,y) \leq \epsilon$ hence $\displaystyle y\in B_{\rho}(x,\epsilon)$ .

Thus in all : $\displaystyle B_d(x,\epsilon) \subset B_{\rho}(x,\epsilon)$.
For the last inequality, why did they divide epsilon by

$\displaystyle B_{\rho}\left(x,\frac{\epsilon}{\sqrt{n}}\right) \subset B_d(x,y)$

6. ## Re: Metric Space

Remember that $\displaystyle d(x,y)\leq \sqrt{n} \cdot \rho (x,y)$, and try to prove the inclusion as I did for the other one.

(that $\displaystyle \sqrt{n}$ multiplying there obliges us to 'contract' the radius a bit)

PS: You meant to say $\displaystyle \epsilon$ and not $\displaystyle y$ in $\displaystyle B_d(x,y)$ , I suppose.