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Here is the question:

Let $X = \mathbb{R}$ and let $\Omega$ consist of the empty set and all infinite subsets of $\mathbb{R}$. Is $\Omega$ a topological structure?

My attempt : I think the answer is No; it is not a topology. Because we have $$ \Omega = \lbrace \varnothing \rbrace \cup \lbrace U \subseteq \mathbb{R} \mid \text{$U$ is infinite} \rbrace $$

If we check the axiom for the finite intersection property then let $U_1, U_2 ..U_n$ be finite elements of $\Omega$ now $\cap_{i=1}^{n} U_i$ might be finite and thus doesn't belong to $\Omega$. Is this explanation correct?

EDIT: Let $U_1$ be set of all positive even numbers and $U_2$ be the set of all primes. Now we know both $U_1$ and $U_2$ are elements of $\Omega$ as they are infinite sets but $ U_1 \cap U_2 = \lbrace 2 \rbrace $ which is finite.

Fabrosi
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Rusty
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  • Yes it is. Maybe you can provide a basic example with two sets $U_1$ and $U_2$. – mathcounterexamples.net Aug 12 '15 at 17:56
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    “might be finite” needs evidence. Can you find two infinite subsets of $\mathbb{R}$ with a finite intersection? – Matthew Leingang Aug 12 '15 at 17:57
  • You're on the right track, but you should cook up a concrete example for your claim. Hint: you can do it with just two sets. – Ian Aug 12 '15 at 17:57
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    Does set of positive even numbers and primes work ? – Rusty Aug 12 '15 at 17:59
  • Perhaps (it's your problem). But you don't need to use well-known sets. – Matthew Leingang Aug 12 '15 at 18:00
  • Can you give another non standard example ? It will help me understand better.Thanks – Rusty Aug 12 '15 at 18:02
  • Try to construct counterexample from any infinite subset $A$ such that $X\setminus A$ is infinite as well. Can it work if $X\setminus A$ is finite? What if you consider all infinite subsets $A$ of $X$ such that $X\setminus A$ is finite, does it form topology then? – Ennar Aug 12 '15 at 18:04
  • ummm Sorry but totally confused now... – Rusty Aug 12 '15 at 18:06
  • Well think about what you need, if infinite sets are to be open, as you said, their finite intersections must be open as well, and in this case it means infinite or empty. So, let's take some infinite $A$. Now, if $X\setminus A$ happens to be infinite, it is open as well. And $A\cap X\setminus A = \emptyset$ which is open. But, can you adjust it a bit so it serves as an counterexample? And my other question is another exercise similar to your question that you should try to work out. – Ennar Aug 12 '15 at 18:12
  • Oh. ok. Thanks for these detailed. I understood now. – Rusty Aug 12 '15 at 18:13

4 Answers4

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No, because two infinite sets may have a non-empty finite intersection (for example, the intersection of prime numbers and even numbers is $\{2\}$). This violates the condition that the intersection of two open sets be open in a topology.

However, if one requires not only that the non-empty “open” sets be infinite but also that their complements be finite (which is a stronger condition), then one has a topology. It is called the cofinite topology or finite-complement topology: $$\Omega=\{\varnothing\}\cup\{U\subseteq R\,|\,U^{\mathsf c}\text{ is finite}\}.$$

triple_sec
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You can let your two sets be the even and odd integers, except the even integers set is augmented to also include the element 1. Then the intersection is just $\{1\}$, which is finite, so closure under intersection of finitely many open sets fails. In general all you need are two disjoint infinite sets and then take a finite number of elements from one set and include them as additional elements in the second set. Then the intersection will be this finite non-empty set of additional elements you added to the second set. So again, closure under intersection of finitely many open sets fails.

user2566092
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Let $U_1$ be the set of even numbers union 1, and let $U_2$ be the set of odd numbers. then..

George
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I think the answer is no,it is not a topology Because it does not satisfied the 1st condition of topology.As taa consist of empty set and all infinite subsets of $\mathbb{R}$ But it does not have $\mathbb{R}$ in it. According to the first condition of topology, empty set and $X${ground set} which is $\mathbb{R}$ belong to taa but it have all infinite subsets of $\mathbb{R}$, not $\mathbb{R}$ itself. That's why in my view's taa is not a topology

MRobinson
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Arslan Sheikh
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