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.
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.
This is just to highlight a point which is essentially already made (in a special case) in Henno Brandsma's answer which also perhaps clarifies the final remark in skyking's answer.
Claim 1: If $(X,d)$ is any metric spaces, then the function $d_b\colon X\times X \to \mathbb R$ given by setting $$ d_b(x_1,x_2) = \min\{d(x_1,x_2),1\}. $$ is a metric on $X$.
Proof: It is clear from the definition that $d_b$ is symmetric if $d$ is and since $d_b(x_1,x_2) =0$ if and only if $d(x_1,x_2)=0$ it follows that $d_b(x_1,x_2)=0$ if and only if $x_1=x_2$. Thus it remains to verify the triangle inequality. Let $x_1,x_2,x_3 \in X$. Since $d_b(x_1,x_3) \leq 1$ the triangle inequality for $d_b$ is trivial if $d_b(x_1,x_2)+d_b(x_2,x_3)\geq 1$. On the other hand, since $d$ and $d_b$ are non-negative, $$ d(x_1,x_2)+d(x_2,x_3) \leq 1 \implies d(x_i,x_j)\leq 1, \quad \forall (i,j)=(1,2),(1,3),(2,3) $$ and the triangle inequality for $d_b$ then coincides with that for $d$.
Claim 2: The metrics $d$ and $d_b$ are weakly equivalent.
Proof: The identity map $\text{id}_X\colon X \to X$ given by $\text{id}_X(a)=a$ for all $a \in X$ induces a homeomorphism $\text{id}_X\colon (X,d)\to (X,d_b)$, since for any $\epsilon>0$ if we set $\delta = \min\{\epsilon,1\}$ then if $d(x,a)<\delta$ then $d_b(x,a)<\epsilon$ and similarly if $d_b(x,a)< \delta$ then $d(x,a)<\epsilon$. It follows that $d$ and $d_b$ are weakly equivalent metrics, and hence we have produced, for any metric spaces, a metric which is weakly equivalent to it.
Claim 3: The metrics $d$ and $d_b$ are strongly equivalent if and only if $d$ is bounded. Thus if $X$ is a normed vector spaces (with $d$ induced from the norm) then $d$ and $d_b$ are not strongy equivalent.
Proof: If $d_1,d_2$ are strongly equivalent metrics, then there exist positive constants $C_1,C_2>0$ such that $d_1(x_1,x_2) \leq C_1 d_2(x_1,x_2)$ and $d_2(x_1,x_2)\leq C_2d_1(x_1,x_2)$. If $d_1,d_2$ are $d$ and $d_b$ respectively then $d_b(x_1,x_2)\leq d(x_1,x_2)$ so that we may take $C_2=1$. If $d$ is bounded, so that $d(x_1,x_2)\leq D$ for some $D \in \mathbb R$ and all $x_1,x_2 \in X$, then clearly setting $C_1=\max\{D,1\}$ it follows that $d(x_1,x_2) \leq C_1d_b(x_1,x_2)$ and hence that $d$ and $d_b$ are strong ly equivalent.
Conversely, if $d,d_b$ are strongly equivalent then there is a constant $B$ such that $d(x_1,x_2) \leq Bd_b(x_1,x_2) \leq B$ so that $d$ is bounded. It follows that if $d$ is not bounded then it is not strongly equivalent to $d_b$. Since the homogeneity property of a norm ensures that the metric associated to it is unbounded, the final sentence also follows.
Finally, elaborating on skywing's final remark. we have:
Claim: If $\|.\|_a$ and $\|.\|_b$ are any two norms on a vector space, then if $\|.\|_a$ is weakly equivalent to $\|.\|_b$ then they are strongly equivalent.
Proof: The two norms are equivalent if and only if $\text{id}_X$ is a homeomorphism from $(X,\|.\|_a)$ to $(X,\|.\|_b)$. But this means that $\|.\|_b = \|.\|_b\circ \text{id}_X$ is a continuous function on $(X,\|.\|_a)$ and hence there is a $\delta>0$ such that if $\|x_1\|_a \leq \delta$ then $\|x_1\|_b \leq 1$. It follows that $\|x\|_b\leq \delta^{-1}\|x\|_a$ and hence if we set $C_b =\delta^{-1}$ then $\|x\|_b \leq C_b\|x\|_a$ for all $x \in X$. Similarly, there is some constant $C_a$ such that $\|x\|_a \leq C_a\|x\|_b$ for all $x \in X$.
