Is the Notation Implied for a Singleton Set or Can It Have Multiple Elements?

[zzmanzz]

Homework Statement

For the following notation,

$$D \subseteq \{ z | \exists x,y \in S (z = x - y ) \}$$

I'm wondering if S can only have one element, or does the notation imply that | S | > 1

Homework Equations

The Attempt at a Solution

For example, if I have

$$S = \{ 3 \}$$

is z = 3 - 3?

Or would it be z = 3 - null = 3?

This formula is part of another problem but I'm just confused about the notation before moving further.
Thanks

 

[Mark44]

Homework Statement

For the following notation,

$$D \subseteq \{ z | \exists x,y \in S (z = x - y ) \}$$

IMO, D is not well defined, since the set notation doesn't say anything about x and y, other than "they exist." What set do x and y belong to? Can you supply some more context?

I'm wondering if S can only have one element, or does the notation imply that | S | > 1

Homework Equations

The Attempt at a Solution

For example, if I have

$$S = \{ 3 \}$$

is z = 3 - 3?

Or would it be z = 3 - null = 3?

I don't think so. Doing arithmetic with null doesn't make sense. My take on set D is that it's sort of the set of distances between x and y things, possibly belonging to two different sets.

This formula is part of another problem but I'm just confused about the notation before moving further.
Thanks

 

[FactChecker]

The notation "∃x,y∈" does not imply that x and y are distinct. So "z = 3 - 3" is valid. "z = 3 - null" is not a valid use of arithmatic subtraction. Null does not have an additive inverse.

 

[nuuskur]

You have constructed a candidate set $\{z\mid (\exists x,y\in S)(z= x-y)\}$. You are collecting all such $z$ which can be represented as a sum defined in the set constructor. Naturally, the representation doesn't have to be unique. An element belongs to the set, if at least one such representation for it exists.

The base set $S$ can be singleton. The statement $\exists x,y \in S$ doesn't require $x\neq y$. $S$ can also be empty.

I advise you to be careful with $\{z\mid P(z)\}$. Write instead $\{z\in T\mid P(z)\}$, this way, it is immediately clear that you end up with a set, provided $P(z)$ is a meaningful statement, since it is already contained as a subset in $T$. (This is a result of the axiom schema of replacement)

Long story short, everything depends on the underlying set $S$. If you don't know what $+$ means, for instance, then your set isn't even well-defined.