Here, what I mean closed and open sets are those in real analysis sense. $[a,b]$ is a closed interval, and $(a,b)$ is an open interval.

My reasoning:

By the axioms for a topology, any infinite or finite union of members in a topology needs to be in the topology. On another hand, only finite intersection of members in the topology needs to be in the topology.

In one dimension, I may construct an infinite union of closed intervals that approaches an non-closed interval. Such as $\cup_{x\in N} [0, (2x-1)/x]$ that approaches $[0,2)?$ For open interval to do the same, we need infinite intersections instead of infinite unions, but infinite intersections are not required to be in the topology. Is this why the closed intervals do not construct a topology but the open intervals do?

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    $\begingroup$ What is a closed set if you don't have a topology defined? $\endgroup$ – H. Gutsche Jun 20 '18 at 14:24
  • $\begingroup$ @H.Gutsche What is the topology if you haven't defined which sets are open (or closed)? What comes first, the chicken or the egg? Cleary OP means closed intervals (or better, finite unions of closed intervals) as closed sets. This is indeed not a topology for the reason he gave $\endgroup$ – Jens Renders Jun 20 '18 at 14:35
  • $\begingroup$ Your reasoning is good. Another method: Let $C$ be the set of closed subsets of $\Bbb R$ with respect to the usual topology on $\Bbb .$ Every $S\subset \Bbb R$ is a union of members of $C$ because $S=\cup_{x\in S}\{x\}.$ ... So if $C $ $ were $ a topology then $C$ would contain every $S\subset \Bbb R.$ .... But $C$ does $not$ contain every $S \subset \Bbb R.$ $\endgroup$ – DanielWainfleet Jun 21 '18 at 1:44

Is this why the closed intervals do not construct a topology but the open intervals do?

Not really. Neither closed nor open intervals form a topology. Is $[0,1]\cup[3,4]$ an interval? Or $(0,1)\cup(3,4)$?

So I assume that what you are asking is "is this why closed subsets do not form a topology unlike open subsets"? I assume that by "analytic sense" you mean "induced by the Euclidean metric". Otherwise a question "why open subsets form a topology?" is a bit weird - by definition.

Then your reasoning is almost correct regarding closed subsets. First of all a closed/open set need not be a closed/open interval. So you got the first part right:

$$\bigcup [0,(2x-1)/x]=[0,2)$$

however here's where you are wrong: you say $[0,2)$ is not an open interval but this doesn't mean that it is not open. For example $(0,1)\cup(3,4)$ is an open subset that is not an open interval.

But $[0,2)$ indeed is not closed (in Euclidean sense) hence it cannot be open in our new topology. It is not closed because the very same sequence $(2x-1)/x$ is convergent outside of it.

  • $\begingroup$ Oh, thank you for reminding me of the definition of interval. I thought that union of intervals is still called interval. If I simply change intervals to be sets in my argument, then my argument is almost mostly correct? $\endgroup$ – Johnny Chen Jun 20 '18 at 14:32
  • $\begingroup$ I don't think that the OP necessarily is thinking about the Euclidean metric (maybe OP doesn't know metric spaces even). Open intervals by themselves are enough, and the are determined by the order on $\mathbb{R}$. It's just that they form a base for a topology and not a topology. That seems to me to be the right way to correct OP's mistake. $\endgroup$ – Jens Renders Jun 20 '18 at 14:39
  • $\begingroup$ @JohnnyChen indeed, unions of open intervals form the open sets for the usual topology on $\mathbb{R}$. That's all need to change. $\endgroup$ – Jens Renders Jun 20 '18 at 14:41
  • $\begingroup$ The open sets in the usual sense forms a topology because we can only have infinite intersection of open sets to be a non-open set, but we can not have infinite union of open set to be non-open set. Infinite intersection of sets in a topology is not required to be also in the topology. Is this part of my argument correct? $\endgroup$ – Johnny Chen Jun 20 '18 at 14:42
  • $\begingroup$ @JensRenders Thank you. You clear my confusion. $\endgroup$ – Johnny Chen Jun 20 '18 at 14:44

The open intervals form a base for the (usual) topology on $\mathbb{R}$, because every open set is a union of open intervals.

A family of subsets of $X$ is the base for a topology on $X$ if and only if

  1. the family is a cover for $X$ (the union of the family is $X$);
  2. the intersection of two members of the family is a union of members of the family.

If a family satisfies these property, then the set of arbitrary unions of members of the family is a topology on $X$.

We clearly see that the open intervals satisfy both conditions (the second almost trivially, because the intersection of two open intervals is either empty or an interval).

Do the closed intervals form a base for a topology? Certainly so, if we admit that singletons are closed interval; but in this case the induced topology is the discrete topology.

To the contrary, the topology having as a base the set of half-open intervals $[a,b)$ is quite interesting: it is called the Sorgenfrey line.


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