Understanding the concepts of topologies and equivalent metrics

Question

Could someone please help me in understanding the concepts of topologies and equivalent metrics. If possible, giving some examples of equivalent metrics.

For example, I don't know why for the Euclidean space, the d1, d2 and d(infinity) metrics are (strongly) equivalent.

I would really appreciate any help! Thanks :)

Answer 1

Let $x$ and $y$ be two point and consider $\delta_j = x_j-y_j$ then the metrics are defined as $$d_1(x,y) = \sum^N |\delta_j|$$ $$d_2(x,y) = \sqrt{\sum^N \delta_j^2}$$ $$d_\infty(x,y) = \max^N|\delta_j|$$

Now we see for example $|\delta_j| < \max|\delta_j|$ so $\sum |\delta_j| < N\max|\delta_j|$, that is $d_1\le Nd_\infty$.

By the square rule we have $\left(\sum |\delta_j|\right)^2 = \sum |\delta_j|^2 + \sum_{j<k}2|\delta_j\delta_k| \ge \sum|\delta_j|^2$. So we have that $d_1^2 \ge d_2$.

Also we have that $\sum \delta_j^2 \ge |\delta_k|^2$ for all $k$ and especially that $\sum \delta_j^2 \ge \left(\max |\delta_j|\right)^2$ so $\delta_2\ge \delta_\infty$.

To summarize we have:

$$N\delta_\infty\ge d_1 \ge d_2 \ge d_\infty$$

The relation between equivalent and strongly equivalent metrics can be seen if we reformulate the definition of strongly equivalent in a way more similar to the definition for weak equivalence. The definition that $L\tilde d\le d\le K\tilde d$ means that $\tilde B_{r/L}(x)\subset B_r(x)\subset B_{r/K}(x)$, compare this to the definition of mere equivalence $\tilde B_{r'}(x) \subset B_r(x)\subset \tilde B_{r''}(x)$. The difference is that in strong equivalence the $r'$ and $r''$ have a fixed dependency to $r$ while in mere equivalence $r'$ and $r''$ may not only depend on $r$ in a more complex way, it may also depend on $x$.

From this we can see that we cannot form a non-strong equivalence that easily. We must either drop translation invariance or the scaling property ofthe norms mentioned.

Answer 2

If you have a metric $d$ on a set $X$, then this defines (often called "induces") a topology on $X$ as well, where a set $O$ is open iff $$\forall x \in O: \exists r>0: B_d(x,r) \subseteq O$$ where $B_d(x,r) = \{p \in X: d(x,p) < r\}$ is the metric ball. I'll call this topology (one can check the above defines a topology, in fact the smallest one where all sets of the form $B_d(x,r), x \in X, r>0$ are open) $\mathcal{T}_d$

If we have two metrics $d$ and $d'$ on the same set $X$, then $d$ is equivalent to $d'$ iff $\mathcal{T}_d = \mathcal{T}_{d'}$, i.e. They give rise to the same topology on $X$.

There is a criterion for this that is often useful: $d$ is equivalent to $d'$ iff the following conditions hold:

1. $\forall x \in X: \forall r>0: \exists r' > 0: B_{d'}(x,r') \subseteq B_d(x,r)$

2. $\forall x \in X: \forall r>0: \exists r' > 0: B_{d}(x,r') \subseteq B_{d'}(x,r)$

Suppose that the topologies are the same, then to see 1. we let $X \in X$, $r>0$, and note that $x$ is in the interior of $B_d(x,r)$ in the $\mathcal{T}_d$ topology, so it also should be an interior point of that set in $\mathcal{T}_{d'}$ as well, which comes down to the existence of some $ r'$ as stated. To see 2. we use the symmetric argument starting from $\mathcal{T}_{d'}$ etc. And if 1. and 2. hold we get that the topologies are the same: let $O$ be open in $\mathcal{T}_d$. Then $O$ is open in $\mathcal{T}_{d'}$, for let $x \in O$. Then we have some $r>0$ with $B_d(x,r) \subseteq O$, and 1. gives us an $r' > 0$ with $B_{d'}(x,r') \subseteq B_d(x,r) \subseteq O$, so we have found a radius for $x$ w.r.t. $d'$ as well. Similarly condition 2 will gives us the other inclusion.

Now a common way to prove these conditions is when we have global inequalities:

Suppose we have $A, B > 0$ such that $$\text{3. } \forall x,y \in X: A\cdot d(x,y) \le d'(x,y) \le B\cdot d(x,y)$$ then we can show 1. and 2. quite easily: for the first, given $r>0$ we take $s = Ar$ and then $d'(p,y) < s$ implies $d(x,y) \le \frac{1}{A}d'(x,y) < \frac{1}{A}\cdot Ar = r$ showing the inclusion of balls. For the second we take $s=\frac{r}{B}$ and note that $d(x,p) < r'$ implies $d'(x,y) \le Bd(x,y) < B\cdot r'= r$ and we are done once again.

When we have this global inequality 3. we call the metric $d$ and $d'$ strongly equivalent. We have just seen that strongly equivalent metrics are indeed equivalent, and this in a uniform way. The usual example of this phenomenon are the metrics defined on $\mathbb{R}^n$, which are related by inequalities. E.g.:

$$(d_2)^2(x,y) = \sum_{i=1}^n (x_i - y_i)^2 \le \sum_{i=1}^n d_{\infty}^2(x,y) = nd_{\infty}^2(x,y), \text{ so } d_2(x,y) \le \sqrt{n} d_{\infty}(x,y)$$ and also $$(d_2)^2(x,y) = \sum_{i=1}^n (x_i -y_i)^2 \ge d^2_\infty(x,y) \text{ hence } d_2(x,y) \ge d_\infty(x,y)$$ which shows that $d_2$ and $d_\infty$ are strongly equivalent for $\mathbb{R}^n$ with constants $1$ and $\sqrt{n}$. Similar inequalities exist between $d_1$ and $d_2$, showing these 2 to be equivalent as well (and that makes them all equivalent of course).

A non-example: if $d(x,y) = |x-y|$ is the standard metric on the reals, then $d_t(x,y) = \min(d(x,y), 1)$ ,the so-called truncated metric on the reals are equivalent but not strongly equivalent. The latter holds because if we assume $A,B$ exist such that $$\forall x,y \in \mathbb{R}: Ad_t(x,y) \le d(x,y) \le Bd_t(x,y)$$ then we note that $Bd_t(x,y)$ is only maximally $B$ while $d(x,y)$ can assume arbitarily large values. So this cannot hold for all $x,y$ at the same time. Equivalence is easy to show using either the definition or the criterion, and I'll leave that for you to figure out.