0
$\begingroup$

I just started treading a topology and I came up with this conjecture, although I'm not sure if it is true:

A topology $X,\tau$ satisfies that all of its open sets are closed if and only if it admits a base of pairwise disjoint sets.

Proof: Let $\tau$ be a topology on $X$ having a base $B$ of pairwise disjoint subsets. Let $U$ be an open set. Then $U=\cup a$ with $a\subseteq B$. We have $X\setminus U=\cup(B\setminus a)$.

Let $X$ be a topology such that every open set is closed, then arbitrary intersections of open sets are open (since the closed sets are the open sets). Define the following equivalence relation on the open sets: $A\sim B\iff A=X\setminus B$.

Let $B=\{\cap a |a\in \prod X/\sim \}$. Then $B$ is a basis for $\tau$

What do you think? Is this correct?

$\endgroup$
  • $\begingroup$ I do not understand the construction of your set $B$. It does not seem that its elements are pairwise disjoint. $\endgroup$ – Mariano Suárez-Álvarez Feb 20 '15 at 23:00
  • $\begingroup$ oh yeah, oops, my bad. I fixed it, it should be intersection. $\endgroup$ – Jorge Fernández Hidalgo Feb 20 '15 at 23:03
  • $\begingroup$ I don't understand $a\subseteq \tau$, don't you mean $a\in\tau$? Also, if $U$ is a union, shouldn't $X\backslash U$ be an intersection? $\endgroup$ – Gregory Grant Feb 20 '15 at 23:14
  • $\begingroup$ @Gregory: The subset relation is fine, but $\tau$ should be the base of pairwise disjoint sets, not the whole topology. If the base is $\beta$, MB wants $a\subseteq\beta$. $\endgroup$ – Brian M. Scott Feb 20 '15 at 23:16
  • $\begingroup$ Oh yeah Brian, thanks, I fixed it. $\endgroup$ – Jorge Fernández Hidalgo Feb 20 '15 at 23:20
1
$\begingroup$

That all open sets be closed does not imply directly that arbitrary intersections of open sets are open.

Under that hypothesis, it is true that an arbitrary intersection of open sets is closed, but to get what you want you would need to know that every closed set is open. If this last statement is true, then you should prove it!

$\endgroup$
  • 1
    $\begingroup$ It is very easy to see that if every open is closed, then every closed is open, and viceversa. $\endgroup$ – Crostul Feb 20 '15 at 23:00
  • 1
    $\begingroup$ Well, my point is that that should be part of the argument. $\endgroup$ – Mariano Suárez-Álvarez Feb 20 '15 at 23:00
  • $\begingroup$ Since intersections of closed sets are closed and closed and open mean the same thing it follows. $\endgroup$ – Jorge Fernández Hidalgo Feb 20 '15 at 23:05
  • 1
    $\begingroup$ Again, my point is that in what you wrote you did not prove that under your hypothesis «closed and open mean the same thing». $\endgroup$ – Mariano Suárez-Álvarez Feb 20 '15 at 23:06
  • 1
    $\begingroup$ This should be a comment, not an answer. Also, it seems an unfriendly (in the sense of opaque to the student) way of suggesting OP include in the proof the detail Crostul alludes to. $\endgroup$ – aes Feb 20 '15 at 23:34
0
$\begingroup$

Say that $x,y\in X$ are in the same class if their closures coincide.

This is an equivalence relation. Its classes are closed and open sets and the collection of classes forms the base of the topology.

Suppose that $x\neq y\in\overline{x}$ then $\overline{y}\subset\overline{x}$. But it can't be smaller otherwise $\overline{x}\setminus\overline{y}$ would be a smaller closed containing $x$.

Pretty sure this is a classic exercise in many topology books.

$\endgroup$
  • 1
    $\begingroup$ The classes are sets, the collections of those sets is the base. $\endgroup$ – Donna Feb 20 '15 at 23:09
  • 3
    $\begingroup$ You really ought to distinguish points from sets: $x\sim y$ iff $\operatorname{cl}\{x\}=\operatorname{cl}\{y\}$, or, if you prefer, $\overline{\{x\}}=\overline{\{y\}}$. $\endgroup$ – Brian M. Scott Feb 20 '15 at 23:14
  • 4
    $\begingroup$ I consider it necessary, especially for those still learning the subject: I’ve seen far too many students confuse themselves badly by not keeping track of what kind of objects they’re dealing with. What you’ve written is simply sloppy — unacceptably so in a pædagogical setting. And what algebraic geometers do is not terribly relevant to me, since I’m a set-theoretic topologist. $\endgroup$ – Brian M. Scott Feb 20 '15 at 23:21
  • 1
    $\begingroup$ I agree with @BrianM.Scott per usual, and it is definitely relevant to you as it is your answer he is commenting on. No need to get unreasonably chippy. $\endgroup$ – Daniel W. Farlow Feb 20 '15 at 23:37
  • 1
    $\begingroup$ Well, the difference between a point and the set of which it is the only element seems to obvious treating it as irrelevant is sloppy unless made explicit; in such a short piece of text, it is quite out of place. This is not notational pedantry, but keeping in mind what things are. I woulf not deduce marks from a student for this, but I would surely make him aware of the issue. $\endgroup$ – Mariano Suárez-Álvarez Feb 21 '15 at 0:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.