# Probability - Set Theory Question

1. Jan 21, 2007

So I'm taking a probability class right now. We are going over elementary set theory, and the professor brought something up which seems non-intuitive to me.

He said that a set must have distinct objects, so...

$$A = \{ 1, \,\, 1, \,\, 2, \,\, 3 \}$$

is not properly defined, because the 1 is repeated. Instead, it must be written as:

$$A = \{ 1, \,\, 2, \,\, 3 \}$$

why is this?

I asked him then what do we do with,
$$A = \{ x^2 | x= -1, \,\, 0, \,\, 1 \}$$

and he said that it would be,
$$A = \{ 1, \,\, 0 \}$$

This seems weird, since in a sense we are losing information.

Would someone be so kind to elaborate on why you can't have repeated objects? I want to know why this is.

(Maybe I should have not posted this in the homework section.)

2. Jan 21, 2007

### robert Ihnot

It has to do with the definition of a set, which is a collection of distinct objects.

"A set, unlike a multiset, cannot contain two or more identical elements. All set operations preserve the property that each element in the set is unique. Similarly, the order in which the elements of a set are listed is irrelevant, unlike a sequence or tuple." http://en.wikipedia.org/wiki/Set

3. Jan 21, 2007

cool, thanks man.

"All set operations preserve the property that each element in the set is unique"

I can live with that :)

4. Jan 21, 2007

### Hurkyl

Staff Emeritus
By definition, the identity of a set is entirely determined by its membership relation.

Your professor is technically wrong on his first point -- in the usual development of set theory {1, 1, 2, 3} is a perfectly good set. It just so happens that

{1, 1, 2, 3} = {1, 2, 3}.​

You can check that the membership relation is identical for both of these: for all x, x is an element of {1, 1, 2, 3} if and only if x is an element of {1, 2, 3}.

5. Jan 21, 2007

So, if I define as set as:
A = {1, 1, 2, 3}

This is technically allowed, and is equivalent to {1, 2, 3}

However, if I perform a set operation such as:

A U NULL_SET = B

Set B would then equal, B = {1, 2, 3} correct?

or is it perfectly acceptable to say that after the set operation B = {1, 1, 2, 3}?

In my example above, lets say I want to count the number of times a function is equal to 1. So I define a set,
A = {x^2=f(x)| x = -1, 0, 1}

6. Jan 21, 2007

### matt grime

Defining

{x^2=f(x)|x=-1,0,1}

doesn't at all help you count the number of times a function is equal to 1. Firstly, x^2=f(x) is an equation (what is f?). I think you meant to write

{x^2 | x=-1,0,1}

But that doesn't helpy ou count the number of times x^2=1 (all we know from what we've written is that there are at least two elemetsn that square to 1, but that had notthing to do with sets at all). You just do it by counting.

7. Jan 21, 2007

Ok, I think I'm just getting confused on how I am applying set notation. Let me be a little more rigorous in how I am defining things.

Lets say we have a function x^2 with the domain such that x is a subset of the integers (is this ok how I am saying it?), thus:

x = -..., -2, -1, 0, 1, 2, ...

Now if we define a set of all possible values of x^2 we have,

A = {x^2 | x = -..., -2, -1, 0, 1, 2, ...}

Now if we define a subset such that,
B = {x | x is_an_element_of A AND x=1}

What I would want to see is, B = {1, 1} so now I can count the elements B and see that there are in fact two occasions where the function returns the value 1 when I feed it integers. But this is not correct right?

My whole problem seems to be that it seems that we are LOSING information, since B would equal {1} not {1,1} as I wrote above. So counting the elements of the set really does not help me.

Let me point out, that yes this problem of counting is trivial... but I'm trying to understand set notation, and apply it. I am guessing I am applying it wrong.

8. Jan 21, 2007

### matt grime

You want

{x in Z | x^2=1}

this is a two element set {-1,1}

Try to bear in mind: why would the set of square numbers remember what you squared to get there? Sets don't have positions, or multiple elements. If you want something else, then use something else.

9. Jan 21, 2007