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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.

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

3 Answers 3

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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.

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

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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.

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  • $\begingroup$ 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 $\endgroup$ Mar 19, 2020 at 9:46
  • $\begingroup$ @AbhijeetVats that's basic maths you should know before any course in college maths. $\endgroup$ Mar 19, 2020 at 10:29
  • $\begingroup$ 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. $\endgroup$ Mar 19, 2020 at 10:35
  • $\begingroup$ @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. $\endgroup$ Mar 19, 2020 at 10:36
  • $\begingroup$ 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. $\endgroup$ Mar 19, 2020 at 10:38

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