# 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:

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

2. 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.

3. 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.

• I think that that's what it meant? I mean, I implicitly worked with that assumption when writing the argument. Mar 19, 2020 at 4:58
• Alright thank you, this subject seems to be just a little more abstract than the others so I'm really just making sure that I'm getting it right. Mar 19, 2020 at 5:00
• Hmmmmmmmmmm $\mathbb{R}^n$? I'm not quite sure if I could use that argument there. Aren't open disks defined differently $\mathbb{R}^n$? Mar 19, 2020 at 5:03
• Ah alright. I haven't gotten to that point so I guess I have to work for it. Mar 19, 2020 at 5:05
• Well, I'd certainly hope so. It'll be a while before I get to study them in any reasonable detail. Mar 19, 2020 at 5:08

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.

• Alright, I'll try to write a proof by going off of this, thank you. Mar 19, 2020 at 7:36

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.

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.

• Thank you.....hmm I should say that my text hasn’t actually introduced any of this stuff formally. Uhhh I guess I’ll just have to learn it elsewhere Mar 19, 2020 at 9:46
• @AbhijeetVats that's basic maths you should know before any course in college maths. Mar 19, 2020 at 10:29
• I mean, I’m using a text that doesn’t pre-suppose any acquaintance with real analysis or anything like that. Also, like, the set theory content of your answer is entirely understandable. It’s just that the definition of a ball hasn’t been stated in my text, for example. Mar 19, 2020 at 10:35
• @AbhijeetVats a(n open) ball is just $B(x,r)=\{y : d(x,y) < r\}$ where $d$ is the distance in the plane. This ought to have been introduced before. Mar 19, 2020 at 10:36
• Nah it wasn’t. The authors went through some set theory, which I went through very quickly cos I know that stuff, but they introduced the definition of a topology and then began giving examples. No balls mentioned. I mean, it’s fine though, the definition is pretty simple to understand. Mar 19, 2020 at 10:38