# I Why is the empty set a proper subset of every set?

1. Sep 16, 2016

### Logical Dog

I know what a proper subset is, but I never understood why every set has the empty set as its subset?

I mean, is the reasoning something primitive like this: if I have x objects, the number of unordered sets of elements I can make are 2^x, including the case where I throw out x objects and dont make anything of them?

2. Sep 16, 2016

### TeethWhitener

S is a subset of A iff all elements of S are elements of A. Since the empty set has no elements, this condition is trivially satisfied: the empty set is a subset of all sets. S is a proper subset of A iff S is a subset of A and S is not equal to A. The empty set is therefore a proper subset of any non-empty set.

3. Sep 16, 2016

### Staff: Mentor

The "non-empty" is important here. The empty set is not a proper subset of the empty set. It is still a subset.

4. Sep 16, 2016

### Logical Dog

I dont understand why

5. Sep 16, 2016

### Logical Dog

A set A is a proper subset of another B set if and only if all elements of A are contained within B, and B has at least one element that is not contained in A.

The ampty set is a proper subset of all sets except itself.

But why?

6. Sep 16, 2016

### TeethWhitener

Are all elements of $\emptyset$ elements of $A$?

7. Sep 16, 2016

### Logical Dog

There is no intersection, it is a disjoing, but A contains as the empty subset as a subset....so there is an intersection? Now i am confused

8. Sep 16, 2016

### TeethWhitener

Let's think about it a different way: If, for all $x \in S$, it's also true that $x$ is an element of $A$, then $S \subseteq A$. So now all you have to do is list out all of the $x \in \emptyset$, and check to make sure that they're all in $A$.

EDIT: this may require a little knowledge of first-order logic. In symbolic terms, the condition for $S \subseteq A$ is
$$(S \subseteq A) \leftrightarrow \forall x (x \in S \rightarrow x \in A)$$
The rule for material implication is that $p \rightarrow q$ is true when $p$ is false, regardless of what $q$ is. So $x \in \emptyset$ is false for all $x$, which means the condition $$\forall x (x \in S \rightarrow x \in A)$$
evaluates to
$$\forall x (F \rightarrow x \in A)$$
$$\forall x (T)$$
$$T$$
where $F$ is "false" and $T$ is "true."

Last edited: Sep 16, 2016
9. Sep 16, 2016

### Staff: Mentor

Let's take a different approach: "Every square number between 5 and 7 is odd". Is this statement correct?

"Every element in ∅ is also element of set A". Is this statement correct?
Same logic.

10. Sep 16, 2016

### Logical Dog

All I know about logic is this:

mISTAKE/typo..UNION Is the OR operator, but yes, this is all I know

11. Sep 16, 2016

### TeethWhitener

We can try a "proof by contradiction" as well (even though this is a definition, not a theorem). Assume that the empty set isn't a subset of $A$. Then there exists some element of the empty set that is not an element of $A$. But this is impossible, because the empty set has no elements. So the empty set has to be a subset of $A$.

12. Sep 16, 2016

### micromass

Let $A$ be a non-empty set. Can you name me an element of $\emptyset$ that is not in $A$?

13. Sep 16, 2016

### Logical Dog

No, and that is why it must be a subset of A.

14. Sep 16, 2016

### Staff: Mentor

Correct.
This is not logic. This is about unions and intersections of sets, which do not matter here.