Is the following set topology? Consider a potential topology $T$ on $\mathbb R^3$ (three dimension space.) $T$ contains sets $$\{(x,y,z) \mid x^2 + y^2 + z^2 \leq r\},$$ where $r\in\mathbb R^{\geq 0}.$ 
Since we know that that arbitrary union of open sets is open, suppose I take union of all open sets $s \in T$ such that $(x,y,z)$ follows $x^2 + y^2 + z^2 ≤ r$ and $r ∈ (0,1)$. Does the resultant set from union belongs to $T$ ? Since overall radius of sphere formed cannot be $1$ in this case but cannot be strictly less than $1$.
 A: Your argument correctly shows that if $T$ consists of the sets $B_r=\{(x,y,z)\mid x^2+y^2+z^2\leq r\}$ for $r\in\mathbb R^{\geq 0},$ and the empty set $\emptyset$ and the whole set $\mathbb R^3,$ then $T$ is not a topology because:
$$\bigcup_{r\in(0,1)} B_r=\{(x,y,z)\mid x^2+y^2+z^2<1\}$$
is a union of elements of $T$ which is not in $T.$ 
The smallest topology $T'$ on $\mathbb R^3$ containing $T$ has the following definition:

Given subset $U\subseteq \mathbb R^3,$ then $U\in T'$ if and only if, for all $(x,y,z)\in U$ and all $(x_0,y_0,z_0)\in\mathbb R^3$ with $x_0^2+y_0^2+z_0^2\leq x^2+y^2+z^2$, you have that $(x_0,y_0,z_0)\in U.$

Alternatively, this can be written as:

$U\in T'$ if and only if $U\in T$ or $U=U_r=\{(x,y,z)\mid x^2+y^2+z^2<r\}$ for some $r\in\mathbb R^{>0}.$

A: If $T$ is a topology on$\Bbb{R^3}$ that contains a set of the form $\{x^2 + y^2 + z^2 < r\}$ for every real number $r$ then the union you describe is the set $\{x^2 + y^2 + z^2 < 1\}$ which is contained in $T$ by the property stated above.
EDIT: this answer is deprecated as it replies to an unedited version of the question. 
