Understanding the concepts of topologies and equivalent metrics
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 :)
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:
$\forall x \in X: \forall r>0: \exists r' > 0: B_{d'}(x,r') \subseteq B_d(x,r)$
$\forall x \in X: \forall r>0: \exists r' > 0: B_{d}(x,r') \subseteq B_{d'}(x,r)$
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.