Difference between closed under countable union and arbitrary union

Question

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.

2. 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?

Answer 1

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 $X$.

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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 $$S=\bigcup_{s\in S}\{s\}.$$

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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.

Answer 2

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.)

Answer 3

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!