Deciding if a statement is true or false for given sets

Question

Given the sets:

$A = \{a,b,c\},\enspace B = \{a,b,A,C\},\enspace C = \{a,c\},\enspace D = \{A,B,C\},\enspace and\enspace G = \{A,B,C,D\}$

How can I determine if the following statement is true or false:

$∃x ∈ D : ∃y ∈ x : y ∉ B ∪ x$

First of all, please correct me if I'm wrong, but it seems to me that the symbols used in the statements are inconsistent. Say, if some element $y$ belongs to $x$, I assume that $x$ should be a set itself. After taking a quick look at the problem, one can conclude that sets are denoted by capital letters. So I expect $x$ to be $X$ instead.

Apart from this nuisance, I wonder if my translation into English is correct. I'm reading the statement as

For some element $x$ (which is actually set $X$) from set $D$, there is some element $y$ from $x$ such that $y$ does not belong to the union of set $B$ with $x$.

So if my interpretation of the mathematical notation is correct, I go on to assume that

$y ∉ B ∪ x\quad ≡\quad y ∉ \{x,a,b,A,C\}$ (all distinct elements combined)

Then I'm really confused. How can $∃y ∈ x$ and at the same time $y ∉ \{x,a,b,A,C\}$? Does it mean that the statement is false?

Answer 1

It isn't necessary to denote every set by capital letters, $\exists x\in D$ is perfectly fine (where $D$ is itself made up of sets). I don't see how you arrived at $y\notin B\cup x\equiv y\notin \{x,a,b,A,C\}$, it doesn't seem to be entirely correct. We know that $D=\{A,B,C\}$ and that $x$ is in $D$, so $x$ is either $A$, $B$ or $C$, i.e. either $\{a,b,c\}$ or $\{a,b,A,C\}$ or $\{a,c\}$. So $B\cup x$ is either $\{c,a,b,A,C\}$ or $\{a,b,A,C\}$ or $\{c,a,b,A,C\}$. In all cases $y\notin B\cup x\equiv y\notin \{x,a,b,A,C\}$ doesn't hold.

So how to determine whether that statement is correct? There are many ways, intuitively, if $y$ is in the set $x$, then it should also be in the set $x\cup \scr B$ (where $\scr B$ is some arbitrary set), as by definition the elements in the set $x$ are contained in the set $x\cup\scr B$. So that statement is most definitely false. Formally, what we stated here is that since $$\forall x\in D:\forall y \in x:y\in x\cup B$$ is true, then its negation is false, and in this case $$\neg\big(\forall x\in D:\forall y \in x:y\in x\cup B\big)\equiv\exists x\in D:\exists y\in x:y\notin x\cup B.$$ You could also note that $$y\notin x\cup B\iff y\notin x\wedge y\notin B,$$ hence it would be impossible to find a $y$ from $x$ such that $y$ isn't in $x$. A much more tedious way would seek to study every possible case $$\begin{array}{|c|c|c|c|}\hline x & A=\{a,b,c\} & B=\{a,b,A,C\} & C=\{a,c\} \\ \hline y & a,b,c & a,b,A,C & a,c \\ \hline B\cup x & \{a,c,b,A,C\} & \{a,b,A,C\} & \{a,c,b,A,C\} \\ \hline y\notin B\cup X & \tt False & \tt False & \tt False \\ \hline \end{array}$$

Answer 2

The notation is correct. maybe that the use of a capital $X$ for the sets can be better, but it's not necessary.

The statement is false: $y \in x \Rightarrow y \in B\cup x$

Answer 3

Is $y$ a member of $x$ and at the same time does not belong to $x \cup B$!? Seems you don't need to know what $B$ or $x$ are to answer this question.

But if you change $y \notin x \cup B $ condition in your question to $y \notin x \cap B$ you get $y=c$ and $x=A$.

Answer 4

Hint

The set $D=\{ A,B,C \}$ is "made of" other sets; thus, it is really :

$D=\{ \{ a,b,c \},\{ a,b,A,C \}, \{ a,c \} \}$.

Thus the "question" starting with : $∃x∈D:∃y∈x \ldots$, makes perfect sense : the variable $x$ may "range over" the values : $A, B$ and $C$, and they all have members.