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Very simple question. If I have two sets A and B that are both non-open in some topological space, is their intersection necessarily non-open?

And similarly, is the intersection of two non-measurable sets non-measurable?

I think the answer is that it could go either way. But I can't find any good discussion of this on the internet, and I am having trouble thinking of counterexamples...

Paul
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6 Answers6

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In both cases, make the intersection empty, which is both open and measurable.

For other examples, $$(1,2]\cap [0,2)=(1,2). $$ And, take disjoint non-measurable sets $E_1,E_2$ and a disjoint (to both) measurable set $D $. Then $$A=E_1\cup D,\ \ \ \ B=E_2\cup D $$ are non-measurable and $$A\cap B=D $$ is measurable.

Martin Argerami
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  • Thank you Martin. One question -- you use in your counterexample the fact that the union of a measurable set and a non-measurable set is necessarily non-measurable. why is that? – Paul Jun 23 '17 at 20:47
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    "Disjoint" is the key here. If $E\cup D $ is measurable, the $E=(E\cup D)\cap D^c=E $ is measurable. – Martin Argerami Jun 23 '17 at 20:54
  • The example $(0,2]\cap [1,2)=(0,2). $ is wrong. – Emilio Novati Jun 23 '17 at 20:55
  • Yes, there was a mixup of zeroes and ones. – Martin Argerami Jun 23 '17 at 20:59
  • martin, would you mind adding that short proof to you answer? I thought it is an interesting (if simple) statement in an of itself. – Paul Jun 23 '17 at 22:55
  • @Paul: there's no proof to be written as far as I can tell. It's just the fact that the measurable sets form a $\sigma$-algebra. – Martin Argerami Jun 23 '17 at 23:07
  • @Martin, could you explain? why does $E \cup D$ non-measurable follow directly from the fact that the measurable sets form a $\sigma -algebra$? It seems like an argument similar to the one you made is necessary – Paul Jun 24 '17 at 00:36
  • I wrote it three comments above. I repeat: if $E\cup D$ and $D$ are measurable, with $E$ and $D$ disjoint, then so is $E=(E\cup D)\cap D^c$. – Martin Argerami Jun 24 '17 at 00:41
  • ok then I understood correctly. By "follows directly" you meant using the single step you wrote in the comment above. Thanks! – Paul Jul 03 '17 at 18:30
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Onsider $A=(0,1)\cup[2,3]$ and $B=(0,1)\cup[4,5]$. They are both not open but their intersection surely is.

Rocket Man
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  • Ah yes, I suppose that is true. I should have started by thinking of the standard topology on the real line! But what about measurable sets? the above argument doesn't hold for them – Paul Jun 23 '17 at 20:42
  • I propose to you to construct a similar argument. – Rocket Man Jun 23 '17 at 20:43
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No. Take $[-1,1)$ and $(0,2]$ They are both non-open in $\mathbb R$ with respect to the usual topology, but their intersection is $(0,1)$, which is open.

And of course, for any two disjoint non-measurable sets, their intersection is empty and therefore measurable.

José Carlos Santos
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In the same style as the counterexample AJ gave for non-openness, you can construct a counterexample for measurability. Let $C$ be a non-measurable subset of $[2,3]$ and $D$ be a non-measurable subset of $[4,5]$, then $A=]0,1[\cup C$ and $B=]0,1[\cup D$ are non-measurable with a clearly measurable intersection.

Henrik supports the community
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For a large swathe of counterexamples, let $U \subset \mathbb{R}^n$ be an open subset, and let $x \ne y \in \mathbb{R}^n-U$ be two points. The subsets $U \cup \{x\}$ and $U \cup \{y\}$ are not open, and their intersection $U$ is open. A similar construction will work in many topological spaces.

Lee Mosher
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Answers of the both question is NO.

  1. Take two disjoint non-open sets. Then their intersection is empty set which is open.

  2. Similarly for non measurable sets, take two disjoint non measurable sets (infact we can choose a non measurable set and it's complement).Then their intersection is empty set which is measurable.

Non trivial examples:

  1. $A=[0, 2) $ and $B=(1, 2]$ . Then both are non-open in the euclidean space but $A\cap B=(1, 2) $ is open.

  2. Let $\mathcal{B}\subset \Bbb{R}$ is a Bernstein set.Then $\mathcal{B}$ is non measurable( see here).

Let $A=\mathcal{B}\cup [0, 1]$ and $B=\mathcal{B}^c \cap [0, 1]$

Then $A\cap B=[0, 1]$ is measurable.

SoG
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