In the second chapter of an introductory topology text, I stumpled upon the following example concerning the finite-complement topology.

Edit. In a previous version, I accidently defined $\mathcal T$ as a topology.

Let $X$ be an infinite set. We, firstly, define the set $\mathcal T$ by $$\mathcal T = \{U\subseteq X\ |\ U = \emptyset\ \text{ or }\ X - U\ \text{ is finite} \}. $$

Consider an arbitrary collection of open sets $\{U_\alpha,\ \alpha \in J\}$.

Edit. We now continue, to establish the second requirement of a topology as mentioned in the supplement.

By DeMorgan's law we have

$$X- \bigcup_{\alpha \in J} U_\alpha = \bigcap_{\alpha \in J}(X-U_\alpha)\ . $$ Each $X-U_\alpha$ is finite or all of $X$, so we have $X-\bigcup_{\alpha \in J} U_\alpha$ is finite or all of $X$.$\quad (*)$

$(*)$ should implicate that $\bigcup_{\alpha \in J} U_\alpha$ is open. I fail to see why.

I could argue that the union of a collection of open sets is open. I also know that a set is closed if its complement is open.


Or is $(*)$ only used to establish that the collection $\bigcup_{\alpha \in J}U_\alpha$ is contained in $\mathcal T$?

So that we seperately use that, indeed, the union of a collection of open sets is open?


This question concerns proving the second requirement of a topology: the union of an arbitrary collection of members of $\mathcal T$ is in $\mathcal T$.

McCleary's definition of a topology.

Let $X$ be a set and $\mathcal T$ a collection of subsets of $X$ called open sets. The collection $\mathcal T$ is called a topology on $X$ if the usual three requirements are met.

  • $\begingroup$ I believe that (*) is a step to show that $\mathcal{T}$ is indeed a topology. Isn't that what the author is trying to do here? $\endgroup$ – essay Mar 10 '15 at 16:13
  • $\begingroup$ Does $(*)$ implicate that $\bigcup_{\alpha \in J} U_\alpha$ is open? Or are we just using that the union of an open collection is open? $\endgroup$ – Mussé Redi Mar 10 '15 at 16:15
  • $\begingroup$ it does, because, by definition, an open set (i.e., an element of $\mathcal{T}$) is a set whose complement is finite, or the empty set. And (*) just means that either $\bigcup_{\alpha \in J} U_\alpha$ has finite complement, or is the empty set. $\endgroup$ – essay Mar 10 '15 at 16:20
  • $\begingroup$ @sylvia Ah really? Okay, that definition of open explains alot. $\endgroup$ – Mussé Redi Mar 10 '15 at 16:23
  • 1
    $\begingroup$ okay, so there's a light abuse here. Indeed, we should not refer to the elements of $\mathcal{T}$ as open sets before we show that $\mathcal{T}$ is a topology; but sometimes, for convenience, people do that. If this is confusing you, then instead of saying "Consider an arvitrary collection of open sets" just say "Consider an arbitrary collection of elements of $\mathcal{T}$", and then go from there. $\endgroup$ – essay Mar 10 '15 at 16:51

As others have pointed out in the comments, this result you are looking at seems to be about proving that the finite complement topology is indeed a topology. So we are given a set $\mathcal{T}$. We do not yet know that it is a topology. As of now, it is just a collection of subsets of $X$, and we must show that $\mathcal{T}$ satisfies the four requirements to be a topology. In particular, one step is to show an arbitrary union of elements in $\mathcal{T}$ is itself an element of $\mathcal{T}$. That is why it is important to show that $\bigcup_{\alpha \in J} U_\alpha \in \mathcal{T}$ without assuming ahead of time that the finite complement topology is a topology.

So, we begin with any collection of subsets, $\{U_\alpha\}_{\alpha \in J}$, such that $U_\alpha \in \mathcal{T}$.

As you say, we have by DeMorgan's law that $$X- \bigcup_{\alpha \in J} U_\alpha = \bigcap_{\alpha \in J}(X-U_\alpha)$$ Notice that $X-U_\alpha$ is finite or empty. We know this because we chose each $U_\alpha$ in our arbitrary collection from $\mathcal{T}$, and by definition $U_\alpha$ is only an element of $\mathcal{T}$ if $X-U_\alpha$ is finite. This means that $$\bigcap_{\alpha \in J}(X-U_\alpha)$$ is an arbitrary intersection of finite sets. Hence, the intersection itself can contain no more than finitely many elements. This should be intuitive, but if it's not, you should prove it formally. Once you have established that $\bigcap_{\alpha \in J}(X-U_\alpha)$ is a finite set or empty set, we know it's equivalent $$X- \bigcup_{\alpha \in J} U_\alpha$$ is also a finite or empty set. Since $\bigcup_{\alpha \in J} U_\alpha$ is a subset of $X$ and $X- \bigcup_{\alpha \in J} U_\alpha$ is finite or empty, this satisfies the necessary criteria for $\bigcup_{\alpha \in J} U_\alpha$ to be an element of $\mathcal{T}$. This establishes that an arbitrary union of elements of $\mathcal{T}$ is itself an element of $\mathcal{T}$, and we did it without assuming any set was open, or that $\mathcal{T}$ was already a topology.

From here, it would be great practice to prove that $\emptyset \in \mathcal{T}, X \in \mathcal{T}$ and a finite intersection of elements of $\mathcal{T}$ is itself an element of $\mathcal{T}$.

  • $\begingroup$ But the definition of a topology requires that the subsets of $X$, the elements of $\mathcal T$, be open sets. Perhaps that's the reason why the author made the assumption that the $U_\alpha$ are open sets? $\endgroup$ – Mussé Redi Mar 10 '15 at 17:08
  • $\begingroup$ Correct, if we already know $\mathcal{T}$ (the finite complement topology) is indeed a topology. The way the question was worded led me to believe you were trying to prove that it is a topology, and not yet allowed to assume that $\mathcal{T}$ is a topology? $\endgroup$ – graydad Mar 10 '15 at 17:11
  • $\begingroup$ McCleary's definition of a topology, assumes that the elements of $\mathcal T$, as a set, are open sets. We then proceed on showing that $\mathcal T$ is a topology. We do not need to know that $\mathcal T$ is a topology. $\endgroup$ – Mussé Redi Mar 10 '15 at 17:19
  • $\begingroup$ Ah, thank you for adding that. I have to disagree with that definition, as I feel like the notion of "open" should come after establishing that $\mathcal{T}$ is a topology on $X$. But you still needed to show that an arbitrary union of open sets is open, right? Does my answer help? Even if I assumed each $U_\alpha$ was already open, I would've said the same thing. $\endgroup$ – graydad Mar 10 '15 at 17:22
  • $\begingroup$ Your answer helped, definitely. $\endgroup$ – Mussé Redi Mar 10 '15 at 17:25

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