I would like to show that the standard topology on $\mathbb{R}$ has a countable basis.
As far as I understand, the standard topology on $\mathbb{R}$ is generated by open intervals. So I can construct one basis as $\{(i,i+1),(i+1/2,i+3/2)\}$ for $i\in\mathbb{Z}$ which is clearly countable.
EDITED: Thanks to Mees de Vries for pointing it out that my example is not a base. So is there any better way to construct a base?
Is it okay to construct one basis to show the existence of a basis? Or is there way to show in general without constructing a specific example?
Could somebody please give some light on this? Thanks.
Best Answer
Your set isn't a basis. If you intersect $(1,2)$ and $(3/2,5/2)$ (using $i=1$) you get $(3/2,2)$ which does not contain any of your "basis" elements.
If you use your set as a sub-basis, it still isn't enough because you can't generate open intervals shorter than 1/2 in length.
The standard approach to finding a countable basis for $\mathbb{R}$ (or $\mathbb{R}^n$) is to use the rational numbers.
Since $\mathbb{Q}$ is countable so is $\mathbb{Q}^2$ (countable cartesian product countable is still countable). Consider open intervals with rational endpoints: $$B = \{ (r,s) \;|\; r,s \in \mathbb{Q} \}$$ Then $B$ has the same cardinality as $\mathbb{Q}^2$ (i.e. it's countable).
This forms a basis for $\mathbb{R}$'s standard topology. To see this notice that these sets are open intervals (thus open in $\mathbb{R}$'s topology). So every open set generated by this basis is a standard open set. Conversely, if $O$ is open in $\mathbb{R}$ and $x \in O$ then there is an open interval $(a,b) \subseteq O$ such that $x \in (a,b)$. Thus $a<x<b$. Pick any rational number between $a$ and $x$ (say $r$) and any rational number between $x$ and $b$ (say $s$). Then $a<r<x<s<b$ and so $x \in (r,s) \subseteq (a,b) \subseteq O$. So $O$ is open relative to our new base $B$. Thus $B$ generates the standard topology.