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From my research, I am confused in differentiating the difference between closed under arbitrary unions and countable unions.

  1. What exactly is the difference between them?

In an example, it states that:

Let $X$ be an uncountable set. Also let $\tau$ = {O $\subseteq X \mid$ O = X or O is at most countable}

For this example, without much explainantion, it states that $\tau$ is closed under finite intersections and under countable unions. But it isn't a topology on X as it isn't closed under arbitrary unions.

  1. How this example is not a topology?

My note on this is that since $\tau$ contains $\emptyset$ and $X$ so it has finite number of elements, hence it is countable. Taking their union will still be in $\tau$, so it is closed under countable unions. Why it is not closed under arbitrary unions? What's the difference?

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  • $\begingroup$ It's about how many sets you're taking the union of. A countable union has to be, well, a union of countably many sets. An arbitrary union is, well, a union of arbitrarily many sets. E.g. $\mathbb{R}$ is an arbitrary union of singletons - $\mathbb{R}=\bigcup_{r\in\mathbb{R}}\{r\}$ - but not a countably union of singletons (since $\mathbb{R}$ is uncountable). $\endgroup$ – Noah Schweber Aug 12 '18 at 16:48
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    $\begingroup$ Incidentally, I don't understand what you mean when you say that $\tau$ is countable (finite, even!). It's definitely not: if $X$ is uncountable, it has uncountably many countable sets. E.g. if $X=\mathbb{R}$ then each set of the form $\{r\}$ for $r\in\mathbb{R}$ is in $\tau$, and there are uncountably many of those. $\endgroup$ – Noah Schweber Aug 12 '18 at 16:49
  • $\begingroup$ Can you please differentiate between term arbitrary and countable here? Because what I am not perceiving here is that $\mathbb{R}$ can not be union of arbitrary sets since it will lose some points. Isn't it? $\endgroup$ – Wasiq Noor Aug 12 '18 at 17:25
  • $\begingroup$ "Arbitrary" just means "as many as you want." E.g. $\bigcup_{r\in\mathbb{R}}\{r\}$ is an arbitrary union (every union is an arbitrary union) but not a countable union (since the number of sets involved is uncountable - it uses "$\mathbb{R}$-many" sets, and $\mathbb{R}$ is uncountable). $\endgroup$ – Noah Schweber Aug 12 '18 at 17:47
  • $\begingroup$ To quote my answer: "The difference between countable unions and arbitrary unions is just how many sets we're allowed to "union together." In a countable union, we're taking the union of only countably many sets; in an arbitrary union, we're taking the union of as many sets as we want" $\endgroup$ – Noah Schweber Aug 12 '18 at 17:48
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First of all, I think you have some confusion about what $\tau$ is, based on your claim that $\tau$ is countable (finite, even!). $\tau$ is definitely uncountable: any uncountable set has uncountably many countable subsets - e.g. the set of singleton subsets is uncountable.

I think you may be conflating $\tau$ - which is a set of subsets of $X$ - with individual subsets of $X$.


Now as to the different types of union, it might help to consider a concrete example. Let $X=\mathbb{R}$, and consider the set $[0,1]$. Then:

  • $[0,1]$ is a union of sets in $\tau$, namely $$[0,1]=\bigcup_{x\in[0,1]}\{x\},$$ so $[0,1]$ had better be in $\tau$ if $\tau$ is to be a topology.

  • But $[0,1]$ isn't all of $X$ and it also isn't countable, so it's not in $\tau$. What's going on here is that while we can write it as a union of a bunch of sets in $\tau$ (as above), we can't write it as a union of only countably many sets in $\tau$.

The difference between countable unions and arbitrary unions is just how many sets we're allowed to "union together." In a countable union, we're taking the union of only countably many sets; in an arbitrary union, we're taking the union of as many sets as we want. For example, a countably union of countable sets is countable (briefly, "$\aleph_0\times\aleph_0=\aleph_0$" if you're familiar with $\aleph$-notation), but an arbitrary union of countable sets can be as big as you want: indeed, any set is an arbitrary union of one-element sets via $$S=\bigcup_{s\in S}\{s\}.$$


As a further exercise, thinking along the lines above we can show:

If $\tau$ is a topology on a set $X$ which contains every one-element subset of $X$, then in fact $\tau$ contains every subset of $X$ (we say $\tau$ is the discrete topology in this case).

