Question about sets and relations

Question

Hello :) After visiting the second Math lecture at my university I have the following questions:

1) Suppose the the relation $\sim$ on the set of all sets is given by $X \sim Y⇔X \cap \bar{Y}=\emptyset$, then it must mean that $X-Y=\emptyset$ and $X=Y$ right? In other words, if the difference of two sets is empty, then those two sets are equal. On top of that, the relation $\sim$ should be reflexive and symmetric. But the other way round (Two sets are equal if and only if their difference is empty) is false, correct?

2) If I understood it correctly, then the set of rational numbers and the set of irrational numbers form a partition of the set of real numbers. Does that also mean that the set of rational numbers and the set of irrational numbers constitute equivalence classes on the set of real numbers? If so, how can the relation be formally written?

Thanks in advance!

Answer 1

1) $X\sim Y \implies X\cap \overline Y=\emptyset\implies$

There are no elements in both $X$ and $\overline Y\implies$

No elements of $X$ are in $\overline Y\implies$

All element of $X$ are in $Y\implies $

$X \subset Y$

And likewise

$X\subset Y\implies$ all elements of $X$ are in $Y\implies $ no elements of $X are in $\overline Y\implies X\cap \overline Y = \emptyset\implies X\sim Y$.

So $X\sim Y \iff X\subset Y$.

This is not an equivalence relationship.

a) it is reflexive as $X \subset X$ for all sets.

b) it is not symmetric as $X\subsetneq Y$ implies $Y\not \subset X$.

c) It is anti-symmetric. $X\sim Y$ and $Y\sim X\iff X\subset Y$ and $Y\subset X\iff X=Y$.

d) and it is transitive. $X\subset Y;Y\subset Z \implies X\subset Z$.

It is not an equivalence but it is a partial order (reflexive, transitive and anti-symmetric).

2) The relationship you want is $a\sim y$ if either $a,b\in \mathbb Q$ r $a,b\in \overline{\mathbb Q}$.

Is this an equivalence relationship.

a) Reflexive. Obviously. Any number is in the same sets of itself.

b) Symmetric. Obviously. If two numbers are in the same set it doesn't matter what order you specify them.

c) Anti-symmetric. Obviously not. Two numbers can be in the same set and not be equal to each other.

d) Transitive. Obviously. If $a$ is in the same set as $b$ is and $c$ is in that same set. Then $a$ and $c$ are in the same set.

So equivalent relationship.

Answer 2

For 1) We do not have that $X \cap \overline{Y}= \emptyset$ if and only if $X = Y$. Counterexample: Take $X = \emptyset$ and $Y = \{ 1 \}$

Answer 3

1) The set of all sets does not exist. Hence $\overline{Y}$ is meaningless, hence the relation you mention is too.

2) For instance: $\;x\sim y\iff \bigl((x\in\mathbf Q)\wedge(y\in\mathbf Q)\bigl)\vee\bigl((x\notin\mathbf Q)\wedge(y\notin\mathbf Q)\bigl)$, which is equivalent to $$x\sim y\iff \bigl((x\in\mathbf Q)\vee(y\in\mathbf Q)\bigl)\implies\bigl((x\in\mathbf Q)\wedge(y\in\mathbf Q)\bigl)$$