Is the following set a topology on $\mathbb{R}$?

Question

So, here's what I'm trying to answer:

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

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In order to check if this is a topological structure, we need to verify that all axioms of topological structure hold:

• Clearly, $\varnothing \in \Omega$. Now, $\mathbb{R}$ is an infinite subset of $\mathbb{R}$ so it is clear that $\mathbb{R} \in \Omega$.

• Now, we need to show that $\Omega$ is closed with respect to the finite intersections of its sets. In particular, these infinite subsets of $\mathbb{R}$ are going to be closed or open intervals in $\mathbb{R}$. The intersections of these intervals will, then, just be equal to the smallest interval in the intersection. So, that's going to be an element of $\Omega$. Furthermore, any intersection with $\mathbb{R}$ and $\varnothing$ will simply produce the smallest interval in the intersection and $\varnothing$ respectively.

• Finally, we consider the arbitrary unions of the sets in $\Omega$. Clearly, these unions will be equal to the largest set used in the unions. This set cannot be larger than $\mathbb{R}$. Hence, $\Omega$ is closed with respect to arbitrary unions.

This proves that $(\mathbb{R},\Omega)$ is a topological space.

Is my proof above correct? It feels kind of hand-wavy because I'm not really using anything from Set Theory appropriately, which is why I'm thinking that it must have a flaw somewhere that I'm not seeing. If someone could give feedback on this, I'd appreciate it.

Answer 1

In particular, these infinite subsets of $\mathbb{R}$ are going to be closed or open intervals in $\mathbb{R}$

This is not the case. You are asking whether a particular family of sets forms a topology. A priori, this has nothing to do with the standard topology on $\mathbb{R}$. For example, $\mathbb{N}$ is an infinite subset of $\mathbb{N}$ but not an interval and it also doesn't contain any interval.

Also, you want to show that for any two $U_1, U_2$ in your family of sets, there exists a $U_3$ s.t. $U_1 \cap U_2 = U_3$. So I'm not sure why you write

The intersections of these intervals will, then, just be equal to the smallest interval in the intersection.

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Also, here is a hint for the solution: Find two infinite subsets of $\mathbb{R}$ s.t. their intersection is not infinite. If you can find such two, then you have shown that the proposed family of sets is not closed under finite intersection and is thus not a topology.

Answer 2

No it is not correct.

In the intersection proof, you use open intervals. But there are infinite subsets of $\mathbb{R}$ that are no open intervals.

And then you see that in fact your set is not a topology, because for example

$$]-\infty, 0]\cap [0,+\infty[=\{0\}$$

is a finite set which is the intersection of infinite sets.

Your reasoning in the union argument is also flawed but not necessary anymore at this point.

Answer 3

Your proof isn’t correct. In particular regarding finite intersections.

Both $A=\mathbb N$ and $B=\{1,0,-1,-2, \dots\}$ are elements of $\Omega$. However $A\cap B=\{1\}$ is finite and therefore doesn’t belong to $\Omega$.

As a global remark, for abstract topology, you need to avoid to only think to intervals!