Is $\Sigma$ a topology on $X$? Here's the question that I'm trying to answer. 
Let $X$ be a plane. Then, let $\Sigma$ be a subset of $P(X)$ such that $X \in \Sigma$, $\varnothing \in \Sigma$ and all open disks centered on the origin are also in $\Sigma$. Is $\Sigma$ a topology on $X$?

So, I think that this is a topology on X. Let me try to prove it by checking the topology axioms:


*

*Clearly, $X \in \Sigma$ and $\varnothing \in \Sigma$. That
satisfies the first axiom. 

*Now, consider any finite intersection of sets in $\Sigma$. We know that $X \cap X = X$ and $X \cap \varnothing = \varnothing$. We also know that $\varnothing \cap \varnothing = \varnothing$. Any intersection of open disks in $\Sigma$ will produce the open disk with the smallest radius in that intersection, so that will still belong to $\Sigma$. Finally, the intersection of any open disk with $X$ will produce the open disk and the intersection of any open disk with $\varnothing$ will produce $\varnothing$. So, $\Sigma$ is closed with respect to the intersection of sets. 

*Finally, we consider any arbitrary union of sets in $\Sigma$. Any arbitrary union consisting of open disks will produce the largest open disk. Furthermore, any arbitrary union that consists of $X$ and $\varnothing$ will just produce $X$. So, $\Sigma$ is closed with respect to unions of sets.
Since $\Sigma$ clearly satisfies all of the axioms of topological structure, it follows that $(X,\Sigma)$ is a topological space. 
Does my proof above work? If it doesn't, how can I improve it so that the argument works? Any feedback would be appreciated. 
 A: There is no largest open disk.
For example, if the radii are 1 - 1/n, 0 < n,
none of them is the largest.  
Let r be the supremum of the radii.
Show the open ball of radius r is the union of the balls. 
A: You are right this is a topology. However, when you claim that the arbitrary union of disks is again a disk or the entire space, you have to include a proof. 
Hint: Denote an open disk with radius $r$ and centered at the origin by $D(r)$. Here we also define $D(0)=\emptyset$ and $D(\infty)=\mathbb{R}^2$. Prove that
$$\bigcup_{r\in I\subseteq [0,\infty]} D(r)= D(\sup I)$$
In general $\sup I \notin I$ so your claim that the union is the largest disk in the collection does not hold, as is also illustrated in the other answer.
A: The idea is right, but you could write it down slightly more formally:
Intersection: (the non-trivial case): if $O_i=B(0, r_i), i=1,\ldots ,n$ are non-trivial open sets, then $$\bigcap_{i=1}^n O_i = B(0, r_0)\in \Sigma$$ where $r_0=\min_{1 \le i \le n} r_i$
And if $O_i = B(0, r_i)$, with $r_i > 0, i \in I$, consider $O=\bigcup_{i \in I} O_i$. If the set $R=\{r_i, i \in I\}$ is unbounded above, we can show that $O=X \in \Sigma$ (if $x \in X$, $d(x,0)$ cannot be an upper bound for $R$ so there is some $r_i > d(x,0)$ and then $x \in B(x,r_i)=O_i$) and if $r:=\sup R$ exists we can show that $O=B(0,r) \in \Sigma$ (not the largest ball, which need not exist). In both cases the union is in $\Sigma$ and we're done again.
