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If describing an element is in a set I can write $x\in S$. What if $x$ could be in $S$, or is possibly in $S$?

I am thinking of the case where there is some set $A$, and there is a subsets of $A$, $B$ and $C$. So I can write that $x$ is an element of $B$, but it could also be an element of $C$, but is not guaranteed.

So far I have these pieces I am trying to work through:

$B \subset A$

$C \subset A$

$x \in A$

$x \in B$

$x$ could also be an element of $C$

Maybe not a perfect example, but in general terms I would think something like $A$ is all real numbers, $B$ is even numbers, $C$ is numbers divisible by 10, and $x$ is in $B$, but could possibly be in $C$.

The reason for trying to do this is when describing the sets in the larger context, it is more clear to have a symbol meaning explicitly what I am trying to say than explain in long sentences every combination. However if there is no symbol, or standard/logical way of doing it, then the answer to my question is just; no.

user-2147482637
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    Maybe using probability? $0<P(x\in C)<1.$ In your case, $P(x\in C|x\in B)=0.2.$ – Adrian Keister Jul 08 '19 at 16:47
  • Please [edit] the question to tell us why you need such a symbol. In context, words might be much better. To capture the meaning of your example just say "An even number might or might not be divisible by $10$" and then perhaps to go on to make an argument that depends on each case. – Ethan Bolker Jul 08 '19 at 16:54
  • @AdrianKeister Perhaps not the best choice, since you then need to first define a probability distribution on $\mathbb Z$. I could easily argue that $P(x\in C\mid x\in B)$ is any value between $0$ and $1$, since the two sets are equinumerous. See this question – Vsotvep Jul 08 '19 at 17:04
  • @Vsotvep: Thanks for the link to that question. I like Mike Ecker's solution to that problem, and it corresponds to both intuition as well as what you would undoubtedly get from a simulation. – Adrian Keister Jul 08 '19 at 17:12
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    You might look up the concept of fuzzy logic and fuzzy sets. – MasB Jul 08 '19 at 17:42
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    It could be interpreted as a statement in modal logic - $\Diamond(x \in S)$ if I remember correctly. – Daniel Schepler Jul 08 '19 at 17:45
  • @DanielSchepler You remember correctly, however, the interpretation of that depends on which modal logic you use, so it is not much more than a syntactic translation. – Vsotvep Jul 08 '19 at 17:47
  • @AdrianKeister it is either in the set or not, and probability would change depending on input. – user-2147482637 Jul 08 '19 at 18:39
  • @DanielSchepler i think this is what I was looking for. i need to do some readings on it now, thank you – user-2147482637 Jul 08 '19 at 18:40

2 Answers2

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I see several ways to interpret "could be an element":

  1. If $B$ and $C$ are unrelated, there is not really a point of saying that $x$ could be an element of $C$. In that case, it's saying that "$x\in C$ or $x\notin C$", but this is a tautological statement (i.e. it's always true), so it doesn't convey any information.

  2. We can interpret "$x\in B$ could be an element of $C$" as that there are both elements of $B$ that are elements of $C$ and that are not elements of $C$. This is effectively saying that $C$ is not a subset of $B$, i.e. $C\not\subseteq B$.

  3. We can interpret "$x\in B$ could be an element of $C$" as that any element not in $B$ could not be an element of $C$. This effectively says that being an element of $B$ is a necessary requirement for being in $C$, or in other words that $B\subseteq C$.

  4. We can interpret it as both of the previous two statements combined: effectively saying that $B$ is a proper subset of $C$: $B\subsetneq C$.

  5. If $x$ is not a variable, but a known quantity, then "could be" can convey the meaning that we haven't been able to show yet whether $x\in C$ or $x\notin C$ is the case. For example, it could be that $\pi^e$ is in the set of algebraic numbers, since we haven't been able yet to prove it's not.

  6. Related to the previous, there are statements that are proven to be undecidable using "usual" mathematics. For example, let $B$ be the set of functions $f:\mathbb R\to\mathbb R$, and let $C$ be the set of functions with a range that is countable or has the same cardinality as $\mathbb R$. Then if $x\in B$ is a function, it could be that $x\in C$. Read this as: it is possible to prove the statement under certain assumptions, and possible to prove the converse under other assumptions. In particular under the assumption that the continuum hypothesis holds / doesn't hold (both of which have been proven to be able to coexist perfectly fine with most of "usual" mathematics).

Vsotvep
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  • option 2 is closest to my intention, with an extra step that it could be equal. So would this be accurate? $C \not \subset B \lor C = B$ – user-2147482637 Jul 08 '19 at 18:41
  • @user-2147482637 I think that is possibly a very confused statement. It would really help if you explained why you want to use such a symbol, as was requested in the comments before. In my opinion writing out in words is a better option when there is no common symbol available. – Vsotvep Jul 08 '19 at 19:38
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It seems that all elements (points) $x$ are elements of a ground set $X$ (this $X$ is called $A$ in the question). Furthermore there are given subsets $B\subset X$, $C\subset X$. Maybe there is some relation among these subsets, say $C\subset B$, as in your example.

In your problem it plays a rôle whether an $x\in X$ under consideration belongs to $B$ or not, and whether it belongs to $C$ or not. A priori there are four different cases to consider: $$x\in B\cap C,\quad x\in B\setminus C,\quad x\in C\setminus B,\quad x\in X\setminus(B\cup C)\ .$$ Depending on circumstances you have to discuss these four cases, or a smaller number of them, separately.

Christian Blatter
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