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I want to write a formula that says two boundary objects/sets, $A$ and $B$, just side by side but not intersecting. In 1D, it's equivalent to $(1 - n, 1)$ and $[1, 1 + n)$, where $n$ is positive number.

Currently I write it like below:

$$ A \cap B = \oslash $$

The problem with formula above is it's also true even if A and B are further apart. So, how can I say it's side by side but not intersecting?

enter image description here

EDIT:

Can I say $A \cap B = \oslash$ and $A \cap \overline B = A$ ?

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    Do you mean that you have $A,B$ such that $A \cap B = \emptyset$ but $\overline{A} \cap \overline{B} \neq \emptyset$? – 2017-01-21
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    @Ian roughly like that I guess. I added a pic to better explain my goal.. – 2017-01-21
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    Try thinking in one dimension first to make it clearer. Are you familiar with intervals and interval notation? $[a,b]$ is the interval notation for the set of all $x$ such that $a\leq x\leq b$ while $(a,b)$ is the notation for the set of all $x$ such that $a$(0,1)$ and $[1,2]$. Note that it is impossible for two closed sets to have the property you want, just as it is impossible for two open sets to have the property you want. One end will need to be open with the other end closed. – 2017-01-21
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    @JMoravitz Yes, you're right. In 1D, I want to get something exactly like $(0,1)$ and $[1,2)$..Then if it's impossible using set-theory, do you have any suggestion how to write the formula for that? – 2017-01-21
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    Should $(0,1)$ and $(1,2)$ work or not? These satisfy what I wrote. – 2017-01-21
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    @Ian Shouldn't work. It should be $(0,1)$ and $[1, 2)$ (or, equivalently, $(2, 3)$), only. – 2017-01-21
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    OK, what about this then: $A \cap B = \emptyset$ and $(\partial A \cap B) \cup (A \cap \partial B) \neq \emptyset$. – 2017-01-21
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    Hmm.. I think that represent second case on the pic, but not the first case.. – 2017-01-21
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    I'm not sure what it's supposed to be if it's neither of these... – 2017-01-21
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    @Ian May be as mentioned by JMoravits, representing "side by side" term in sets-theory is rather impossible.. Actually, now I'm thinking your equations may contradict each other. :) – 2017-01-21
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    My two suggestions are different, yes. One is overlapping closures (where it could be that the overlap is inside neither set). The other is one overlapping the closure of the other (so the overlap is in one of the original sets). – 2017-01-21
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    @Bla... I guess it would help if you clarified your intuition of sets that "just touch." Do you mean that they are disjoint, but their union is (path) connected? – 2017-01-21
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    @FabioSomenzi Just touch meaning side by side. In 1D, it's equivalent to $(1 - n, 1)$ and $[1, 1 + n)$, where $n$ is positive number. – 2017-01-21
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    Good luck finding a definition of "side by side" in a topology textbook. It's a good idea to start from $\mathbb{R}$, but notice that while connectedness implies path connectedness in $\mathbb{R}$, it does not in $\mathbb{R}^n$ for $n>1$. – 2017-01-21
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    Actually that second picture that you marked no *is* okay and your first picture is impossible in a continuous space. You have to draw your sets with solid and/or dashed borders. The to get what you want you *do* want their borders to touch but only when for the shared border each point only belongs to one or the other of the sets but not both (informally, if on set has a dashed border at the point and the other has a solid border at that point.) – 2017-01-21
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    You may find this interesting - [Dedekind Cut](https://en.wikipedia.org/wiki/Dedekind_cut) – 2017-01-21
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    @fleablood the way I say "touch" can be a bit misleading, instead it should be "side by side" – 2017-01-21
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    Right. But (0,1) and (1,2) are "side by side" but there is a one point " gap" between them. I assume you mean no gap. That'd be picture 2 with one border fuzzy and the other border solid. – 2017-01-21
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    Which, if either, is a better description of what you want and why? (0,1) and (1,2) or (0,1) and [1,2)? Or are both equally appropriate. – 2017-01-21
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    The *borders* must touch but the sets themselves do not. The borders need not be part of the set. In fact they can't both be to do what you want. – 2017-01-21
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    @fleablood I mentioned in the post and comment, that I want (0,1) and [1,2). – 2017-01-22
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    Right. So the boundary of (0,1) is {0,1} and the boundary of [1,2) is {1,2} and the boundaries have the point {1} in common. To "be right next to each other" is to have at least one point of boundary in common but to be disjoint. – 2017-01-22
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    @fleablood [1,2) is {2} not {1, 2} – 2017-01-23
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    No. The boundary of [1,2) is {1,2}. 1 is a limit/endpoint ands so is 2. 1 is in the set but 2 is not. Boundary points may or may not be in the set. – 2017-01-23
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    An open set is one with all of it's boundary points. A closed set is one with none of its boundary points. A set is neither open or closed if it has some but not all of its boundary points. A set is both open and closed if it has no boundary points. In R, R and empty set are both open and closed. They are the only two that are. – 2017-01-23

2 Answers 2

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Two sets $A$ and $B$ are called separated in topology iff $\overline{A} \cap B = A \cap \overline{B} = \emptyset$. This is obeyed by $(0,1)$ and $(1,2)$, e.g. Having disjoint closures, so $\overline{A} \cap \overline{B} = \emptyset$ is even stronger a form of being apart. So you could say two sets are disjoint but non-separated for cases like $A = (0,1)$ and $B = [1,2)$ etc.

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    Just to be thourough, the direct conditions for disjoint but non-separated is $A \cap B =\emptyset$ but either $A\cap \overline B $ is non empty or $\overline A \cap B $ is. – 2017-01-21
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    Can I say $A \cap B = \oslash$ and $A \cap \overline B = A$ ? – 2017-01-25
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    No, like @fleablood said is fine. – 2017-01-25
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    $A \cap \overline B =A$ means $A \subset \overline B$ which is clearly not a requirement. if $A\cap B =\emptyset$ then $A$ consists only of border points of B outside B, You are over thinking to confusion. You want "next to each other" so they share border points. There is no gap between them so the shared border points belong to one of the sets. And they are disjoint so the shared border points belong to only one of the sets. i.e. $A\cap B=\emptyset$ and $A\cap \overline B\ne\emptyset$ or $\overline A\cap B \ne \emptyset$. – 2017-01-25
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Actually what you want is i) $\partial A \cap \partial B \ne \emptyset $ ii) $A \cap B = \emptyset $ and iii) $\partial A \cap \partial B \subset A \cup B$.

i) says they are "right next to each other". ii) says they are disjoint. iii) says that where they are "right next to each other" there isn't an infinitisimal "gap" between them.

...or in other words...

a) $A \cap B = \emptyset $ b) either $A\cap \overline B \ne \emptyset $ or $\overline A \cap B \ne \emptyset $.

.....

Remember, unless $A $ is closed, $\partial A \not \subset A $. I think you are forgetting that.

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    Can I say $A \cap B = \oslash$ and $A \cap \overline B = A$ ? – 2017-01-25