Once you're comfortable with the proof of this fact, I think you'll understand the issues above perfectly.

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  • $\begingroup$ I understand the example of $[0, 1]$ here. When I relate this to the example given in the question, which denies $\tau$ to be topology, I perceive that: like your example of $[0, 1]$ you can get a topology if $[0, 1]$ is in $\tau$. Since my question allow to have $X$ in the $\tau$, we can have a topology if $X$ is included? $\endgroup$ – Wasiq Noor Aug 12 '18 at 20:43
  • $\begingroup$ No: just having $X$ isn't enough, you need everything that can be written as a union of arbitrarily many elements of $\tau$. In fact, you'll need every subset of $X$, since any subset of $X$ can be written as a union of countable - indeed, one-element - sets: $$A=\bigcup_{a\in A}\{a\}.$$ $\endgroup$ – Noah Schweber Aug 12 '18 at 20:45
  • $\begingroup$ Yes, I intended to say that we can get a topology in the case where we have every subset of $X$ including $X$ in $\tau$. But obviously this is not always the case due to the OR in the question statement. $\endgroup$ – Wasiq Noor Aug 12 '18 at 20:49
  • $\begingroup$ Exactly, that's right. Now you've got it! $\endgroup$ – Noah Schweber Aug 12 '18 at 20:54
  • $\begingroup$ One more question: If this special case is considered, we are closed under arbitrary unions. Can we also say that we are closed under countable unions? $\endgroup$ – Wasiq Noor Aug 12 '18 at 20:59
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Countable union means the union of countably many sets in the system (has nothing to do with the cardinality of the sets themselves). So this system is clearly closed under countable unions by the following case distinction: Case 1: If $X$ is in the family of subsets whose union is taken, then the union is $X$, which is in $\tau$. Case 2: If $X$ is not in the family of subsets whose union is taken, then you take the union of countably many countable subsets of $X$, which yields a countable subset of $X$, which is in $\tau$.

However, $\tau$ is not closed under arbitrary unions. E.g., if $X=\mathbb{R}$, then the set of positive numbers is the union of singleton sets. Singleton sets are countable, so the set of positive numbers can be written as a union of sets in $\tau$, but it is not in $\tau$. (There are uncountably many positive real numbers, and the set of positive real numbers is not $X$ itself.)

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  • $\begingroup$ Why do we need to consider 2 scenarios here? Why do we explicitly include $X$ here to prove our point? $\endgroup$ – Wasiq Noor Aug 12 '18 at 20:22
  • $\begingroup$ The set $\tau$ is also given by a case distinction: a set is in it either because it is countable or because it is equal to the whole underlying set $X$. So it is not very surprising that when proving something about $\tau$, you need to make a case distinction along these line. Of course, there are ways to reformulate the proof, but essentially, you need to distinguish between these two cases somehow. $\endgroup$ – A. Pongrácz Aug 12 '18 at 20:26
  • $\begingroup$ In case 1: we are taking uncountable union of many countable sets to get $X$. Do we still name this as colosed under countable unions? $\endgroup$ – Wasiq Noor Aug 12 '18 at 21:03
  • $\begingroup$ @WasiqNoor I don't understand what you're asking. We're not taking the union of uncountably many countable sets; rather, in case 1 we're considering a countable union of sets, one of which is $X$. Do you understand why this is a countable union, and is all of $X$? $\endgroup$ – Noah Schweber Aug 12 '18 at 21:21
  • $\begingroup$ Consider the case when $\tau$ has all the subsets of $X$ as well as $X$ in it. Now since $X$ has uncountable subsets, when we union all of these subsets from $\tau$, we get $X$ which is in $\tau$. Now actually we have union uncounably many countable subsets. Right? So aren't we violating by considering it a countable union? $\endgroup$ – Wasiq Noor Aug 12 '18 at 21:31
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You’re given that $X$ is an uncountable set. For every $x\in X$, set $O_x = {x}$. Certainly, $O_x$ is a finite set thus $O_x\in\tau$. Now, let $A\subset X$ be any subset. Therefore, the set

$$O_A = \bigcup_{x\in A} O_x$$

is exactly equal to $A$, but it is not always the case that $A\in\tau$. Therefore, it is not true that $\tau$ is closed under arbitrary unions.

Note that this breaks down exactly when $A$ is uncountable, i.e. when we take an uncountable union!

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