what is a usual topology for R

Question

i'm studying topology. Let T be a usual topology for R(real) generated by usual metric. first, i know that elements of T are open sets in R. i wonder form of T(topology). Under given condition, Can T have many form? or is T unique family?

Answer 1

Expanding on what Martín-Blas Pérez Pinilla said in a comment, “the usual metric” for $\Bbb R$ is the metric $$d(x,y) = \lvert x-y \rvert,$$ and “the usual topology” for $\Bbb R$ is the one where the basic open sets have the form $$(a,b) = \{ x\in \Bbb R\mid a < x < b \}.$$ (“Basic” here means that the open sets are unions of basic sets.)

There are many different possible topologies for $\Bbb R$, but only this one is called the “usual” topology. There are many metrics one can put on $\Bbb R$, but only the $\lvert x-y\rvert $ metric is called the “usual” metric.

Answer 2

In general it is possible to place more than one topology on $\mathbb R$. For example, the following topology (the trivial topology) is a perfectly fine topology for $\mathbb R$: $$ \{\varnothing,\mathbb R\}. $$ (You should verify that it satisfies the axioms for a topology.) Here's another example, the discrete topology, which consists of all subsets of $\mathbb R$: $$ 2^\mathbb R. $$ However, the topology you have in mind -- the standard topology on $\mathbb R$, which has as a base the open intervals $(a,b)$ -- is uniquely defined, although there is more than one way to describe it.

Answer 3

Let $(\Bbb R, d_{|.|})$ the usual (or Eucledean) real metric space, where $d_{|.|}$ denotes the usual (or Eucledean) metric $\Bbb R^2\mapsto\ \Bbb R_0^+$ and satisfies

$$\forall x,y \in \Bbb R : d_{|.|}(x,y):=|x-y|.$$

All real subsets that are $d_{|.|}$-open (i.e. open with the usual metric) can be collected in class

$$\mathcal T_{d_{|.|}}:=\{A\subseteq \Bbb R|A\space d_{|.|}-\rm open\}.$$ This class of real subsets forms a topology induced by usual metric, and the pair $(\Bbb R, \mathcal T_{d_{|.|}})$ is called usual topological space.

It is obvious that every open interval belongs to $\mathcal T_{d_{|.|}}$, for it is a $d_{|.|}$-open subset of $\Bbb R$.