Topology of Metric Spaces Why is the open interval $(-\infty,+\infty)$ not an open sphere with usual metric? 
We can find a radius such that the open sphere is subset of real line same as we find that for any open interval.
I am considering the usual metric on $\mathbb{R}$.
 A: In general, if a topology $\mathcal T$ on a space $X$ is generated by a basis $\mathcal B$ (i.e. $\mathcal T$ is the intersection of all topologies $X$ that contain $\mathcal B$), then a set $U\subset$ is open iff 


*

*For each $x\in U$, there exists $B\in\mathcal B$ such that $x\in B\subset U$.

*For each $x\in U$, if there exist $B_1,B_2\in\mathcal B$ such that $x\in B_1\cap B_2$, then there exists $B_3\in\mathcal B$ such that $x\in B_3\subset(B_1\cap B_2)$.


The usual topology $\mathcal T$ on $\mathbb R$ has basis
$$\mathcal B = \{(a,b) : a<b\}. $$
From introductory analysis we recall that a set $U\subset\mathbb R$ is said to be open if for each $x\in U$, there exists $\varepsilon_x>0$ such that $|x-y|<\varepsilon_x$ implies $y\in\mathbb R$. This is equivalent to the first condition above, as $x\in(x-x_\varepsilon, x+\varepsilon)\in\mathcal B$. Now suppose $x\in U$ with $x\in (a,b)\cap (c,d)$. Then
$$(a,b)\cap(c,d) = \left(\max\{a,c\}, \min\{b,d\} \right)\in\mathcal B, $$
so the second condition is satisfied, and we see that the abstract topological definition coincides with our understanding of open sets from analysis (as it should, since metric spaces are topological spaces!).
Now, it is not hard to prove that a set $U$ is open iff it is the union of basis elements (this is a good exercise). Then
$$\mathbb R = (-\infty, \infty) = \bigcup_{n=1}^\infty (-n,n), $$
so that $(-\infty, \infty)$ is an open set. (In fact this is being a bit pedantic since the definition of a topological space necessitates that the underlying space itself is open, but that is a digression). However, $(-\infty, \infty)$ is not an open interval, in the sense that it isn't a member of the set $\mathcal B$ defined above.
