Why is it that when working in the reals, we can choose our metric to be the absolute value in proving things pertaining to convergence

Question

From reading a basic real analysis text, it's never occurred to me to ask why convergence is defined using the absolute value as a metric. When I started reading Rudin, i've noticed that for more general metric spaces, we would also want to use a general metric. This make sense, as I assume there might be some metrics $d(x,y)$ that wouldn't be metrics had $x,y$ come from a different set perhaps. But why is it that in the reals, we are free to choose our metric to be the absolute value in proving things pertaining to convergence?

I am inclined to think that we are free to choose other metrics that make sense for the metric space we are dealing depending on what we are trying to achieve.

Answer 1

I assume you are familiar with the definition of a metric. There are certainly metrics which are very different from $|x-y|$. Just take any (crazy) bijection $\phi:\Bbb R\to\Bbb R$ and define the metric $d(x,y)=|\phi(x)-\phi(y)|$. But usually we make some natural assumptions, e.g. the compatibility of the metric with the other structures on $\Bbb R$, e.g. the field structure or order structure.

If our goal is to create a homogeneous metric space (one in which all points are essentially equal, there are no special points), then a natural assumptions is

• translational invariance, i.e. $d(x+\lambda,y+\lambda) = d(x,y)$ for all $\lambda\in\Bbb R$.

This means thats the distance of two numbers does not change when both numbers are shifted by the same amount. It is not hard to see that any translational invariant metric is given by

$$d(x,y) = \mu(|x-y|)$$

for some function $\mu:\Bbb R^{\ge 0}\to \Bbb R^{\ge0}$. The other axioms of a metric enforce $\mu(0)=0$ and subadditivity, i.e. $\mu(x+y)\le \mu(x)+\mu(y)$.

There are crazy subadditive functions which are not continuous (in the usual sense), and they all need the axiom of choice for construction. However, usually one adds another natural assumption. If we want the resulting topological space to be scale invariant, we can add the assumption of

• homogeneity, i.e. $d(\lambda x, \lambda y)=|\lambda|d(x,y)$ for all $\lambda\in\Bbb R$.

This means that when we scale the space by a factor of $\lambda$, then all distances scale by the same factor. But this assumption already forces $\mu(x)=kx$ for some $k>0$. And this means we have

$$d(x,y)=k|x-y|.$$

So you see that only these two natural assumptions do not leave us with much choice. However, you are free to drop either and look what other metrics are there. But it should be clear why $|x-y|$ might be a natural choice.

Answer 2

Of course that the choice of the metric depends upon what we are trying to achieve. Take, for instance, the spacee $\mathcal{C}\bigl([0,1]\bigr)$ of all continuous functions from $[0,1]$ into $\mathbb R$. If, in the context that you're working with, your idea of one function being close no another one is that, for each $x\in[0,1]$, $f(x)$ is close to $g(x)$, then you consider the metric$$(f,g)\mapsto\sup_{x\in[0,1]}\bigl|f(x)-g(x)\bigr|.$$And if being close means that the area limited by their graphs is small, then you consider the metric$$(f,g)\mapsto\int_0^1\bigl|f(x)-g(x)\bigr|\,\mathrm dx.$$

Answer 3

As a norm is absolutely homogeneous and definite, one can easily show, that every norm on $\mathbb{R}$ is the absolute value times a positive constant. So at least for all induced metrices is the convergence the same, when you use the absolut value.

Note: This would also follow immediately by the norm equivalence in finite dimensional spaces. I am not sure, if one can extend this statement for all metrices.