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Question:

Let $X$ be a set. Let $\tau$ consists of $\varnothing$ , and any set $U \subseteq X$ such that $X\setminus U$ is either finite or countable. Show that $\tau$ is a topology on $X$.

Looking only for $\mathbf{hints}$.

Let $\tau=\left \{ \varnothing,U \right \}$. Obviously, $X \subseteq X$ so $X \in \tau$

Now, $\tau=\left \{ \varnothing, \left \{ U_{i} \right \}_{i \in I} \right \}$ with $X \in \left \{ U_{i} \right \}_{i \in I}: \exists i \in I$

By hypothesis, $X\setminus U$ is countable so it is expected that $\exists$ a bijection

$f:\mathbb{Z}^{+}\rightarrow X\setminus U$

I would like to take this further. My intuition says that if I can show that the elements in $U$ are open sets, then the elements in $U$ are elements of $\tau$ which then satisfies one of the axiom for topological space. However, I am unable to do so.

Any Hint is appreciated.

Thanks in advance.

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Mathematicing
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2 Answers2

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This $\tau$ is the topology of co-countable sets.

Recall the axioms:

  1. Empty set and full space are in $\tau$. (You've shown this.)
  2. Arbitrary union of open sets are open.
  3. Finite intersection of open sets are open.

The second is simple. If you have two set that contain all but countably many elements, then their union must as well since the complement cant get any larger. This extends to arbitrary unions.

The third is where you need to get a bit more precise, but it essentially boils down to that a finite (or even countably infinite) union of countable sets is countable. (These countable sets are the complements of the sets in your topology.)

Alexis Olson
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  • There is one part of the definition where I am still unclear, perhaps due to overthinking. One of the axiom states that "for any collection of elements on $\tau$, the union of the elements must be in $\tau$. It then proceed to claim that elements in $\tau$ are open sets. Is it only after having satisfied the fact that the union of the elements are in$\tau$ do we then call the elements open sets or are we to assume first that the collection of elements are open sets at the outset? – Mathematicing Aug 18 '16 at 06:13
  • "Open sets" is just another way of referring to the elements in a topology. The topology is the collection of open sets. – Alexis Olson Aug 18 '16 at 06:15
  • How you extends the argument that complement of countable union of countable sets is countable to arbitrary unions? – Rafael Aug 18 '16 at 06:23
  • It's not true for arbitrary unions (e.g. $\cup_{x \in \mathbb{R}} {x} = \mathbb{R}$), but it is for finite unions which is what you need since these are the complements of open sets. Intersecting open sets is equivlent unioning their complements. – Alexis Olson Aug 18 '16 at 06:30
  • Sorry, but I cannot understand your argument. You are saying that, in this case, the notion of arbitrary union and finite union are equals? – Rafael Aug 18 '16 at 07:01
  • I believe you are mixing up 2 and 3. 2 involves arbitrary unions whereas 3 involves finite intersections (equivalently finite unions of their complement). I may have misunderstood what you are asking if your first comment was referring to the second axiom. The extension in 2 follows from the fact that any intersection of countable sets is countable. – Alexis Olson Aug 18 '16 at 07:09
  • Yes, my question arose here: "The second is simple. If you have two set that contain all but countably many elements, then their union must as well since the complement cant get any larger. This extends to arbitrary unions." – Rafael Aug 18 '16 at 07:11
  • Basically, any union of sets is bigger than any of the individual sets. Since each individual set contains all but countably many elements, the union must as well. – Alexis Olson Aug 18 '16 at 07:15
  • Ok. Can you prove it "looking only for hints"? I would learn it. – Rafael Aug 18 '16 at 07:27
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Use De Morgan's Laws and that the countable union of countable sets is countable.

$\tau$ define a topology in $X$ if:

$i)$ $\emptyset,X\in\tau$;

$ii)$ Arbitrary union of sets in $\tau$ is in $\tau$;

$iii)$ Finite intersection of sets in $\tau$ is in $\tau$.

$i)$ is trivial. Now, if $\{U_{\lambda}\}_{\lambda\in \Lambda}$ is a family of element of $\tau$, from De Morgan's Laws, $$X\setminus\bigcup_{\lambda\in\Lambda}U_{\lambda\in\Lambda}=\bigcap_{\lambda\in\Lambda}(X\setminus U_{\lambda})\subseteq X\setminus U_{\mu},$$ for a $\mu\in\Lambda$ fixed. Since $X\setminus U_{\mu}$ is either finite or countable, $X\setminus\bigcup_{\lambda\in\Lambda}U_{\lambda\in\Lambda}$ is too, i.e, $\bigcup_{\lambda\in\Lambda}U_{\lambda\in\Lambda}\in\tau.$ Then $ii)$ holds.

For $iii)$, if $U_{1},...U_{n}\in\tau$, we have $$X\setminus\bigcap_{i=1}^{n}U_{i}=\bigcup_{i=1}^{n}(X\setminus U_{i}),$$which is either finite or countable because countable union of countable sets is countable. The proof is done.

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Rafael
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